3B

Physics Question Bank · 物理题库

Ch. 6 Light & EM Waves · Ch. 7 Waves & Sound · 中英对照

Chapter Type Level
202 questions

How to use · 使用方法

1Ch.6 第6章MC 选择题Easy 易2 marksPHY13061101

To obtain an obvious interference pattern of light through two slits, the slits should be very narrow. This is because light

Atravels at a high speed.
Bhas high frequencies.
Chas short wavelengths.
Dis made up of different colours.
Chinese · 中文对照

要用双缝看到明显的光的干涉（interference）图样，两条缝必须做得很窄。这是因为光

interference 干涉slit 缝comparable with 与……差不多diffraction 衍射
Correct answer 正确答案C

Light wave is diffracted at the two slits and interference occurs to form the interference pattern. For diffraction of light to be significant, the width of the slits should be comparable with the wavelength of the light.

中文解析

光波在两条缝处发生衍射（diffraction），并产生干涉，从而形成干涉图样。要让光的衍射效果显著，缝的宽度必须与光的波长相差不多。

2Ch.6 第6章MC 选择题Easy 易2 marksPHY13061102

While travelling through a vacuum, all electromagnetic waves have the same

(1)speed.
(2)wavelength.
(3)frequency.
A(1) only
B(3) only
C(1) and (2) only
D(2) and (3) only
Chinese · 中文对照

在真空中传播时，所有电磁波（electromagnetic waves）都具有相同的

(1) 速度。

(2) 波长。

(3) 频率。

vacuum 真空wavelength 波长frequency 频率constant 恒定的
Correct answer 正确答案A

In a vacuum, all electromagnetic waves travel at a constant speed of 3.00 × 108 m s−1. They have various wavelengths and frequencies.

中文解析

在真空中，所有电磁波都以恒定的速度 3.00 × 108 m s−1 传播。但它们的波长和频率各不相同。

3Ch.6 第6章MC 选择题Easy 易2 marksPHY13061103

The wavelength of a microwave is 20 mm. What is its frequency? (1 MHz = 106 Hz, 1 GHz = 109 Hz)

A100 MHz
B400 MHz
C15 GHz
D73 GHz
Chinese · 中文对照

微波（microwave）的波长是 20 mm。它的频率是多少？（1 MHz = 106 Hz，1 GHz = 109 Hz）

wavelength 波长frequency 频率microwave 微波
Correct answer 正确答案C

By v = fλ,

3.00 × 108    = f × (20 × 10−3)

f    = 1.5 × 1010 Hz

= 15 × 109 Hz

= 15 GHz

中文解析

由 v = fλ，

3.00 × 108 = f × (20 × 10−3)

f = 1.5 × 1010 Hz

= 15 × 109 Hz

= 15 GHz

4Ch.6 第6章MC 选择题Easy 易2 marksPHY13061104

A blue beam of light falls on two narrow slits and produces an interference pattern on a screen. If instead of blue light, a red beam of light was used in the same experiment, which change to the interference pattern can be observed?

AThe other fringes become closer to the central bright fringe.
BThe other fringes become farther away from the central bright fringe.
CThere is no change in the interference pattern.
DBright fringes are replaced by dark fringes.
Chinese · 中文对照

一束蓝光照射到两条窄缝上，在屏幕上形成干涉图样。如果这个实验改用一束红光，可以观察到干涉图样发生什么变化？

beam 光束fringe 条纹separation 间距central bright fringe 中央亮纹
Correct answer 正确答案B

The wavelength of red light is longer than that of blue light. The fringe separation increases when the wavelength of light increases.

中文解析

红光的波长比蓝光长。光的波长增加时，条纹间距会变大。

5Ch.6 第6章MC 选择题Average 中等2 marksPHY13061105

If the Young's double slit experiment is performed with the whole set-up immersed in a liquid of refractive index n, what is the ratio ∆y'∆y? Δy’ and Δy are the fringe separations of the interference pattern of the same set-up in the liquid and in air respectively.

An
B1n
Cn-1n
Dnn-1
Chinese · 中文对照

如果把整个杨氏双缝（Young's double slit）实验装置浸在折射率（refractive index）为 n 的液体中，比值 (∆y')/(∆y) 是多少？Δy’ 和 Δy 分别是同一装置在液体中和在空气中的干涉图样的条纹间距（fringe separation）。

refractive index 折射率immersed 浸在……中fringe separation 条纹间距liquid 液体
Correct answer 正确答案B

By v = cn and v = fλ,

λ = vf = cnf

Fringe separation Δy = λDa = Dcnfa µ 1n

Δy'Δy = nanl = 1n

中文解析

由 v = (c)/(n) 和 v = fλ，

得 λ = (v)/(f) = (c)/(nf)

条纹间距 Δy = (λD)/(a) = (Dc)/(nfa) ∝ (1)/(n)

所以 (Δy')/(Δy) = (n_a)/(n_l) = (1)/(n)

6Ch.6 第6章MC 选择题Average 中等2 marksPHY13061106

Light of wavelength λ passes through a diffraction grating with grating spacing d. A series of bright fringes is observed on a screen. What is the angle between the two first-order bright fringes?

Asin−1(λ2d)
Bsin−1(λd)
C2sin−1(λ2d)
D2 sin−1(λd)
Chinese · 中文对照

波长为 λ 的光通过缝距（grating spacing）为 d 的衍射光栅（diffraction grating），在屏幕上看到一系列亮纹。两条第一级（first-order）亮纹之间的夹角是多少？

diffraction grating 衍射光栅grating spacing 光栅缝距bright fringe 亮纹first-order 第一级
Correct answer 正确答案D

By d sin θn = nλ,

d sin θ1 = 1 × λ

θ1 = sin−1λd

Angle between the two first-order bright fringes = 2θ1 = 2 sin−1λd

中文解析

由 d sin θ_n = nλ，

d sin θ_1 = 1 × λ

θ_1 = sin−1(λ / d)

两条第一级亮纹之间的夹角 = 2θ_1 = 2 sin−1(λ / d)

7Ch.6 第6章MC 选择题Average 中等2 marksPHY13061107

Which of the following statements about the properties of light is/are correct?

(1)The speed of light is always 3.00 × 108 m s−1 in all kinds of medium.
(2)The wavelength of light changes when it passes through different media.
(3)Light is a kind of transverse wave.
A(1) only
B(3) only
C(1) and (2) only
D(2) and (3) only
Chinese · 中文对照

以下关于光性质的陈述，哪项／哪些是正确的？

(1) 光在任何介质中的速度总是 3.00 × 108 m s−1。

(2) 光通过不同介质时，波长会改变。

(3) 光是一种横波（transverse wave）。

transverse wave 横波medium 介质property 性质statement 陈述
Correct answer 正确答案D

The speed of light is 3.00 × 108 m s–1 in a vacuum and is lower in other media.

∴ (1) is incorrect.

The speed and the wavelength of light change when it passes through different media.

∴ (2) is correct.

Light is an electromagnetic wave, which is a transverse wave.

∴ (3) is correct.

中文解析

光在真空中的速度是 3.00 × 108 m s−1，在其他介质中更小。

∴ (1) 不正确。

光通过不同介质时，速度和波长都会改变。

∴ (2) 正确。

光是电磁波，属于横波。

∴ (3) 正确。

8Ch.6 第6章MC 选择题Easy 易2 marksPHY13061108

Which of the following is not an application of microwaves in daily life?

ARadar
BCooking
CExamining fetus
DSatellite communication
Chinese · 中文对照

以下哪一项不是微波在日常生活里的应用？

radar 雷达fetus 胎儿satellite communication 卫星通信application 应用
Correct answer 正确答案C

Ultrasound is usually used in examining fetus in the mother’s body.

中文解析

检查母亲体内的胎儿通常是用超声波（ultrasound）。

9Ch.6 第6章MC 选择题Difficult 难2 marksPHY13061109

A beam of white light is incident normally on a plane transmission grating and a pattern with some continuous spectra is formed on the screen behind the grating. The figure below shows how the boundary of the 1st order spectrum is formed. The grating has 500 lines per mm. The white light covers visible light with wavelength 4 × 10−7 m to 7 × 10−7 m.

What is the angle θ subtended by the 1st order spectrum?

A8.95°
B11.5°
C20.5°
D20.8°
Chinese · 中文对照

一束白光垂直入射到平面透射光栅（plane transmission grating）上，在光栅后面的屏幕上形成一些连续光谱（continuous spectra）。下图显示第一级光谱的边界是怎样形成的（见图）。这块光栅每毫米有 500 条线。白光包含波长由 4 × 10−7 m 到 7 × 10−7 m 的可见光。第一级光谱所张的角 θ 是多少？

continuous spectra 连续光谱grating 光栅boundary 边界subtended by 所张的（角）
Correct answer 正确答案A

Grating spacing d = 1×10−3500 = 2 × 10−6 m

Apply d sin θn = nλ.

For the light with the maximum wavelength in white light,

(2 × 10−6) × sin θ1    = 1 × (7 × 10−7)

sin θ1    = 0.35

θ1    = 20.487°

For the light with the minimum wavelength,

(2 × 10−6) × sin θ1    = 1 × (4 × 10−7)

sin θ1    = 0.2

θ1    = 11.537°

Angle θ subtended by the 1st order spectrum = 20.487° − 11.537° = 8.95°

中文解析

光栅缝距 d = (1 × 10−3)/(500) = 2 × 10−6 m

应用 d sin θ_n = nλ。

对白光中波长最长的那种光：

(2 × 10−6) × sin θ_1 = 1 × (7 × 10−7)

sin θ_1 = 0.35

θ_1 = 20.487°

对波长最短的那种光：

(2 × 10−6) × sin θ_1 = 1 × (4 × 10−7)

sin θ_1 = 0.2

θ_1 = 11.537°

第一级光谱所张的角 θ = 20.487° − 11.537° = 8.95°

10Ch.6 第6章MC 选择题Easy 易2 marksPHY13061110

The Young’s double slit experiment provides evidence that

Awaves can undergo diffraction.
Bwaves can undergo interference.
Clight is a wave.
Dlight is a kind of electromagnetic wave.
Chinese · 中文对照

杨氏双缝实验提供了什么证据？

evidence 证据diffraction 衍射interference 干涉
Correct answer 正确答案C

The experiment shows that light exhibits diffraction and interference. This provides evidence that light is a wave.

中文解析

这个实验显示光会表现出衍射和干涉。这就提供了证据，说明光是一种波。

11Ch.6 第6章MC 选择题Easy 易2 marksPHY13061111

In a Young’s double slit experiment, the fringe separation on the screen can be increased by

(1)using a double slit with a larger separation.
(2)using a light with a larger wavelength.
(3)increasing the separation between the screen and the double slit.
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

在杨氏双缝实验中，要增大屏幕上的条纹间距，可以

(1) 使用缝距更大的双缝。

(2) 使用波长更长的光。

(3) 增大屏幕与双缝之间的距离。

fringe separation 条纹间距double slit 双缝slit separation 缝距
Correct answer 正确答案C

In the interference pattern formed by a double slit, the fringe separation increases when the slit separation decreases.

∴ (1) is incorrect.

中文解析

双缝形成的干涉图样中，缝距减小，条纹间距就会增大。

∴ (1) 不正确。

12Ch.6 第6章MC 选择题Easy 易2 marksPHY13061112

A beam of monochromatic light falls onto a plane transmission grating and forms a fringe pattern on a screen. The separation between bright fringes will increase if

(1)a grating with more lines per unit length is used.
(2)a light beam with a larger frequency is used.
(3)the screen is moved farther away from the grating.
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

一束单色光（monochromatic light）照射到平面透射光栅上，在屏幕上形成条纹图样。如果出现以下哪种情况，亮纹之间的间距会增大？

(1) 使用每单位长度有更多条线的光栅。

(2) 使用频率更大的光束。

(3) 把屏幕移得离光栅更远。

monochromatic light 单色光per unit length 每单位长度grating 光栅frequency 频率
Correct answer 正确答案B

In the interference pattern formed by a diffraction grating, the fringe separation increases when:

(i)    the grating spacing decreases (i.e. more lines per unit length);

(ii)    the wavelength of light increases (i.e. frequency decreases);

(iii)    the distance between the grating and the screen increases.

∴ (1) and (3) are correct. (2) is incorrect.

中文解析

在衍射光栅形成的干涉图样中，条纹间距会在以下情况增大：

(i) 光栅缝距减小（即每单位长度的线数增多）；

(ii) 光的波长增大（即频率减小）；

(iii) 光栅与屏幕之间的距离增大。

∴ (1) 和 (3) 正确，(2) 不正确。

13Ch.6 第6章MC 选择题Easy 易2 marksPHY13061113

Which of the following statements about electromagnetic waves is incorrect?

AThey can transmit energy through a vacuum.
BThey travel with the same speed in all media.
CThey are transverse waves in the electric and magnetic fields.
DThey all show reflection, refraction, interference and diffraction.
Chinese · 中文对照

以下关于电磁波的陈述，哪一项是不正确的？

transverse wave 横波vacuum 真空reflection 反射refraction 折射
Correct answer 正确答案B

An electromagnetic wave travels with different speeds in different media. Different electromagnetic waves travel with different speeds in the same medium and also in different media.

中文解析

电磁波在不同介质中的传播速度不同。不同的电磁波在同一种介质中，以及在不同介质中的传播速度也各不相同。

14Ch.6 第6章MC 选择题Easy 易2 marksPHY13061114

Which of the following gives the best approximation of the wavelengths of violet light and red light in a vacuum?

AViolet light · 400 nm
Red light · 550 nm
BViolet light · 400 nm
Red light · 700 nm
CViolet light · 550 nm
Red light · 400 nm
DViolet light · 700 nm
Red light · 400 nm
Chinese · 中文对照

以下哪一项最能近似表示真空中紫光和红光的波长？

violet light 紫光red light 红光approximation 近似值visible light 可见光
Correct answer 正确答案B

The wavelength of visible light ranges from 400 nm to 700 nm. Red light is at the end of long wavelength while violet light is at the end of short wavelength.

中文解析

可见光的波长范围是 400 nm 到 700 nm。红光在长波长的一端，紫光在短波长的一端。

15Ch.6 第6章MC 选择题Easy 易2 marksPHY13061115

The distance between Pluto and the Sun is about 5.91 × 1012 m. How long does it take for the light to travel from the Sun to Pluto? (Given: speed of light in a vacuum = 3 × 108 m s−1)

A1.97 s
B197 s
C19 700 s
D1.97 × 108 s
Chinese · 中文对照

冥王星（Pluto）与太阳的距离约为 5.91 × 1012 m。光从太阳走到冥王星需要多长时间？（已知：真空中的光速 = 3 × 108 m s−1）

Pluto 冥王星speed of light 光速travel 传播、走
Correct answer 正确答案C

Time taken

中文解析

所用时间（见图，原解析用图写出计算过程）= 距离 ÷ 速率 = (5.91 × 1012) ÷ (3 × 108) = 1.97 × 104 s = 19 700 s

16Ch.6 第6章MC 选择题Easy 易2 marksPHY13061116

Which of the following statements about microwaves is incorrect?

AThey can be used in satellite telecommunications.
BThey can travel in a vacuum at a speed of 3 × 108 m s–1.
CThey can be used in measuring body temperature.
DThey can be used to heat food.
Chinese · 中文对照

以下关于微波的陈述，哪一项是不正确的？

microwave 微波body temperature 体温telecommunications 电信
Correct answer 正确答案C

Infra-red radiation is usually used in measuring body temperature.

中文解析

量度体温通常是用红外线（infra-red radiation）。

17Ch.6 第6章MC 选择题Easy 易2 marksPHY13061117

The broadcasting frequency of CRHK 1 is 88.1 MHz. What is its corresponding broadcasting wavelength? (Given: 1 MHz = 106 Hz)

A3.41 nm
B3.41 mm
C3.41 m
D341 m
Chinese · 中文对照

香港商业电台第一台（CRHK 1）的广播频率是 88.1 MHz。它对应的广播波长是多少？（已知：1 MHz = 106 Hz）

broadcasting frequency 广播频率broadcasting wavelength 广播波长corresponding 对应的
Correct answer 正确答案C

By v = fλ,

λ = vf = 3.00×10888.1×106 = 3.41 m

中文解析

由 v = fλ，

λ = (v)/(f) = (3.00 × 108)/(88.1 × 106) = 3.41 m

18Ch.6 第6章MC 选择题Easy 易2 marksPHY13061118

When electromagnetic waves travel from one medium to another, which of the following properties of the waves will change?

(1)Frequency
(2)Speed
(3)Wavelength
A(1) only
B(2) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

当电磁波从一种介质传播到另一种介质时，波的以下哪些性质会改变？

(1) 频率

(2) 速度

(3) 波长

medium 介质wavelength 波长property 性质
Correct answer 正确答案C

When an electromagnetic wave travels from one medium to another, its speed changes while its frequency remains unchanged.

∴ (2) is correct. (1) is incorrect.

By v = fλ, the wavelength of the electromagnetic wave also changes.

∴ (3) is correct.

中文解析

电磁波从一种介质传播到另一种介质时，速度改变，但频率保持不变。

∴ (2) 正确，(1) 不正确。

由 v = fλ，电磁波的波长也会改变。

∴ (3) 正确。

19Ch.6 第6章MC 选择题Easy 易2 marksPHY13061119

When a light wave passes through a narrow slit, which of the following wave properties remain(s) unchanged?

(1)Frequency
(2)Speed
(3)Wavelength
A(1) only
B(2) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

光波通过一条窄缝时，以下哪些波的性质保持不变？

(1) 频率

(2) 速度

(3) 波长

narrow slit 窄缝unchanged 保持不变diffraction 衍射
Correct answer 正确答案D

When diffraction occurs, the speed, frequency and wavelength of the light wave remain unchanged.

中文解析

发生衍射时，光波的速度、频率和波长都保持不变。

20Ch.6 第6章MC 选择题Easy 易2 marksPHY13061120

Which of the following electromagnetic waves has the longest wavelength?

AMicrowaves
BInfra-red radiation
CRed light
DX-rays
Chinese · 中文对照

以下哪种电磁波的波长最长？

microwaves 微波infra-red radiation 红外线X-rays X 射线
Correct answer 正确答案A

Wavelength of microwaves > wavelength of infra-red radiation > wavelength of red light > wavelength of X-rays

中文解析

微波的波长 > 红外线的波长 > 红光的波长 > X 射线的波长

21Ch.6 第6章MC 选择题Easy 易2 marksPHY13061121

Arrange the following in descending order of frequency: (1) infra-red radiation, (2) microwaves, (3) radio waves, (4) ultra-violet radiation, (5) X-rays.

A(5), (1), (2), (3), (4)
B(4), (1), (3), (5), (2)
C(3), (2), (4), (1), (5)
D(5), (4), (1), (2), (3)
Chinese · 中文对照

把以下各项按频率由大到小排列：(1) 红外线、(2) 微波、(3) 无线电波、(4) 紫外线、(5) X 射线。

descending order 由大到小排列ultra-violet radiation 紫外线radio waves 无线电波
Correct answer 正确答案D
中文解析

电磁波的频率由低到高依次是：无线电波 < 微波 < 红外线 < 可见光 < 紫外线 < X 射线。

所以按频率由大到小排列为：(5) X 射线 > (4) 紫外线 > (1) 红外线 > (2) 微波 > (3) 无线电波。

∴ 选 D。（原题未附文字解析，此顺序由电磁波谱（electromagnetic spectrum）决定。）

22Ch.6 第6章MC 选择题Easy 易2 marksPHY13061122

The following figure shows part of the electromagnetic spectrum.

Microwaves

A

Visible light

B

X-rays

Which of the following statements is/are correct?

(1)The wavelength of A is longer than that of B.
(2)B is infra-red radiation.
(3)A is blue light.
A(1) only
B(3) only
C(1) and (3) only
D(2) and (3) only
Chinese · 中文对照

下图显示电磁波谱的一部分（见图）：微波 | A | 可见光 | B | X 射线。以下哪些陈述正确？

(1) A 的波长比 B 的波长长。

(2) B 是红外线。

(3) A 是蓝光。

electromagnetic spectrum 电磁波谱visible light 可见光infra-red radiation 红外线
Correct answer 正确答案A

Wavelengths of microwaves > wavelengths of visible light > wavelengths of X-rays

∴ (1) is correct.

The wavelength of infra-red radiation is longer than that of visible light, so B could not be infra-red radiation.

∴ (2) is incorrect.

Blue light belongs to visible light.

∴ (3) is incorrect.

中文解析

微波的波长 > 可见光的波长 > X 射线的波长。

∴ (1) 正确。

红外线的波长比可见光长，所以 B 不可能是红外线。

∴ (2) 不正确。

蓝光属于可见光。

∴ (3) 不正确。

23Ch.6 第6章MC 选择题Easy 易2 marksPHY13061123

A person is looking at his passbook. He can see the signature on the passbook because a ray X travels from the passbook to his eye. What is X?

AInfra-red radiation
BVisible light
CUltra-violet radiation
DX-ray
Chinese · 中文对照

一个人正在看他的存折。他能看见存折上的签名，是因为有一束光线 X 从存折射入他的眼睛。X 是什么？（见图）

passbook 存折signature 签名ray 光线
Correct answer 正确答案B

A person can only see visible light.

中文解析

人只能看见可见光（visible light）。

24Ch.6 第6章MC 选择题Easy 易2 marksPHY13061124

Which of the following is not a use of ultraviolet radiation?

ADetection of counterfeit banknotes (偽鈔)
BProduction of vitamin D
CSterilization of drinking water
DSecurity check for luggage
Chinese · 中文对照

以下哪一项不是紫外线的用途？

counterfeit banknotes 伪钞sterilization 消毒、杀菌security check 安全检查ultraviolet radiation 紫外线
Correct answer 正确答案D

X-rays are usually used in security check for luggage.

中文解析

行李安全检查通常是用 X 射线。

25Ch.6 第6章MC 选择题Average 中等2 marksPHY13061125

A narrow beam of white light is incident to a prism at X with incident angle θ and a visible spectrum is formed on a screen. The following figure shows the spectrum and its yellow component.

A monochromatic light beam of wavelength greater than that of yellow light is now incident at X with the same incident angle θ. The emergent beam hits the screen at a position

Aabove the yellow light.
Bwhich is the same as the yellow light.
Cbelow the yellow light.
Dthat cannot be determined.
Chinese · 中文对照

一束窄白光以入射角 θ 从 X 点射入棱镜（prism），在屏幕上形成可见光谱。下图显示该光谱及其中的黄色成分（见图）。现在改以一束波长比黄光更长的单色光，用相同的入射角 θ 从 X 点射入。出射光束会打在屏幕上的哪个位置？

prism 棱镜spectrum 光谱emergent beam 出射光束bends 偏折
Correct answer 正确答案A

Among the components of white light, red light bends least while violet light bends most when travelling from air to prism and from prism to air. Light beam of wavelength greater than that of yellow light is closer to the red end, so it bends less than yellow light and hits a position above the yellow light.

中文解析

在白光的各种成分中，光从空气进入棱镜以及从棱镜射入空气时，红光偏折最少，紫光偏折最多。波长比黄光更长的光束更靠近红光一端，所以它比黄光偏折得更少，会打在黄光的上方。

26Ch.6 第6章MC 选择题Average 中等2 marksPHY13061126

For a given transparent material, the refractive index for a visible light

Ais lower if the wavelength is longer.
Bis higher if the wavelength is longer.
Cdoes not depend on the wavelength.
Ddepends on the angle of incidence.
Chinese · 中文对照

对某一种透明材料来说，可见光的折射率

refractive index 折射率transparent material 透明材料angle of incidence 入射角
Correct answer 正确答案A

Among the components of white light, red light bends least while violet light bends most when travelling from air to prism, showing that the refractive index for red light is lowest and the refractive index for violet light is highest in the material. On the other hand, the wavelength of red light is the longest while that of violet light is the shortest.

∴ A is correct.

中文解析

在白光的各种成分中，光从空气射入棱镜时红光偏折最少、紫光偏折最多，这说明该材料对红光的折射率最低，对紫光的折射率最高。另一方面，红光的波长最长，紫光的波长最短。

∴ A 正确。

27Ch.6 第6章MC 选择题Average 中等2 marksPHY13061127

A beam of light emitted from a filament lamp is incident on a plane transmission grating with 500 lines per mm. A fringe pattern is observed on a screen placed at a distance 50 cm from the grating. Find the distance between the 1st-order red fringe and the 2nd-order violet fringe on the screen. It is given that the wavelengths of red and violet light are respectively 700 nm and 400 nm.

A3.14 cm
B8.48 cm
C27.2 cm
D38.8 cm
Chinese · 中文对照

一盏钨丝灯（filament lamp）发出的光束照射到每毫米有 500 条线的平面透射光栅上。在与光栅相距 50 cm 的屏幕上观察到条纹图样。求屏幕上第一级红色条纹与第二级紫色条纹之间的距离。已知红光和紫光的波长分别为 700 nm 和 400 nm。

filament lamp 钨丝灯plane transmission grating 平面透射光栅1st-order 第一级2nd-order 第二级
Correct answer 正确答案A

Grating spacing d = 1×10−3500 = 2 × 10−6 m

Apply d sin θn = nλ.

For 1st-order red fringe,

(2 × 10−6) × sin θ1    = 1 × (700 × 10−9)

sin θ1    = 0.35

θ1    = 20.49°

Distance from central bright fringe = 0.5 × tan 20.49° = 0.1868 m

For 2nd-order violet fringe,

(2 × 10−6) × sin θ2    = 2 × (400 × 10−9)

sin θ2    = 0.4

θ2    = 23.58°

Distance from central bright fringe = 0.5 × tan 23.58° = 0.2182 m

Distance between the two fringes = 0.2182 − 0.1868 = 0.0314 m = 3.14 cm

中文解析

光栅缝距 d = (1 × 10−3)/(500) = 2 × 10−6 m

应用 d sin θ_n = nλ。

对第一级红色条纹：

(2 × 10−6) × sin θ_1 = 1 × (700 × 10−9)

sin θ_1 = 0.35

θ_1 = 20.49°

离中央亮纹的距离 = 0.5 × tan 20.49° = 0.1868 m

对第二级紫色条纹：

(2 × 10−6) × sin θ_2 = 2 × (400 × 10−9)

sin θ_2 = 0.4

θ_2 = 23.58°

离中央亮纹的距离 = 0.5 × tan 23.58° = 0.2182 m

两条条纹之间的距离 = 0.2182 − 0.1868 = 0.0314 m = 3.14 cm

28Ch.6 第6章MC 选择题Average 中等2 marksPHY13061128

A monochromatic light of wavelength 550 nm is incident on a diffraction grating with 200 lines per mm and a fringe pattern is observed on a screen behind the grating. If the screen is moved away from the grating by 0.5 m, how will the separation between the central fringe and the 1st-order fringe change?

AIncrease by 5.53 cm
BIncrease by 11.1 cm
CIncrease to 1.11 times of the original
DCannot be determined
Chinese · 中文对照

波长为 550 nm 的单色光照射到每毫米有 200 条线的衍射光栅上，在光栅后面的屏幕上观察到条纹图样。如果把屏幕移离光栅 0.5 m，中央条纹与第一级条纹之间的间距会怎样变化？

monochromatic light 单色光diffraction grating 衍射光栅separation 间距
Correct answer 正确答案A

Grating spacing d = 1×10−3200 = 5 × 10−6 m

By d sin θn = nλ,

(5 × 10−6) × sin θ1    = 1 × (550 × 10−9)

sin θ1    = 0.11

θ1    = 6.315°

Let D be the original distance between the grating and the screen.

Increase in separation between the central fringe and the 1st-order fringe

= (D + 0.5) × tan 6.315° − D × tan 6.315°

= 0.5 × tan 6.315°

= 0.0553 m

= 5.53 cm

中文解析

光栅缝距 d = (1 × 10−3)/(200) = 5 × 10−6 m

由 d sin θ_n = nλ，

(5 × 10−6) × sin θ_1 = 1 × (550 × 10−9)

sin θ_1 = 0.11

θ_1 = 6.315°

设 D 为光栅与屏幕原来的距离。

中央条纹与第一级条纹间距的增大量

= (D + 0.5) × tan 6.315° − D × tan 6.315°

= 0.5 × tan 6.315°

= 0.0553 m

= 5.53 cm

29Ch.6 第6章MC 选择题Average 中等2 marksPHY13061129

When a beam of white light falls onto a plane transmission grating, continuous spectra can be observed. This is because

Alight of different colours are refracted by different amounts by the grating.
Blight of different colours are diffracted by different amounts by the grating.
Cthe angles for bright fringes to form are different for light of different colours.
DNone of above
Chinese · 中文对照

当一束白光照射到平面透射光栅上时，可以看到连续光谱。这是因为

continuous spectra 连续光谱refracted 折射diffracted 衍射by different amounts 程度不同
Correct answer 正确答案C

The continuous spectra are formed by interference of light.

White light is composed of a mixture of a whole range of coloured lights. Light of different colours have different wavelengths. By d sin θn = nλ, the angles θn for bright fringes to form are different for light of different colours.

中文解析

连续光谱是由光的干涉形成的。

白光由一整段范围的各种色光混合而成。不同颜色的光波长不同。由 d sin θ_n = nλ，不同颜色的光形成亮纹的角度 θ_n 各不相同。

30Ch.6 第6章MC 选择题Average 中等2 marksPHY13061130

Which of the following support that light is a kind of wave?

(1)When light is reflected, its angle of incidence equals its angle of reflection.
(2)A light beam spreads after passing a narrow slit.
(3)Fringes can be observed when light passes through a double slit.
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

以下哪些支持「光是一种波」？

(1) 光被反射时，入射角等于反射角。

(2) 光束通过窄缝后会散开。

(3) 光通过双缝时可以观察到条纹。

narrow slit 窄缝fringe 条纹angle of reflection 反射角spreads 散开
Correct answer 正确答案C

Only waves exhibit diffraction and interference.

∴ (2) and (3) are correct.

Apart from waves, other objects also exhibit reflection.

∴ (1) is incorrect.

中文解析

只有波才会表现出衍射和干涉。

∴ (2) 和 (3) 正确。

除了波以外，其他物体也会发生反射。

∴ (1) 不正确。

31Ch.6 第6章MC 选择题Average 中等2 marksPHY13061131

Which of the following can be explained by the particle theory of light?

(1)Light travel in straight lines.
(2)Light can be reflected by mirrors.
(3)Regular reflection does not occur if a light beam falls onto a rough surface.
A(1) only
B(2) only
C(1) and (2) only
D(1), (2) and (3)
Chinese · 中文对照

以下哪些可以用光的微粒说（particle theory of light）解释？

(1) 光沿直线传播。

(2) 光可以被镜子反射。

(3) 光束照射到粗糙表面时不会发生规则反射（regular reflection）。

particle theory 微粒说regular reflection 规则反射rough surface 粗糙表面
Correct answer 正确答案D

Particles travel in straight lines and can be reflected.

∴ (1) and (2) are correct.

When particles collide with a rough surface, they are not reflected to the same direction.

∴ (3) is correct.

中文解析

微粒沿直线运动，而且可以被反射。

∴ (1) 和 (2) 正确。

微粒撞到粗糙表面时，不会都被反射到同一个方向。

∴ (3) 正确。

32Ch.6 第6章MC 选择题Easy 易2 marksPHY13061132

Which of the following electromagnetic waves can be applied to medical treatments?

(1)X-rays
(2)Gamma rays
(3)Infra-red radiation
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

以下哪些电磁波可以用于医学治疗？

(1) X 射线

(2) 伽马射线（gamma rays）

(3) 红外线

medical treatments 医学治疗medical imaging 医学成像gamma rays 伽马射线
Correct answer 正确答案D

X-rays and gamma rays can be used in medical imaging.

∴ (1) and (2) are correct.

An infra-red thermometer measures the body temperature of a person by detecting the infra-red radiation emitted by the person.

∴ (3) is correct.

中文解析

X 射线和伽马射线可用于医学成像。

∴ (1) 和 (2) 正确。

红外线温度计通过探测人体发出的红外线来测量体温。

∴ (3) 正确。

33Ch.6 第6章MC 选择题Average 中等2 marksPHY13061133

Diffraction of visible light is seldom noticed in daily life because

(1)the wavelength of visible light is very short.
(2)the light rays that we see in daily life have different wavelengths.
(3)the speed of light is very high.
A(1) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

日常生活中很少察觉到可见光的衍射，因为

(1) 可见光的波长很短。

(2) 我们在日常生活中看到的光线波长各不相同。

(3) 光速很高。

seldom noticed 很少被察觉comparable to 与……差不多gap 缝隙
Correct answer 正确答案A

The diffraction of a wave is observable only when the width of the gap is comparable to the wavelength of the wave. Since the wavelength of light is much shorter than the gaps we encounter in daily life, we seldom observe the diffraction of light.

∴ (1) is correct.

中文解析

只有当缝隙的宽度与波的波长相差不多时，波的衍射才容易被观察到。由于光的波长比日常生活中遇到的缝隙短得多，我们很少观察到光的衍射。

∴ (1) 正确。

34Ch.6 第6章MC 选择题Average 中等2 marksPHY13061134

A beam of violet light travels from a vacuum into a piece of glass. Which of the following statements is/are correct?

(1)The wavelength of the beam decreases.
(2)The frequency of the beam decreases.
(3)The speed of the beam decreases.
A(2) only
B(3) only
C(1) and (3) only
D(2) and (3) only
Chinese · 中文对照

一束紫光从真空射入一块玻璃。以下哪些陈述正确？

(1) 光束的波长减小。

(2) 光束的频率减小。

(3) 光束的速度减小。

violet light 紫光refracted 被折射medium 介质
Correct answer 正确答案C

The frequency of light remains unchanged when the light is refracted.

∴ (2) is incorrect.

When light travels from a vacuum to any medium, its speed v decreases. By v = fλ, its wavelength λ also decreases.

∴ (1) and (3) are correct.

中文解析

光发生折射时，频率保持不变。

∴ (2) 不正确。

光从真空进入任何介质时，速度 v 都会减小。由 v = fλ，它的波长 λ 也随之减小。

∴ (1) 和 (3) 正确。

35Ch.6 第6章MC 选择题Average 中等2 marksPHY13061135

In the Young's double slit experiment, a light wave travels towards a double slit. Which of the following quantities remain unchanged after the wave passes through the slits?

(1)Wavelength
(2)Frequency
(3)Speed
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

在杨氏双缝实验中，一个光波射向双缝。波通过双缝后，以下哪些物理量保持不变？

(1) 波长

(2) 频率

(3) 速度

remain unchanged 保持不变quantity 物理量double slit 双缝
Correct answer 正确答案D

The wavelength, frequency and speed of light remain unchanged when the light is diffracted at the slits.

中文解析

光在缝处发生衍射时，它的波长、频率和速度都保持不变。

36Ch.6 第6章MC 选择题Average 中等2 marksPHY13061136

In the Young’s double slit experiment, a beam of light of wavelength 0.4 μm passes through a double slit and falls onto a screen.

If the path difference at point P on the screen is 3 μm, which of the following are correct?

(1)The 8th-order dark fringe is formed at P.
(2)The fringe separation on the screen increases if a red light source is used.
(3)A bright fringe is formed at P if light of wavelength 500 nm is used.
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

在杨氏双缝实验中，一束波长为 0.4 μm 的光通过双缝后照射到屏幕上（见图）。如果屏幕上 P 点的光程差（path difference）是 3 μm，以下哪些正确？

(1) 在 P 点形成第 8 级暗纹。

(2) 如果改用红色光源，屏幕上的条纹间距会增大。

(3) 如果改用波长 500 nm 的光，在 P 点会形成亮纹。

path difference 光程差dark fringe 暗纹constructive interference 相长干涉destructive interference 相消干涉
Correct answer 正确答案D

Path difference at P = 3 μm = 7.5 × 0.4 μm = 7.5λ

Destructive interference occurs at P, so a dark fringe is formed there.

The 1st-order dark fringe corresponds to a path difference of 0.5λ.

The 2nd-order dark fringe corresponds to a path difference of 1.5λ.

¼

The 8th-order dark fringe corresponds to a path difference of 7.5λ.

∴ (1) is correct.

The fringe separation increases when the wavelength of light increases. The wavelength of red light is about 700 nm (0.7 μm).

∴ (2) is correct.

If light of wavelength 500 nm is used,

path difference at P = 3 μm = 6 × 0.5 μm = 6λ

Constructive interference occurs at P, so a bright fringe is formed there.

∴ (3) is correct.

中文解析

P 点的光程差 = 3 μm = 7.5 × 0.4 μm = 7.5λ

P 点发生相消干涉，所以在那里形成暗纹。

第 1 级暗纹对应的光程差是 0.5λ。

第 2 级暗纹对应的光程差是 1.5λ。

第 8 级暗纹对应的光程差是 7.5λ。

∴ (1) 正确。

光的波长增大时，条纹间距会增大。红光的波长约为 700 nm（0.7 μm）。

∴ (2) 正确。

如果改用波长 500 nm 的光，

P 点的光程差 = 3 μm = 6 × 0.5 μm = 6λ

P 点发生相长干涉，所以在那里形成亮纹。

∴ (3) 正确。

37Ch.6 第6章MC 选择题Average 中等2 marksPHY13061137

In a diffraction grating experiment, the distance between adjacent slits is d, the distance from the grating to the screen is s, and the wavelength of the monochromatic light used is λ. Can the separation between any two bright fringes and the highest order of bright fringe be obtained from the given information?

ASeparation between any two bright fringes · yes
Highest order of bright fringe · yes
BSeparation between any two bright fringes · yes
Highest order of bright fringe · no
CSeparation between any two bright fringes · no
Highest order of bright fringe · yes
DSeparation between any two bright fringes · no
Highest order of bright fringe · no
Chinese · 中文对照

在衍射光栅实验中，相邻两条缝之间的距离是 d，光栅到屏幕的距离是 s，所用单色光的波长是 λ。根据这些已知资料，能否求出任意两条亮纹之间的间距，以及最高级亮纹的级数？

highest order 最高级数given information 已知资料obtained 求出
Correct answer 正确答案A

The angle of diffraction θn for any bright fringe can be obtained by d sin θn = nλ, where n is the order of fringe. The distance yn of the bright fringe from the central bright fringe can then be obtained by yn = d tan θn. By finding yn for any two bright fringes, their separation can be obtained.

The highest order of bright fringe is equal to the largest integer n satisfying n ≤ dλ.

中文解析

由 d sin θ_n = nλ 可以求出任意一条亮纹的衍射角 θ_n，其中 n 是该条纹的级数。然后用 y_n = d tan θ_n 就可以求出该亮纹与中央亮纹之间的距离 y_n。求出任意两条亮纹的 y_n 之后，就能得到它们的间距。

最高级亮纹的级数等于满足 n ≤ (d)/(λ) 的最大整数 n。

38Ch.6 第6章MC 选择题Average 中等2 marksPHY13061138

In the following figure, the Young’s double slit experiment is set up and yellow fringes are found on the screen.

If the yellow filter is replaced with a red or a blue one, how will the fringe separation change?

ARed filter · increase
Blue filter · increase
BRed filter · increase
Blue filter · decrease
CRed filter · decrease
Blue filter · increase
DRed filter · decrease
Blue filter · decrease
Chinese · 中文对照

在下图的杨氏双缝实验装置中，屏幕上看到黄色条纹（见图）。如果把黄色滤光片（filter）分别换成红色和蓝色的滤光片，条纹间距会怎样变化？

yellow filter 黄色滤光片fringe separation 条纹间距replace 更换
Correct answer 正确答案B

The fringe separation increases when the wavelength of light increases.

Since red light has a longer wavelength than yellow light, the fringe separation increases when red filter is used.

Since blue light has a shorter wavelength than yellow light, the fringe separation decreases when blue filter is used.

中文解析

光的波长增大时，条纹间距会增大。

由于红光的波长比黄光长，使用红色滤光片时条纹间距增大。

由于蓝光的波长比黄光短，使用蓝色滤光片时条纹间距减小。

39Ch.6 第6章MC 选择题Average 中等2 marksPHY13061139

Which of the following can reduce the brightness of the fringes in Young’s double slit experiment?

(1)Use a light source with the same power but lower frequency.
(2)Use a double slit with a smaller slit separation.
(3)Place the screen further away from the light source.
A(1) only
B(3) only
C(1) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

以下哪些做法会降低杨氏双缝实验中条纹的亮度？

(1) 使用功率相同但频率较低的光源。

(2) 使用缝距较小的双缝。

(3) 把屏幕放得离光源更远。

brightness 亮度slit separation 缝距power 功率
Correct answer 正确答案B

The frequency of the light source and the slit separation are not related to the brightness of the fringes.

∴ (1) and (2) are incorrect.

When the screen is placed farther away from the light source, light is spread over a larger area on the screen and hence the brightness of the fringes decreases.

∴ (3) is correct.

中文解析

光源的频率和缝距都与条纹的亮度无关。

∴ (1) 和 (2) 不正确。

屏幕放得离光源更远时，光散布在更大的面积上，所以条纹的亮度降低。

∴ (3) 正确。

40Ch.6 第6章MC 选择题Average 中等2 marksPHY13061140

When a monochromatic light of wavelength 4 × 10–5 cm is directed normally to a plane transmission grating, the second order bright fringe is formed at an angle of 30° to the normal of the grating. What is the number of lines per cm of the grating?

A3125
B6250
C13 000
D25 000
Chinese · 中文对照

当波长为 4 × 10−5 cm 的单色光垂直照射到平面透射光栅上时，第二级亮纹出现在与光栅法线成 30° 的方向上。这块光栅每厘米有多少条线？

directed normally 垂直照射second order bright fringe 第二级亮纹normal to the grating 光栅的法线lines per cm 每厘米的线数
Correct answer 正确答案B

By d sin θn = nλ,

d sin 30° = 2 × (4 × 10−5 × 10−2)

d = 1.6 × 10−6 m

Number of lines per cm = 0.011.6×10−6 = 6250

中文解析

由 d sin θ_n = nλ，

d sin 30° = 2 × (4 × 10−5 × 10−2)

d = 1.6 × 10−6 m

每厘米的线数 = (0.01)/(1.6 × 10−6) = 6250

41Ch.6 第6章MC 选择题Average 中等2 marksPHY13061141

If monochromatic light of wavelength λ is directed normally at a plane transmission grating, what is the path difference from two adjacent slits at the 1st-order bright fringe?

A0
B
C
Dλ
Chinese · 中文对照

如果波长为 λ 的单色光垂直照射到平面透射光栅上，在第一级亮纹处，来自相邻两条缝的光的光程差是多少？

path difference 光程差adjacent slits 相邻的缝constructive interference 相长干涉
Correct answer 正确答案D

At positions where the path difference is equal to whole number of wavelengths (0, λ, 2λ…), constructive interference occurs and hence bright fringe is formed.

The central bright fringe corresponds to a path difference of 0.

The 1st-order bright fringe corresponds to a path difference of λ.

The nth-order bright fringe corresponds to a path difference of nλ.

中文解析

在光程差等于波长整数倍（0、λ、2λ……）的位置，会发生相长干涉，因此形成亮纹。

中央亮纹对应的光程差是 0。

第一级亮纹对应的光程差是 λ。

第 n 级亮纹对应的光程差是 nλ。

42Ch.6 第6章MC 选择题Average 中等2 marksPHY13061142

In the figure below, a transmitter produces circular electromagnetic waves of wavelength 3 cm which is reflected by a metal plate.

The receiver moves along the line PQ. Which of the following graphs best shows how the reading of the meter varies in the process?

A
B
C
D
Chinese · 中文对照

下图中，发射器（transmitter）产生波长为 3 cm 的圆形电磁波，这些波被一块金属板反射（见图）。接收器沿 PQ 线移动。以下哪个图像最能表示过程中电表读数的变化？

transmitter 发射器receiver 接收器metal plate 金属板amplitude 振幅
Correct answer 正确答案B

The electromagnetic waves travelling directly from the transmitter and those reflected by the metal plate meet and therefore interference occurs. As the path difference varies along PQ, constructive interference and destructive interference occur alternately along PQ. As a result, maximum and minimum readings are recorded alternately.

Note that the amplitude of the reflected waves are smaller than that of the waves directly from the transmitter, so the minimum readings are not zero.

Since

中文解析

直接从发射器传来的电磁波与被金属板反射的电磁波相遇，因此发生干涉。由于沿 PQ 线移动时光程差不断变化，沿 PQ 线会交替出现相长干涉和相消干涉。结果电表会交替记录到最大和最小读数。

注意：反射波的振幅小于直接从发射器传来的波的振幅，所以最小读数不是零。

由于（原解析在这里被截断，未写完。）

43Ch.6 第6章MC 选择题Average 中等2 marksPHY13061143

Which of the following statements concerning visible light and microwaves is correct?

AVisible light is electromagnetic wave while microwaves are not.
BVisible light and microwaves travel with the same speed in glass.
CVisible light can be diffracted while microwaves cannot.
DNone of above
Chinese · 中文对照

以下关于可见光和微波的陈述，哪一项正确？

visible light 可见光diffracted 衍射medium 介质
Correct answer 正确答案D

Both visible light and microwaves are electromagnetic waves.

∴ (1) is incorrect.

Electromagnetic waves with different wavelengths travel at different speeds in any media except in a vacuum.

∴ (2) is incorrect.

All electromagnetic waves can be diffracted.

∴ (3) is incorrect.

中文解析

可见光和微波都是电磁波。

∴ (1)（即 A）不正确。

除真空外，在任何介质中，波长不同的电磁波传播速度都不同。

∴ (2)（即 B）不正确。

所有电磁波都能发生衍射。

∴ (3)（即 C）不正确。

44Ch.6 第6章MC 选择题Average 中等2 marksPHY13061144

Which of the following statements about electromagnetic waves is/are correct?

(1)All kinds of electromagnetic wave travel at the same speed in any media.
(2)All kinds of electromagnetic wave can travel in a vacuum.
(3)All electromagnetic waves are transverse waves.
A(2) only
B(1) and (2) only
C(1) and (3) only
D(2) and (3) only
Chinese · 中文对照

以下关于电磁波的陈述，哪项／哪些正确？

(1) 所有种类的电磁波在任何介质中的传播速度都相同。

(2) 所有种类的电磁波都能在真空中传播。

(3) 所有电磁波都是横波。

transverse waves 横波vacuum 真空any media 任何介质
Correct answer 正确答案D

Electromagnetic waves with different wavelengths travel at different speeds in any media except in a vacuum.

∴ (1) is incorrect.

中文解析

除真空外，在任何介质中，波长不同的电磁波传播速度都不同。

∴ (1) 不正确。

所有电磁波都能在真空中传播，而且都是横波，所以 (2) 和 (3) 正确。

45Ch.6 第6章MC 选择题Average 中等2 marksPHY13061145

Arrange the following electromagnetic waves in ascending order of frequency: infra-red radiation (IR), ultra-violet radiation (UV), visible light, radio waves.

AIR < visible light < UV < radio waves
BUV < IR < visible light < radio waves
CRadio waves < IR < visible light < UV
DIR < visible light < radio waves < UV
Chinese · 中文对照

把以下电磁波按频率由小到大排列：红外线（IR）、紫外线（UV）、可见光、无线电波。

ascending order 由小到大排列infra-red radiation 红外线frequency 频率
Correct answer 正确答案C
中文解析

电磁波的频率由低到高依次是：无线电波 < 微波 < 红外线 < 可见光 < 紫外线 < X 射线。

所以按频率由小到大排列是：无线电波 < IR（红外线）< 可见光 < UV（紫外线）。

∴ 选 C。（原题未附文字解析，此顺序由电磁波谱决定。）

46Ch.6 第6章MC 选择题Difficult 难2 marksPHY13061146

A plane transmission grating ruled with 5000 lines per cm is illuminated normally by a beam of white light. If the wavelengths for yellow light and violet light are 0.6 μm and 0.4 μm respectively, which of the following statements is incorrect?

AThe central fringe is white.
BThe violet end of the first-order spectrum is closer to the central fringe than the red end of the first-order spectrum.
CThe 2nd-order bright fringe of yellow light coincides with the 3rd-order bright fringe of violet light.
DThere is no fourth-order bright fringe for violet light.
Chinese · 中文对照

一块每厘米刻有 5000 条线的平面透射光栅被一束白光垂直照射。如果黄光和紫光的波长分别是 0.6 μm 和 0.4 μm，以下哪一项是不正确的？

coincides with 与……重合violet end 紫端incorrect 不正确overlap 重叠
Correct answer 正确答案D

As constructive interference occurs at the central fringe for all wavelengths, the central fringe is white.

∴ A is correct.

By d sin θn = nλ, the longer the wavelength λ, the larger the angle of diffraction θn for the nth-order bright fringe. As red light has a longer wavelength than violet light, the 1st-order bright fringe of red light is farther away from the central fringe than the 1st-order bright fringe of violet light.

∴ B is correct.

d = 0.015000 = 2 × 10−6 m

Apply d sin θn = nλ.

For the 2nd-order bright fringe of yellow light,

(2 × 10−6) × sin θ2 = 2 × (0.6 × 10−6)

sin θ2 = 0.6

For the 3rd-order bright fringe of violet light,

(2 × 10−6) × sin θ3 = 3 × (0.4 × 10−6)

sin θ3 = 0.6

∴ C is correct.

n ≤ dλ = 2×10−60.4×10−6 = 5

The highest order of bright fringe for violet light is 5.

∴ D is incorrect.

中文解析

由于所有波长在中央条纹处都发生相长干涉，所以中央条纹是白色的。

∴ A 正确。

由 d sin θ_n = nλ，波长 λ 越长，第 n 级亮纹的衍射角 θ_n 就越大。由于红光的波长比紫光长，红光的第一级亮纹比紫光的第一级亮纹离中央条纹更远。

∴ B 正确。

d = (0.01)/(5000) = 2 × 10−6 m

应用 d sin θ_n = nλ。

对黄光的第二级亮纹：

(2 × 10−6) × sin θ_2 = 2 × (0.6 × 10−6)

sin θ_2 = 0.6

对紫光的第三级亮纹：

(2 × 10−6) × sin θ_3 = 3 × (0.4 × 10−6)

sin θ_3 = 0.6

∴ C 正确。

n ≤ (d)/(λ) = (2 × 10−6)/(0.4 × 10−6) = 5

紫光的最高级亮纹是第 5 级。

∴ D 不正确。

47Ch.6 第6章MC 选择题Difficult 难2 marksPHY13061147

A beam of light emitted by a filament bulb is directed towards a plane transmission grating. Which of the following statements must be correct?

(1)Both ends of the central fringe are violet in colour.
(2)The end of the 1st-order spectrum closer to the central fringe is violet in colour.
(3)The 1st-order spectrum will not overlap with the 2nd-order spectrum.
A(2) only
B(1) and (2) only
C(1) and (3) only
D(2) and (3) only
Chinese · 中文对照

一盏钨丝灯泡发出的光束照射到平面透射光栅上。以下哪些陈述一定正确？

(1) 中央条纹的两端都是紫色的。

(2) 第一级光谱中较靠近中央条纹的一端是紫色的。

(3) 第一级光谱不会与第二级光谱重叠。

filament bulb 钨丝灯泡must be correct 一定正确overlap 重叠central fringe 中央条纹
Correct answer 正确答案D

The central bright fringes for light of all wavelengths form at the same position.

∴ (1) is incorrect.

By d sin θn = nλ, the shorter the wavelength λ, the smaller the angle of diffraction θn for the nth-order bright fringe. As violet light has the shortest wavelength among the visible light, the 1st-order bright fringe of violet light is the closest to the central fringe than the other colours.

∴ (2) is correct.

From (2), we can see that light of short wavelength (violet) is at the end closer to the central fringe, and light of long wavelength (red) is at the end farther away from the central fringe.

Apply d sin θn = nλ.

For the 1st-order bright fringe of light with wavelength 700 nm,

d sin θ1 = 1 × (700 × 10−9)

sin θ1 = 7×10−7d

For the 2nd-order bright fringe of light with wavelength 400 nm,

d sin θ2 = 2 × (400 × 10−9)

sin θ2 = 8×10−7d

The violet end of the 2nd-order spectrum is closer to the central fringe than the red end of the 1st-order spectrum.

∴ (3) is correct.

中文解析

所有波长的光的中央亮纹都形成在同一个位置。

∴ (1) 不正确。

由 d sin θ_n = nλ，波长 λ 越短，第 n 级亮纹的衍射角 θ_n 就越小。由于紫光在可见光中波长最短，紫光的第一级亮纹比其他颜色更靠近中央条纹。

∴ (2) 正确。

由 (2) 可知，短波长（紫光）的光在较靠近中央条纹的一端，长波长（红光）的光在较远离中央条纹的一端。

应用 d sin θ_n = nλ。

对波长 700 nm 的光的第一级亮纹：

d sin θ_1 = 1 × (700 × 10−9)

sin θ_1 = (7 × 10−7)/(d)

对波长 400 nm 的光的第二级亮纹：

d sin θ_2 = 2 × (400 × 10−9)

sin θ_2 = (8 × 10−7)/(d)

第二级光谱的紫端比第一级光谱的红端更靠近中央条纹。

∴ (3) 正确。

1Ch.6 第6章Long Q 长题目Difficult 难10 marksPHY13063101

(a) The figure below shows a set-up with two identical sodium vapour lamps generating monochromatic light. A screen is placed at a certain distance away from the lamps. O is the point on the screen and it is equidistant from the two lamps. The intensity of light at O is I0.

In the figure below, sketch a graph showing how the intensity of light at different points on the screen varies with the distance y of the points from O. (1 mark)

(b) One lamp is then removed and the other lamp is placed at the mid-point of the original positions of the two lamps. A double slit is placed in front of the lamp. The distance between the slits is 8 × 10−4 m and the distance between the screen and the slits is 4 m. A series of dark and bright fringes appears on the screen.

(i) Explain why fringes cannot be observed in the set-up with two lamps. (1 mark)

(ii) Explain why dark fringes are formed. (3 marks)

(iii) On the screen, the distance between the 2nd bright fringe and the 5th bright fringe is 8.8 mm. Estimate the wavelength of the light emitted by the lamp. (3 marks)

(iv) State the advantage of measuring the distance between fringes far away rather than two neighbouring fringes in estimating wavelength of light. (1 mark)

(v) State an observable change on the screen if the slit separation is increased. (1 mark)

Chinese translation · 中文对照

(a) 下图（见图）的装置中有两盏完全相同的钠蒸气灯，它们发出单色光（monochromatic light）。在离灯一定距离处放一个屏。O 是屏上的一点，它到两盏灯的距离相等。O 点的光强（intensity）为 I_0。

在图（见图）中画出图像，显示屏上各点的光强如何随该点到 O 点的距离 y 变化。（1 分）

(b) 然后拿走一盏灯，把另一盏灯放在原来两盏灯位置的中点。在灯前面放一个双缝（double slit）。两缝间距为 8 × 10−4 m，屏与缝的距离为 4 m。屏上出现一系列明暗条纹（fringe）。

(i) 解释为什么用两盏灯的装置看不到条纹。（1 分）

(ii) 解释为什么会形成暗纹。（3 分）

(iii) 屏上第 2 条明纹与第 5 条明纹的距离为 8.8 mm。估算灯发出光的波长（wavelength）。（3 分）

(iv) 说明在估算光波长时，量相距较远的条纹之间的距离，比量两条相邻条纹之间的距离有什么好处。（1 分）

(v) 如果增大缝间距，指出屏上可观察到的变化。（1 分）

Approach · 思路提示

  • Two separate lamps are independent sources, so they are not coherent: the phase difference keeps changing randomly and the fringes shift so fast that they average out. The graph is therefore a nearly flat horizontal line at about I_0 (no fringes).
  • Always use 'path difference' to explain fringes: bright when path difference = nλ (n = 0, 1, 2...), dark when path difference = (n + 1/2)λ, where crests meet troughs and cancel (destructive interference).
  • From the 2nd to the 5th bright fringe there are 3 fringe spacings, not 5: Δy = 8.8 mm / 3 = 2.93 × 10−3 m.
  • Apply Δy = λD / a → λ = aΔy / D = (8 × 10−4 × 2.93 × 10−3) / 4 ≈ 5.9 × 10−7 m (about 590 nm, sodium yellow).
  • Measuring across many fringes makes Δy bigger, so the percentage error is smaller; but a larger a gives a smaller Δy = λD/a, i.e. fringes closer together and harder to see.
  • 两盏灯是两个独立的源，不是相干（coherent）的：相位差随机变化，条纹位置不停移动并相互叠加平均，所以看不到稳定条纹。图像应该画成一条接近水平、高度约 I_0 的直线（没有起伏）。
  • 解释条纹一律用「光程差（path difference）」：光程差 = nλ 时相长干涉（明纹）；光程差 = (n + 1/2)λ 时波峰与波谷相遇、互相抵消，相消干涉（暗纹）。
  • 从第 2 条明纹到第 5 条明纹只隔 3 个条纹间距（不是 5 个）：Δy = 8.8 mm ÷ 3 = 2.93 × 10−3 m。
  • 代入 Δy = λD / a，得 λ = aΔy / D = (8 × 10−4 × 2.93 × 10−3) ÷ 4 ≈ 5.9 × 10−7 m（约 590 nm，正是钠黄光）。
  • 量很多条条纹的总距离，Δy 变大，百分误差（percentage error）就变小，结果更准；相反，缝间距 a 增大时 Δy = λD/a 变小，条纹更密、更难看清。
monochromatic 单色的equidistant 等距的fringe 条纹slit separation 缝间距
Suggested answer 参考答案

Solutions

Marks

(a)

(Correct shape of the curve)

1A

(b) (i) The two lamps are not coherent sources.

1A

(ii) When light waves pass through the two slits, diffraction occurs.

1A

The light waves through the two slits arrive at some positions on the screen in antiphase.

1A

Destructive interference occurs at these positions and dark fringes are formed.

1A

(iii) Fringe separation Δy == 2.933 × 10−3 m

1M

By Δy =,

1M

λ === 5.87 × 10−7 m

1A

The wavelength of the light emitted by the lamp is 5.87 × 10−7 m.

(iv) The percentage error of the measurement can be reduced.

1A

(v) The separation between the fringes would increase.

1A

2Ch.6 第6章Long Q 长题目Difficult 难12 marksPHY13063102

(a) In the figure below, yellow light from a sodium lamp is directed normally onto a double slit. The wavelength of the light is 590 nm. Bright and dark fringes are observed on a screen at a distance of 2.3 m from the double slit.

(i) The variation of the intensity of light on the screen is shown in the figure below.

Explain why there are bright fringes and dark fringes shown on the screen. (2 marks)

(ii) The separation between the first and the sixth dark fringes is 6.5 cm. Estimate the slit separation of the double slit. (3 marks)

(b) The double slit is then replaced by a plane transmission grating which has 600 lines per mm. The variation of the intensity of light on the screen is shown in the figure below.

(i) State and explain two differences between the patterns observed on the screen produced by the double slit and the plane transmission grating. (4 marks)

(ii) Describe the changes in the peaks of intensity in the pattern when each of the following change is made independently.

(1) A plane transmission grating with 400 lines per mm is used.

(2) Monochromatic green light is used.

(3 marks)

Chinese translation · 中文对照

(a) 在下图（见图）中，钠灯发出的黄光垂直射到双缝上。光的波长为 590 nm。在与双缝相距 2.3 m 的屏上可观察到明纹和暗纹。

(i) 屏上光强的变化如下图所示（见图）。解释为什么屏上会出现明纹和暗纹。（2 分）

(ii) 第 1 条暗纹与第 6 条暗纹之间的距离为 6.5 cm。估算双缝的缝间距。（3 分）

(b) 然后把双缝换成每毫米 600 条刻线的平面透射光栅（plane transmission grating）。屏上光强的变化如下图所示（见图）。

(i) 写出并解释双缝与平面透射光栅在屏上产生的图样（pattern）的两个不同点。（4 分）

(ii) 描述下列每个改变（各自独立进行）会令图样中的光强峰（peak）发生什么变化。

(1) 改用每毫米 400 条刻线的平面透射光栅。

(2) 改用单色绿光。（3 分）

Approach · 思路提示

  • Explain with path difference from the two slits: bright fringe when path difference = nλ, dark fringe when path difference = (n + 1/2)λ (complete cancellation).
  • From the 1st to the 6th dark fringe there are 5 fringe spacings: Δy = 6.5 cm / 5 = 1.3 cm = 1.3 × 10−2 m.
  • Use Δy = λD / a → a = λD / Δy = (590 × 10−9 × 2.3) / (1.3 × 10−2) ≈ 1.0 × 10−4 m (about 0.10 mm).
  • Grating spacing d = 1 mm / 600 = 1.67 × 10−6 m. Grating peaks are very narrow/sharp and far apart with wide dark regions; double-slit fringes are broad and evenly spaced.
  • 400 lines/mm gives a larger d, so the maxima move closer together, more maxima appear and each is brighter; green light has a shorter λ, so all maxima move towards the centre and more of them fit on the screen.
  • 用两缝的光程差解释：光程差 = nλ 时为明纹；光程差 = (n + 1/2)λ 时光完全抵消，为暗纹。
  • 第 1 条暗纹到第 6 条暗纹之间有 5 个条纹间距：Δy = 6.5 cm ÷ 5 = 1.3 cm = 1.3 × 10−2 m。
  • 用 Δy = λD / a 反解：a = λD / Δy = (590 × 10−9 × 2.3) ÷ (1.3 × 10−2) ≈ 1.0 × 10−4 m（约 0.10 mm）。
  • 光栅常数 d = 1 mm ÷ 600 = 1.67 × 10−6 m。光栅的峰很窄很尖、彼此分得很开，中间大片是暗的；双缝的条纹很宽而且等距。
  • 400 条/mm 时 d 变大 → 相邻峰靠得更近、屏上峰的数目变多、每个峰更亮；绿光波长较短 → 所有峰向中央靠拢，屏上能看到的峰更多。
grating 光栅fringe 条纹peak 峰pattern 图样
Suggested answer 参考答案

Solutions

Marks

(a) (i) Light waves through the two slits arrive at some positions on the screen in phase and some positions in antiphase.

1A

Constructive interference occurs at the former and destructive interference occurs at the latter,

1A

forming bright fringes and dark fringes.

(ii) Fringe separation Δy == 1.3 × 10−2 m

1M

By Δy =,

1M

a === 1.04 × 10−4 m

1A

The slit separation of the double slit is 1.04 × 10−4 m.

(b) (i) The fringes produced by grating are further apart

1A

because the separation of neighbouring slits in the grating is much smaller than that in the double slit.

1A

The bright fringes produced by grating are brighter

1A

because the grating allows more light to pass through than the double slit does.

1A

(ii) (1) The peaks become further apart

1A

and the intensity of the peaks is higher.

1A

(2) The peaks become closer.

1A

3Ch.6 第6章Long Q 长题目Average 中等10 marksPHY13063103

The figure below shows the regions in the electromagnetic spectrum.

(a) State an estimation to the range of wavelength of visible light. (1 mark)

(b) Name regions A and B. (2 marks)

(c) In the set-up shown below, microwave emitted from the transmitter passes through a single narrow slit.

(i) When a receiver moves along PQ, the intensity of the microwave varies with positions along PQ as in the figure below.

The width of the slit is then reduced and the intensity along PQ is measured again. Sketch a new graph in the figure. Explain your answer. (4 marks)

(ii) The frequency of the microwave is 2.45 GHz. Find the wavelength of the microwave. (Given: 1 GHz = 109 Hz) (2 marks)

(d) Microwave can be used in microwave oven to cook food. During the cooking procedure, standing wave is formed in the microwave oven. Suggest why a rotating plane is required in a microwave oven. (1 mark)

Chinese translation · 中文对照

下图（见图）显示电磁波谱（electromagnetic spectrum）中的各个区域。

(a) 说出你估计的可见光波长范围。（1 分）

(b) 写出区域 A 和区域 B 的名称。（2 分）

(c) 在下图所示的装置（见图）中，发射器（transmitter）发出的微波穿过一条狭窄的单缝（single slit）。

(i) 当接收器（receiver）沿 PQ 移动时，微波强度沿 PQ 随位置的变化如下图所示（见图）。然后把缝宽减小，再次测量沿 PQ 的强度。在图中画出新的曲线，并解释你的答案。（4 分）

(ii) 微波的频率为 2.45 GHz。求微波的波长。（已知：1 GHz = 109 Hz）（2 分）

(d) 微波可用于微波炉煮食。煮食过程中炉内会形成驻波（standing wave）。建议为什么微波炉需要一块转盘（rotating plate）。（1 分）

Approach · 思路提示

  • Visible light is roughly 400 nm to 700 nm (about 4 × 10−7 m to 7 × 10−7 m).
  • Read the positions of A and B from the figure: the two regions next to visible light are infrared (longer λ side) and ultraviolet (shorter λ side); check which side each letter is on in the figure.
  • A narrower slit gives wider diffraction spreading: the new central maximum is broader and lower, and the side maxima also spread out and get weaker.
  • For the wave: λ = c / f = (3 × 108) / (2.45 × 109) ≈ 0.12 m (12 cm).
  • Standing waves create fixed hot spots (antinodes) and cold spots (nodes); the rotating plate turns the food so every part passes through the antinodes and is heated evenly.
  • 可见光波长大约由 400 nm 到 700 nm（即约 4 × 10−7 m 至 7 × 10−7 m）。
  • A、B 的名称要「从图中读出」（见图）：紧邻可见光两侧的通常是长波长一侧的红外线（infrared）和短波长一侧的紫外线（ultraviolet），看清楚图中 A、B 各在哪一侧。
  • 缝宽减小 → 衍射更显著：新的中央极大变得更宽但高度更低，两侧的次极大同样向外伸展、强度更弱（画出更矮更宽的曲线）。
  • 用 λ = c / f = (3 × 108) ÷ (2.45 × 109) ≈ 0.12 m（约 12 cm）。
  • 驻波会形成固定的热点（波腹 antinode）和冷点（波节 node）；转盘让食物各部分轮流经过波腹，所以受热均匀。
electromagnetic spectrum 电磁波谱slit 狭缝standing wave 驻波receiver 接收器
Suggested answer 参考答案

Solutions

Marks

(a) 400 nm to 700 nm (Accept ±50 nm)

1A

(b) Region A is ultra-violet radiation.

1A

Region B is infra-red radiation.

1A

(c) (i)

(Peak with lower intensity)

1A

(Spread out more horizontally)

1A

The intensity of the peak is lower because less microwave passes through the slit.

1A

The curve spreads out more horizontally because the degree of diffraction is greater for a narrower slit.

1A

(ii) By v = fλ,

1M

λ === 0.122 m

1A

The wavelength of the microwave is 0.122 m.

(d) A rotating plane can avoid the nodes of the stationary waves to stay at fixed positions of the food.

1A

4Ch.6 第6章Long Q 长题目Difficult 难10 marksPHY13063104

In the set-up shown below, monochromatic light emitted from a laser pointer passes through a double slit. A sensor is placed on the other side of the double-slit. Its output voltage increases when the intensity of light falling on it increases.

When the sensor moves along AC with a uniform speed, a graph of the output voltage against time as shown below is obtained. AC is 4.5 m away from the double slit. The sensor is at point B at time t = 20 ms.

(a) Explain why there is a series of peaks in the graph. (2 marks)

(b) Explain why the peak at t = 20 ms is the highest among all the peaks. (2 marks)

(c) The wavelength of the light is 540 nm and the slits are separated by 1.5 mm. Estimate the speed of the sensor when it moves along AC. (3 marks)

(d) Kitty wants to detect the speed of a car.

(i) Suggest a reason why the above set-up is not a desirable method. (1 mark)

(ii) A laser gun can be used to measure the speed of a car by measuring the distances from the car at two very close instants. Assume a pulse is emitted from a laser gun towards an object and the echo from the object is received after 9 × 10−7 s, find the distance of the object from the laser gun at that moment. (2 marks)

Chinese translation · 中文对照

在下图所示的装置（见图）中，激光笔发出的单色光穿过双缝。双缝的另一边放一个传感器（sensor），当照射到它的光强增大时，它的输出电压就增大。

当传感器沿 AC 以匀速移动时，得到下图（见图）所示输出电压随时间变化的图像。AC 与双缝相距 4.5 m。在 t = 20 ms 时传感器位于 B 点。

(a) 解释为什么图像中会出现一系列的峰（peak）。（2 分）

(b) 解释为什么 t = 20 ms 的那个峰是全部峰中最高的。（2 分）

(c) 光的波长为 540 nm，两缝相距 1.5 mm。估算传感器沿 AC 移动的速率（speed）。（3 分）

(d) Kitty 想测量一辆汽车的速率。

(i) 建议一个上述装置不合用的理由。（1 分）

(ii) 可以用激光枪（laser gun）在两个非常接近的时刻测量汽车的距离，从而测出汽车的速率。假设从激光枪向物体发出一个脉冲，9 × 10−7 s 后收到来自物体的回声，求此刻物体与激光枪的距离。（2 分）

Approach · 思路提示

  • As the sensor moves, the path difference from the two slits changes, so it alternately meets constructive interference (bright) and destructive interference (dark). The bright fringes give the peaks in the voltage graph.
  • At t = 20 ms the sensor is at B, on the central axis where the path difference is zero, so the waves always arrive in phase (central maximum). The central fringe is also the peak of the single-slit diffraction envelope, so it is the brightest.
  • Fringe separation: Δy = λD / a = (540 × 10−9 × 4.5) / (1.5 × 10−3) = 1.62 × 10−3 m.
  • The sensor moves at constant speed, so read the time Δt between two neighbouring peaks from the graph (见图), then v = Δy / Δt.
  • For the laser gun: the pulse travels there and back, so 2d = ct = 3 × 108 × 9 × 10−7 = 270 m, giving d = 135 m. Remember to halve the distance.
  • 传感器移动时，来自两缝的光的光程差不断改变，于是它轮流遇到相长干涉（亮纹，电压出现峰）和相消干涉（暗纹，电压下降）。
  • t = 20 ms 时传感器在 B 点，位于中央轴线上，光程差为零，两束光永远同相到达，所以是中央极大；而中央条纹又刚好在单缝衍射包络的中心，强度最大，因此这个峰最高。
  • 先求条纹间距：Δy = λD / a = (540 × 10−9 × 4.5) ÷ (1.5 × 10−3) = 1.62 × 10−3 m（约 1.6 mm）。
  • 传感器匀速移动，所以要从图中读出相邻两个峰之间的时间间隔 Δt（见图），再用 v = Δy ÷ Δt 求速率。
  • 激光枪那题：脉冲去一趟再回来，所以 2d = ct = 3 × 108 × 9 × 10−7 = 270 m，得 d = 135 m；切记最后的距离要除以 2。
sensor 传感器output voltage 输出电压peak 峰echo 回声
Suggested answer 参考答案

Solutions

Marks

(a) Light waves through the two slits arrive at some positions along AC in phase.

1A

Constructive interference occurs at these positions and the intensity measured by the sensor peaks at these positions.

1A

(b) The intensity of light decreases with the travelling distance.

1A

As the distance of point B from the two slits is the shortest among the positions where output voltage at peaks is measured,

1A

the peak corresponding to B is the highest among all the peaks.

(c) Fringe separation Δy =

1M

=

= 1.62 × 10−3 m

Speed v =

1M

=

= 0.216 m s−1

1A

(d) (i) The detected signal can be affected by the light from the surroundings.

1A

(ii) Distance from the object =

1M

=

= 135 m

1A

5Ch.6 第6章Long Q 长题目Difficult 难10 marksPHY13063105

A beam of white light is incident normally on a plane transmission grating and a pattern with some continuous spectra is formed on the screen behind the grating. The grating has 600 lines per mm.

(a) What is white light composed of? Which wave phenomenon does the white light occur when it passes through the plane transmission grating? (2 marks)

(b) Find the angular position of the 2nd order maximum of red light of wavelength 650 nm.
(3 marks)

(c) Light of a certain wavelength in the 3rd order spectrum overlaps with maximum of red light in (b). Find the wavelength of this light and its colour. (3 marks)

(d) Peter wanted to differentiate the different orders of spectra which overlap with each other. He suggested to replace the grating by one with smaller grating spacing. Determine if this method works and explain. (2 marks)

Chinese translation · 中文对照

一束白光垂直射到平面透射光栅（plane transmission grating）上，在光栅后面的屏上形成一些连续光谱（continuous spectrum）。光栅每毫米有 600 条刻线。

(a) 白光由什么组成？白光通过平面透射光栅时发生的是哪种波动现象（wave phenomenon）？（2 分）

(b) 求波长为 650 nm 的红光的第 2 级极大值（2nd order maximum）的角位置（angular position）。（3 分）

(c) 第 3 级光谱中某一波长的光，与 (b) 中红光的极大值重叠。求这种光的波长及其颜色。（3 分）

(d) Peter 想把互相重叠的不同级光谱区分开。他建议把光栅换成刻线间距更小的光栅。判断这个方法是否有效，并解释。（2 分）

Approach · 思路提示

  • White light is a mixture of light of many wavelengths (colours), from red to violet. Through a grating each wavelength is diffracted through its own angle — this wave phenomenon is diffraction (dispersion into a spectrum).
  • Grating spacing: d = 1 mm / 600 = 1.67 × 10−6 m.
  • Use d sin θ = nλ for n = 2: sin θ = (2 × 650 × 10−9) / (1.67 × 10−6) = 0.78, so θ ≈ 51°.
  • Overlap means the same angle: 3λ' = 2 × 650 nm, so λ' ≈ 433 nm, which is blue-violet.
  • The (d) method does not work: sin θ = nλ/d, so two orders overlap when nλ1 = (n + 1)λ2. The d cancels out — overlap depends only on the wavelengths, not on the grating spacing.
  • 白光由许多不同波长（颜色）的光混合而成，由红到紫连续分布。经过光栅时每种波长被衍射到不同的角度，这个波动现象就是衍射（diffraction），也就是色散成光谱。
  • 先求光栅常数：d = 1 mm ÷ 600 = 1.67 × 10−6 m。
  • 用 d sin θ = nλ，取 n = 2：sin θ = (2 × 650 × 10−9) ÷ (1.67 × 10−6) = 0.78，所以 θ ≈ 51°。
  • 「重叠」就是同一个角度：3λ' = 2 × 650 nm，得 λ' ≈ 433 nm，属于蓝紫色。
  • (d) 这个方法无效：由 sin θ = nλ/d 可知，两级光谱重叠的条件是 nλ1 = (n + 1)λ2，d 会被约掉，所以重叠与否只由波长决定，与光栅间距无关。
continuous spectrum 连续光谱angular position 角位置overlap 重叠grating spacing 光栅间距
Suggested answer 参考答案

Solutions

Marks

(a) White light is composed of a mixture of different colours of light.

1A

Diffraction occurs when white light passes through the plane transmission grating.

1A

(b) Grating spacing d == 1.667 × 10−6 m

1M

By d sin θ = nλ,

1M

sin θ === 0.780

⇒ θ = 51.3°

1A

The angular position is 51.3°.

(c) Apply d sin θ = nλ.

3λ = d sin θ = 2(650 × 10−9)

1M

⇒ λ = 4.33 × 10−7 m

1A

The colour of this light is violet.

1A

(d) This method does not work.

1A

As the grating spacing d changes, the angular positions of the two maxima in (c) changes but still equal to each other.

1A

6Ch.6 第6章Long Q 长题目Average 中等9 marksPHY13063106

A beam consisting of red and violet lights is shone normally on a plane transmission grating that has ruling of 500 lines per mm. The figure below shows the bright fringes of the first 2 orders.

The angular separation between the first-order fringe at W and the 0th-order fringe is 11.5o. The angular separation between the first-order fringe at X and the second-order fringe at Y is 4.5o.

(a) What is the colour of the fringe W? (1 mark)

(b) Find the wavelength of the violet light. (3 marks)

(c) Find the wavelength of the red light. (3 marks)

(d) Find the maximum order of fringe of the red light. (2 marks)

Chinese translation · 中文对照

一束由红光和紫光组成的光垂直射到每毫米 500 条刻线的平面透射光栅上。下图（见图）显示前 2 级的明纹。

W 处的第 1 级明纹与第 0 级明纹的角间距为 11.5°。X 处的第 1 级明纹与 Y 处的第 2 级明纹的角间距为 4.5°。

(a) W 处明纹是什么颜色？（1 分）

(b) 求紫光的波长。（3 分）

(c) 求红光的波长。（3 分）

(d) 求红光能显示的最高级数（maximum order）。（2 分）

Approach · 思路提示

  • First find the grating spacing: d = 1 mm / 500 = 2 × 10−6 m.
  • Longer wavelength is diffracted through a bigger angle, so the first-order fringe at the smaller angle is the violet one — W is violet.
  • For violet: λ = d sin θ / n = (2 × 10−6 × sin 11.5°) / 1 ≈ 3.99 × 10−7 m (about 399 nm).
  • Second-order violet: sin θ = 2λ/d = 0.399, θ ≈ 23.5°. Red first order is 4.5° less: θ ≈ 19.0°, so λ = d sin 19.0° ≈ 6.51 × 10−7 m (about 651 nm).
  • Highest order: n must satisfy nλ ≤ d, so n < d/λ = 2 × 10−6 / 6.51 × 10−7 ≈ 3.07, giving n_max = 3.
  • 先求光栅常数：d = 1 mm ÷ 500 = 2 × 10−6 m。
  • 波长越长，衍射角越大，所以同一级里角度小的那条是紫光 → W 是紫色。
  • 紫光：λ = d sin θ / n = (2 × 10−6 × sin 11.5°) ÷ 1 ≈ 3.99 × 10−7 m（约 399 nm）。
  • 紫光第 2 级：sin θ = 2λ/d = 0.399，θ ≈ 23.5°。红光第 1 级比它小 4.5°，θ ≈ 19.0°，所以 λ = d sin 19.0° ≈ 6.51 × 10−7 m（约 651 nm）。
  • 最高级数要满足 nλ ≤ d，即 n < d/λ = 2 × 10−6 ÷ 6.51 × 10−7 ≈ 3.07，取整数 n_max = 3。
order 级数angular separation 角间距violet 紫色grating 光栅
Suggested answer 参考答案

Solutions

Marks

(a) Violet

1A

(b) Grating spacing == 2 × 10−6 m

1M

By d sin θ = nλ,

1M

wavelength === 3.99 × 10−7 m

1A

(c) Consider the 2nd order bright fringe of violet light.

d sin θ = nλ,

2 × 10−6 × sin θ = 2 × 3.99 × 10−7

θ = 23.5°

1M

Angle of the 1st order red light = 23.5° − 4.5° = 19.0°

1M

Wavelength of red light === 6.51 × 10−7 m

1A

(d) nmax ≤

1M

=

= 3.07

\ The maximum order is 3.

1A

7Ch.6 第6章Long Q 长题目Average 中等10 marksPHY13063107

F E

A monochromatic light of wavelength λ falls onto two slits S1 and S2 and forms interference fringes on a screen at a distance D from the slits. The separation between S1 and S2 is a. It is given that D is much greater than a. P is a point on the screen and M is the mid-point of S1 and S2.

(a) It is given that λ is of the order 10–7 m. Suggest suitable values for a and D when setting up the experiment. (2 marks)

(b) Let N and θ be the point and angle as shown in the above figure. Show that S2N = a sin θ. (2 marks)

(c) (i) Derive an equation for λ, a and θ if constructive interference occurs at P. (2 marks)

(ii) Express the fringe width in terms of λ, a and D. You may assume that θ is very small so that sin θ ≈ tan θ. (3 marks)

(d) Explain why the fringe separation differs from your result in (c) (ii) for fringes far away from the central fringe. (1 mark)

Chinese translation · 中文对照

一束波长为 λ 的单色光射到两条缝 S_1 和 S_2 上，在与缝相距 D 的屏上形成干涉条纹。S_1 与 S_2 的间距为 a。已知 D 比 a 大得多。P 是屏上的一点，M 是 S_1 与 S_2 的中点。

（见图）

(a) 已知 λ 的数量级（order of magnitude）为 10−7 m。建议做这个实验时 a 和 D 的合适数值。（2 分）

(b) 设 N 和 θ 为上图（见图）中所示的点和角。证明 S_2N = a sin θ。（2 分）

(c) (i) 如果 P 点发生相长干涉（constructive interference），推导 λ、a 与 θ 之间的关系式。（2 分）

(ii) 用 λ、a 和 D 表示条纹宽度（fringe width）。可假设 θ 很小，因此 sin θ ≈ tan θ。（3 分）

(d) 解释为什么远离中央条纹时，条纹间距（fringe separation）会与你 (c)(ii) 的结果不同。（1 分）

Approach · 思路提示

  • To see clear fringes you need Δy = λD/a of a few millimetres: with λ ≈ 5 × 10−7 m, a ≈ 0.1 mm and D ≈ 1 m gives Δy ≈ 5 mm — a good practical choice.
  • For the proof: from S_1 drop a perpendicular to meet S_2P at N. In the right-angled triangle S_1S_2N the hypotenuse is S_1S_2 = a and the angle at S_1 is θ (because the screen is far away), so S_2N = a sin θ.
  • S_2N is the extra distance travelled by the ray from S_2. Constructive interference at P means this path difference equals a whole number of wavelengths: a sin θ = nλ.
  • For small θ, tan θ = y/D and sin θ ≈ tan θ, so y = nλD/a. The fringe width is the spacing between neighbouring fringes: Δy = λD/a.
  • Far from the centre θ is no longer small, sin θ ≈ tan θ fails and the real fringe spacing becomes slightly smaller than λD/a; the intensity of the fringes also drops.
  • 要看清条纹，Δy = λD/a 要有几毫米：取 λ ≈ 5 × 10−7 m、a ≈ 0.1 mm、D ≈ 1 m，则 Δy ≈ 5 mm，合情合理。
  • 证明的关键：由 S_1 向 S_2P 作垂线，垂足为 N。在直角三角形 S_1S_2N 中，斜边 S_1S_2 = a，S_1 处的角就是 θ（因为屏很远），所以 S_2N = a sin θ。
  • S_2N 就是来自 S_2 的光多走的那段路程。P 点相长干涉的条件是光程差等于整数个波长：a sin θ = nλ。
  • θ 很小时 tan θ = y/D，且 sin θ ≈ tan θ，于是 y = nλD/a。相邻两条明纹的距离就是条纹宽度 Δy = λD/a。
  • 离开中央很远时 θ 不再小，sin θ ≈ tan θ 不再成立，真实条纹间距会比 λD/a 略小；而且条纹的亮度也会下降。
order of magnitude 数量级constructive interference 相长干涉fringe width 条纹宽度perpendicular 垂直的
Suggested answer 参考答案

Solutions

Marks

(a) a = 0.2 mm

1A

D = 1.5 m

1A

(Or other reasonable values)

(b)

Since D is much greater than a,
it can be assumed that S1P is parallel to S2P.

1M

Therefore, α = 90° – θ

and ∠S2S1N = 90° – α = θ

1A

and hence S2N = a sin θ.

(c) (i) Constructive interference occurs at P only if the path difference at P is an integral multiple of the wavelength.

1A

i.e. S2N = nλ for some integer n

a sin θ = nλ

1A

(ii) Take n = 1, then the fringe width Δx is the distance between P and the central fringe.

1M

tan θ =

Since θ is very small, sin θ ≈ tan θ =

1M

From (c) (i), a sin θ = nλ

a × = 1 × λ

Δx =

1A

(d) The assumption that θ is small is invalid for fringes far away from the central fringe.

1A

8Ch.6 第6章Long Q 长题目Average 中等10 marksPHY13063108

(a) (i) Explain what diffraction is. (1 mark)

(ii) Explain why diffraction of light is an important evidence supporting the wave theory of light. (2 marks)

(b) A beam of monochromatic light is diffracted by a narrow slit and falls onto a screen. A fringe pattern is seen on the screen as shown below.

(i) Sketch a curve to show the intensity of the light against the position on the screen.
(2 marks)

(ii) Describe the fringe pattern shown on the screen. (2 marks)

(iii) Describe and explain what will be seen on the screen if the slit is replaced with a much wider one. (3 marks)

Chinese translation · 中文对照

(a) (i) 解释什么是衍射（diffraction）。（1 分）

(ii) 解释为什么光的衍射是支持光的波动说（wave theory of light）的重要证据。（2 分）

(b) 一束单色光被一条狭缝衍射后落在屏上。屏上看到的条纹图样如下图所示（见图）。

(i) 画出显示光强随屏上位置变化的曲线。（2 分）

(ii) 描述屏上显示的条纹图样。（2 分）

(iii) 描述并解释如果把狭缝换成一个宽得多的缝，屏上会看到什么。（3 分）

Approach · 思路提示

  • Diffraction is the spreading (bending) of waves after they pass through a narrow slit or around an obstacle.
  • Light can bend and spread out and form a fringe pattern after a narrow slit; a stream of particles would travel in straight lines and never do this, and the pattern depends on wavelength — strong support for the wave theory.
  • For the sketch: a tall, wide central maximum with narrower, much lower secondary maxima on both sides (each side maximum is only about a few percent of the central height).
  • Central bright fringe is the widest and brightest; the side fringes are narrower and become dimmer the further they are from the centre.
  • A much wider slit diffracts much less: the pattern becomes narrower with fringes closer together; when the slit is much wider than the wavelength the spreading is negligible and light just goes straight, so the fringes cannot be seen.
  • 衍射就是波经过狭窄的缝或绕过障碍物后，向外散开（弯折）的现象。
  • 光能绕过狭缝散开并形成条纹；如果光是微粒，就只会直线前进，不会这样散开；而且图样与波长有关 —— 这些都是波动说的有力证据。
  • 画图要点：中央极大又高又宽，两侧的次极大又窄又矮（旁边的极大只有中央的百分之几高度）。
  • 中央亮纹最宽最亮；两侧亮纹较窄，离中央越远越暗。
  • 缝宽得多时衍射不明显：图样变窄、条纹靠得更近；当缝宽远大于波长，散开可以忽略，光近似直线传播，就看不清条纹了。
diffraction 衍射slit 狭缝fringe pattern 条纹图样wave theory 波动说
Suggested answer 参考答案

Solutions

Marks

(a) (i) Diffraction refers to the spreading or bending of waves when they travel past the edges of objects.

1A

(ii) Diffraction is shown by waves but not by particles.

1A

The diffraction of light shows that light is not a stream of particles

1A

and the wave theory of light better explains the behaviours of light than other theories.

(b) (i)

(High peak at the central)

1A

(Smaller peaks on the two sides of the central peak)

1A

(ii) The fringe pattern consists of a central bright fringe

1A

and dimmer fringes on the two sides of the central fringe.

1A

(iii) The central fringe will be wider

1A

since the slit is wider and more light passes through the slit.

The edge of the central fringe will be less blurred

1A

and the dimmer fringes will not be observed

because diffraction is prominent only when the slit width is comparable to the wavelength of the light.

1A

9Ch.6 第6章Long Q 长题目Average 中等11 marksPHY13063109

(a) State the wave properties of light that relate to its colour and intensity. (2 marks)

(b) Refer to the following schematic diagram below. Interference fringes are formed on the screen when a monochromatic light is incident onto the double slit. The slit separation is 0.2 mm. The separation of the 6th-order bright fringe from the central bright fringe on the screen is 30 mm.

(i) Calculate the wavelength of the monochromatic light. (2 marks)

(ii) State any assumption in the above calculations. (1 mark)

(iii) What is the colour of the monochromatic light? (1 mark)

(c) (i) Describe what would be observed if the monochromatic light in (b) is replaced by a white light emitted from a filament bulb. (3 marks)

(ii) Explain the observations. (2 marks)

Chinese translation · 中文对照

(a) 说出与光的颜色和强度有关的波动性质（wave properties）。（2 分）

(b) 参考下面的示意图（见图）。当单色光射到双缝上时，屏上形成干涉条纹。缝间距为 0.2 mm。屏上第 6 级明纹与中央明纹的距离为 30 mm。

(i) 计算这种单色光的波长。（2 分）

(ii) 说出上述计算中的一个假设。（1 分）

(iii) 这种单色光是什么颜色？（1 分）

(c) (i) 描述如果把 (b) 中的单色光换成灯丝灯泡发出的白光，会观察到什么。（3 分）

(ii) 解释这些观察结果。（2 分）

Approach · 思路提示

  • Colour is related to wavelength (or frequency); intensity (brightness) is related to amplitude.
  • With a = 0.2 mm: the 6th bright fringe from the centre = 6 fringe spacings, so Δy = 30 mm / 6 = 5.0 × 10−3 m.
  • λ = aΔy / D = (0.2 × 10−3 × 5.0 × 10−3) / D = 1.0 × 10−6 / D — read the screen-to-slits distance D from the figure (见图) and substitute it.
  • Assumption: D is much greater than a (or: the angle is so small that sin θ ≈ tan θ), and the two slits are coherent and emit the same wavelength.
  • White light: the central fringe stays white (all colours overlap there), and on both sides there are coloured fringes with violet nearest the centre and red farthest away; higher orders overlap into white-ish mixtures.
  • 颜色与波长（或频率）有关；强度（亮度）与振幅（amplitude）有关。
  • 已知 a = 0.2 mm：第 6 级明纹离中央 6 个条纹间距，所以 Δy = 30 mm ÷ 6 = 5.0 × 10−3 m。
  • λ = aΔy / D = (0.2 × 10−3 × 5.0 × 10−3) ÷ D = 1.0 × 10−6 ÷ D，其中 D 要「从图中读出」（见图）再代入。
  • 假设：D 比 a 大得多（等价于 θ 很小、sin θ ≈ tan θ），而且两条缝是相干的、发出同一波长。
  • 换成白光后：中央条纹仍是白色（所有颜色在这里重合），两侧出现彩色条纹，靠中央是紫、最外侧是红；更高级数时不同颜色的条纹互相重叠，看起来混成白色。
wave property 波动性质amplitude 振幅fringe 条纹filament bulb 灯丝灯泡
Suggested answer 参考答案

Solutions

Marks

(a) The colour of the light is related to the wavelength.

1A

The intensity of the light is related to the amplitude.

1A

(b) (i) Δy =

1M

λ === 6.67 × 10–7 m

1A

(ii) The distance between the double slit and the screen is much greater than the slit width.

1A

(iii) Red

1A

(c) (i) The central fringe is white

1A

while other fringes show a continuous spectrum

1A

with the side nearer to the central fringe being violet and the other side being red.

1A

(ii) Light emitted from a filament bulb is a mixture of light with different colours (wavelengths).

1A

For different colours (wavelengths), the positions on the screen for constructive interference to occur are different.

1A

10Ch.6 第6章Long Q 长题目Average 中等11 marksPHY13063110

(a) Electromagnetic waves are varying electric and magnetic fields propagating in space.

(i) State three properties of electromagnetic waves. (3 marks)

(ii) State the two kinds of electromagnetic waves having the highest frequency ranges. Give one application for each of them. (4 marks)

(b) In the figure below, a transmitter emits radio signals to the surrounding areas.

(i) Diffraction of waves occurs during the transmission. Explain whether waves with a longer or shorter wavelength are advantageous for reception in the radio shadows as indicated. (2 marks)

(ii) Explain how low-flying aeroplanes may affect the reception of radio signals.
(2 marks)

Chinese translation · 中文对照

(a) 电磁波是在空间中传播的变化的电场和磁场。

（见图）

(i) 写出电磁波的三个性质。（3 分）

(ii) 写出频率范围最高的两种电磁波，并各举一个应用例子。（4 分）

(b) 在下图（见图）中，发射器向周围地区发出无线电信号。

(i) 信号传播时会发生波的衍射。解释波长较长还是较短的波，更有利于在所示的电波阴影（radio shadow）区接收。（2 分）

(ii) 解释低飞的飞机可能怎样影响无线电信号的接收。（2 分）

Approach · 思路提示

  • Three properties (any three): transverse waves; the electric and magnetic fields are perpendicular to each other and to the direction of travel; they travel through vacuum at 3 × 108 m/s; they carry energy; they show reflection, refraction, diffraction, interference and polarisation.
  • Highest frequency ranges: X-rays and gamma rays. Applications: X-rays for medical imaging/security scanning; gamma rays for sterilising medical equipment or treating cancer.
  • Radio shadow: a longer wavelength is diffracted (spread) more, so it bends further around buildings and hills and reaches the shadow region — longer waves are better there.
  • Low-flying aircraft can block the signal (extending the shadow), and they also reflect the waves, so the receiver gets both the direct wave and the reflected wave — this causes interference and the signal strength varies.
  • 任写三个性质：横波；电场、磁场互相垂直，且都垂直于传播方向；可在真空中以 3 × 108 m/s 传播；携带能量；能反射、折射、衍射、干涉和偏振（polarisation）。
  • 频率最高的两种是 X 射线（X-rays）和伽马射线（gamma rays）。应用：X 射线用于医学造影、机场安检；伽马射线用于消毒医疗器材或治疗癌症。
  • 电波阴影区：波长越长，衍射（散开）越明显，能绕过建筑物和山丘弯折得更远、进入阴影区，所以较长的波更有利。
  • 低飞的飞机会挡住信号（令阴影区扩大），也会把电波反射到接收器，使接收器同时收到直达波和反射波，产生干涉（干扰），信号强度时强时弱。
radio shadow 电波阴影diffraction 衍射reception 接收gamma rays 伽马射线
Suggested answer 参考答案

Solutions

Marks

(a) (i) Any three of the following:

3 × 1A

They are transverse waves.

They do not require a medium to travel.

They have a speed 3 × 108 m s–1 in a vacuum.

They obey the relation v = fλ.

They show reflection, refraction, diffraction and interference.

They carry energy.

(Or other reasonable answers)

(ii) X-rays and gamma rays

1A + 1A

X-rays can be used for medical imaging.

1A

Gamma rays can be used to sterilize food.

1A

(Or other reasonable answers)

(b) (i) Waves with longer wavelengths are advantageous

1A

as they have larger degrees of diffraction.

1A

(ii) A low-flying aeroplane may reflect the radio waves emitted by the transmitter.

1A

The reflected waves interfere with the waves emitted directly from the transmitter and may cause a poor reception.

1A

11Ch.6 第6章Long Q 长题目Average 中等11 marksPHY13063111

(a) In Hong Kong, TV signals are broadcasted on UHF (ultra high frequency) at about 500 MHz. Some TV stations in other countries, such as Australia, use signals on VHF (very high frequency) at about 50 MHz.

(Given: speed of electromagnetic wave in air = 3 × 108 m s−1)

(i) Name the electromagnetic waves used in TV broadcasting. (1 mark)

(ii) Calculate the wavelengths of VHF and UHF waves at the frequencies given above.
(3 marks)

(iii) Will it be easier to receive strong signals on VHF or UHF in an urban area full of tall buildings? Explain your answer. (3 marks)

(iv) For a user 25 km away from a broadcasting station, what is the shortest time the signals take to reach the user? (2 marks)

(b) The table below shows the broadcasting frequencies of some local radio stations.

Radio station

FM

AM

RTHK 1

92.6 – 94.4 MHz

—

RTHK 2

94.8 – 96.9 MHz

—

RTHK 3

—

567 kHz

RTHK 4

97.6 – 98.9 MHz

—

RTHK 5

—

783 kHz

CRHK 1

88.1 – 89.5 MHz

—

CRHK 2

90.3 – 92.1 MHz

—

Metro Radio

99.7 MHz

—

Metro Finance

104 MHz

—

Metro Plus

—

1044 kHz

A radio station uses a wave of wavelength 3.4 m in broadcasting.

(i) Find the broadcasting frequency of the radio station. (1 mark)

(ii) Which radio station is it? (1 mark)

Chinese translation · 中文对照

(a) 在香港，电视信号以大约 500 MHz 的超高频（UHF，ultra high frequency）广播。一些外国的电视台（例如澳洲）则用大约 50 MHz 的甚高频（VHF，very high frequency）广播。（已知：电磁波在空气中的速率 = 3 × 108 m s−1）

(i) 写出电视广播所用的电磁波名称。（1 分）

(ii) 计算上述频率下 VHF 波和 UHF 波的波长。（3 分）

(iii) 在满布高楼大厦的市区，VHF 还是 UHF 更容易接收到强信号？解释你的答案。（3 分）

(iv) 对于离广播站 25 km 远的用户，信号到达该用户所需的最短时间是多少？（2 分）

(b) 下表显示一些本地电台的广播频率。

表格：

电台 || FM || AM

RTHK 1 || 92.6 – 94.4 MHz || —

RTHK 2 || 94.8 – 96.9 MHz || —

RTHK 3 || — || 567 kHz

RTHK 4 || 97.6 – 98.9 MHz || —

RTHK 5 || — || 783 kHz

CRHK 1 || 88.1 – 89.5 MHz || —

CRHK 2 || 90.3 – 92.1 MHz || —

Metro Radio || 99.7 MHz || —

Metro Finance || 104 MHz || —

Metro Plus || — || 1044 kHz

有一个电台用波长 3.4 m 的波广播。

(i) 求这个电台的广播频率。（1 分）

(ii) 这个电台是哪一个？（1 分）

Approach · 思路提示

  • TV broadcasting uses radio waves (UHF and VHF are just frequency bands of radio waves).
  • Use c = fλ: VHF λ = 3 × 108 / (50 × 106) = 6 m; UHF λ = 3 × 108 / (500 × 106) = 0.6 m.
  • In a city of tall buildings the longer wave diffracts more and bends around the buildings, so VHF (6 m) is easier to receive strongly; the 0.6 m UHF is blocked more easily and needs roughly line of sight.
  • Shortest time: t = d / c = 25 000 / (3 × 108) ≈ 8.3 × 10−5 s.
  • For 3.4 m: f = c / λ = 3 × 108 / 3.4 ≈ 8.82 × 107 Hz = 88.2 MHz, which falls in 88.1 – 89.5 MHz → CRHK 1.
  • 电视广播用的是无线电波（radio waves）；UHF、VHF 只是无线电波里的两个频段。
  • 用 c = fλ：VHF 波长 = 3 × 108 ÷ (50 × 106) = 6 m；UHF 波长 = 3 × 108 ÷ (500 × 106) = 0.6 m。
  • 市区高楼多，波长较长的波衍射更明显、能绕过建筑物，所以 VHF（6 m）较易收到强信号；0.6 m 的 UHF 容易被挡住，大致要直线传播（line of sight）。
  • 最短时间：t = d ÷ c = 25 000 ÷ (3 × 108) ≈ 8.3 × 10−5 s。
  • 由 3.4 m 求频率：f = c ÷ λ = 3 × 108 ÷ 3.4 ≈ 8.82 × 107 Hz = 88.2 MHz，落在 88.1 – 89.5 MHz 之内 → 是 CRHK 1。
broadcast 广播ultra high frequency 超高频wavelength 波长urban area 市区
Suggested answer 参考答案

Solutions

Marks

(a) (i) Radio wave

1A

(ii) By v = fλ,

1M

wavelength of VHF === 6 m

1A

wavelength of UHF === 0.6 m

1A

(iii) VHF

1A

VHF waves have longer wavelengths

1A

and diffract more when they encounter buildings.

1A

(iv) Shortest time ==

1M

= 8.33 × 10−5 s

1A

(b) (i) Frequency === 88.2 × 106 Hz

1A

(ii) CRHK 1

1A

12Ch.6 第6章Long Q 长题目Average 中等10 marksPHY13063112

(a) You are provided with a laser pointer, a double slit and a translucent screen.

(i) Describe, with the aid of a diagram, how you obtain an interference pattern. (3 marks)

(ii) Describe what you expect to observe in the experiment. Explain the pattern observed briefly. (3 marks)

(b) The double slit is replaced by a single slit.

(i) What kind of wave phenomenon will occur? Describe what you expect to observe in this case. (2 marks)

(ii) Explain why it is difficult to observe this phenomenon of light in daily life. (2 marks)

Chinese translation · 中文对照

(a) 给你一支激光笔、一个双缝和一块半透明屏（translucent screen）。

(i) 配合示意图，描述你怎样得到干涉图样（interference pattern）。（3 分）

(ii) 描述你预期在实验中观察到什么，并简要解释所观察到的图样。（3 分）

(b) 把双缝换成单缝。

(i) 会发生哪种波动现象？描述你预期在这情况下观察到什么。（2 分）

(ii) 解释为什么在日常生活中很难观察到光的这种现象。（2 分）

Approach · 思路提示

  • Set-up needed in the description and diagram: laser → double slit → screen some distance away, all in a straight line, with the laser beam hitting the slits squarely and the screen far from the slits.
  • Expected pattern: a series of equally spaced bright and dark fringes, the central bright fringe being the widest and brightest, getting dimmer away from the centre.
  • Explain with path difference: bright where the path difference is nλ (in phase), dark where it is (n + 1/2)λ (out of phase and cancels).
  • Single slit gives diffraction: a wide, bright central band with narrower, dimmer bands on both sides, and the central band is about twice as wide as the others.
  • Hard to see in daily life because everyday gaps/obstacles are much larger than the wavelength of light (about 10−7 m), and ordinary light sources are not monochromatic or coherent, so the pattern washes out.
  • 描述和画图都要有：激光笔 → 双缝 → 隔一段距离放屏，三者大致在同一直线上，激光要垂直照到双缝，屏要放得够远。
  • 预期观察：一系列明暗相间、间距相等的条纹，中央明纹最宽最亮，离中央越远越暗。
  • 用光程差解释：光程差 = nλ 处两光同相 → 明纹；光程差 = (n + 1/2)λ 处反相、互相抵消 → 暗纹。
  • 单缝的情况是衍射（diffraction）：中央是一条又宽又亮的亮带，两侧是较窄较暗的亮带，而中央亮带的宽度约为旁边亮带的两倍。
  • 日常生活中难观察到，是因为日常的缝或障碍物比光的波长（约 10−7 m）大得多，衍射很弱；而且普通光源不是单色、不相干，图样会模糊掉。
translucent 半透明的interference pattern 干涉图样laser pointer 激光笔phenomenon 现象
Suggested answer 参考答案

Solutions

Marks

(a) (i)

(Correct set-up)

1A

Direct a laser beam normally on the double slit.

1A

Place the screen at a distance behind the double slit to capture the interference pattern.

1A

(ii) A series of evenly spaced dark and bright lines called fringes will be produced.

1A

The bright fringes are formed due to constructive interference

1A

and the dark fringes are formed due to destructive interference.

1A

(b) (i) Diffraction of light occurs.

1A

When the laser passes through the slit, a wide central bright fringe and some dimmer narrow fringes on its two sides will be observed on the screen.

1A

(ii) Diffraction is prominent only if the size of the gap is comparable to the wavelength of the wave.

1A

Since the wavelength of light is in the order of 10–7 m, diffraction can be observed only when it passes through a very narrow slit.

1A

13Ch.6 第6章Long Q 长题目Average 中等11 marksPHY13063113

A student uses the following set-up to find the wavelength of a laser beam.

The laser beam is shone normally on a plane transmission grating which has ruling of 100 lines per mm. An interference pattern appears on a screen 0.5 m behind the grating. The student records the separation x between two bright fringes of the each order in the following table.

Order of fringe, n

1

2

3

4

x / cm

5.7

11.5

17.4

23.4

θ

λ / nm

(a) State one safety precaution of using laser. (1 mark)

(b) For each order, calculate the values of θ and λ in the table. (4 marks)

(c) Give a conclusion for the experiment. (1 mark)

(d) The width of the screen is 50 cm. Find the highest order of bright fringe shown on the screen. (4 marks)

(e) Using the same apparatus, how can the number of bright fringes formed on the screen be increased? (1 mark)

Chinese translation · 中文对照

一名学生用下面的装置（见图）求激光束的波长。激光束垂直射到每毫米 100 条刻线的平面透射光栅上。在光栅后面 0.5 m 处的屏上出现干涉图样。学生把每一级两条明纹之间的距离 x 记录在下表。

表格：

条纹级数 n || 1 || 2 || 3 || 4

x / cm || 5.7 || 11.5 || 17.4 || 23.4

θ || || || ||

λ / nm || || || ||

(a) 写出一个使用激光的安全措施。（1 分）

(b) 对每一级，计算表中的 θ 和 λ 数值。（4 分）

(c) 写出这个实验的结论。（1 分）

(d) 屏的宽度为 50 cm。求屏上能显示的最高明纹级数。（4 分）

(e) 用同一套仪器，怎样可以增加屏上形成的明纹数目？（1 分）

Approach · 思路提示

  • Safety: never point the laser at anyone's eyes (do not look directly into the beam); wear laser goggles and keep the beam below eye level.
  • Grating spacing d = 1 mm / 100 = 1.0 × 10−5 m. x is the distance between the two fringes of the same order, so y = x/2 and tan θ = y / 0.5 m.
  • Values: θ ≈ 3.3°, 6.6°, 9.9°, 13.2°; then λ = d sin θ / n gives about 569, 571, 571, 570 nm.
  • Conclusion: the wavelength of the laser is about 5.7 × 10−7 m (570 nm), constant for all orders, so the grating equation is verified.
  • For (d) the largest y that fits is 25 cm: tan θ = 0.25/0.5 = 0.5. Use nλ ≤ d to get n ≤ 7.8, but n = 8 would land at y ≈ 25.6 cm which is off the 50 cm wide screen, so the highest order shown is 7. For (e), use a grating with fewer lines per mm (larger d) or move the screen closer.
  • 安全措施：绝不要把激光射向别人的眼睛（不要直视激光）；戴激光护目镜，并让光束低于眼睛的高度。
  • 光栅常数 d = 1 mm ÷ 100 = 1.0 × 10−5 m。x 是同一级左右两条明纹的总距离，所以 y = x/2，tan θ = y ÷ 0.5 m。
  • 算得 θ ≈ 3.3°、6.6°、9.9°、13.2°；再用 λ = d sin θ / n，得约 569、571、571、570 nm。
  • 结论：激光的波长约为 5.7 × 10−7 m（570 nm），各级算出的值基本一致，说明光栅公式成立。
  • (d) 屏上最大 y = 25 cm，tan θ = 0.25 ÷ 0.5 = 0.5。由 nλ ≤ d 得 n ≤ 7.8，但 n = 8 时落在 y ≈ 25.6 cm，已经超出 50 cm 宽的屏，所以最高是第 7 级。(e) 可换用每毫米刻线较少的（d 较大的）光栅，或把屏移近光栅。
ruling 刻线order of fringe 条纹级数safety precaution 安全措施grating 光栅
Suggested answer 参考答案

Solutions

Marks

(a) Do not direct the laser beam to the eyes.

1A

(Or other reasonable answer)

(b) θ : 3.26°, 6.56°, 9.87°, 13.2°

4 × 0.5A

λ: 568, 571, 571, 570

4 × 0.5A

(c) Wavelength of the laser == 570 nm

1A

(d) tan θ =

θ = 26.57°

1M

By d sin θ = nλ,

1M

n ===7.85

1M

\ The highest order is 7.

1A

(e) Move the screen towards the grating.

1A

14Ch.6 第6章Long Q 长题目Average 中等11 marksPHY13063114

A student uses the following set-up to find the wavelength of a laser beam.

He places a screen at a distance D from a double slit of slit separation 0.2 mm. Then, he measures the separation x between the bright fringes of the third order. By setting D to different values, the following results are obtained.

D / m

0.5

1.0

1.5

2.0

2.5

x / cm

1.1

1.9

3.1

3.9

5.1

(a) State one safety precaution of the experiment. (1 mark)

(b) (i) Plot a graph of x against D on graph paper. (3 marks)

(ii) Express the slope of the graph in (b)(i) in terms of λ. (2 marks)

(iii) Estimate the wavelength λ of the laser. (2 marks)

(c) Comment on the following statement: ‘Using a double slit of wider slit separation will improve the accuracy of the result.’ (3 marks)

Chinese translation · 中文对照

一名学生用下面的装置（见图）求激光束的波长。他把屏放在离缝间距为 0.2 mm 的双缝距离 D 处，然后测量第 3 级明纹之间的距离 x。改变 D 的数值，得到以下结果。

表格：

D / m || 0.5 || 1.0 || 1.5 || 2.0 || 2.5

x / cm || 1.1 || 1.9 || 3.1 || 3.9 || 5.1

(a) 写出这个实验的一个安全措施。（1 分）

(b) (i) 在方格纸上画出 x 对 D 的图像。（3 分）

(ii) 用 λ 表示 (b)(i) 中图像的斜率（slope）。（2 分）

(iii) 估算激光的波长 λ。（2 分）

(c) 评论以下说法：「使用缝间距较大的双缝会提高结果的准确度（accuracy）。」（3 分）

Approach · 思路提示

  • Safety: do not point the laser beam at anyone's eyes (and do not look into the beam directly).
  • The graph of x against D is a straight line through the origin; the slope is x/D (in cm per m, convert units carefully).
  • x here is the distance between the two third-order fringes, i.e. 6 fringe widths: x = 6λD/a, so the slope = 6λ/a.
  • From the table the slope is about 2.0 cm per m = 0.020 (in the same units). Then λ = slope × a / 6 = 0.020 × 0.2 × 10−3 / 6 ≈ 6.7 × 10−7 m (about 670 nm, a red laser).
  • The statement is wrong: a wider slit separation gives a smaller fringe spacing (Δy = λD/a), so x is smaller and the percentage error in measuring x is larger, making the result less accurate. Better to use a larger D and measure across many fringes.
  • 安全措施：不要把激光射向别人的眼睛，也不要直视激光束。
  • x 对 D 的图像是一条过原点的直线，斜率就是 x/D（注意单位：表中 x 用 cm、D 用 m）。
  • 这里的 x 是左右两条第 3 级明纹的距离，即 6 个条纹间距：x = 6λD/a，所以斜率 = 6λ/a。
  • 由表算得斜率约为每米 2.0 cm，即 0.020（单位要统一）。于是 λ = 斜率 × a ÷ 6 = 0.020 × 0.2 × 10−3 ÷ 6 ≈ 6.7 × 10−7 m（约 670 nm，属红色激光）。
  • 这个说法不对：缝间距 a 增大时，由 Δy = λD/a 可知条纹间距变小，x 也随之变小，量 x 的百分误差反而变大，准确度更差。要提高准确度，应增大 D 并量取很多条条纹的总距离。
slit separation 缝间距slope 斜率accuracy 准确度graph paper 方格纸
Suggested answer 参考答案

Solutions

Marks

(a) Do not direct the laser beam to the eyes.

1A

(Or other reasonable answer)

(b) (i)

(Correct axes)

1A

(Correct data points)

1A

(Straight line graph passing through origin)

1A

(ii) By Δy =,

1M

=

Slope ==== 3 × 104 λ

1A

(iii) Slope = 3 × 104 λ

= 3 × 104 λ

1M

λ = 6.67 × 10−7 m

1A

\ The wavelength of the laser is 6.67 × 10−7 m.

(c) The statement is incorrect.

1A

When the slit separation decreases, the fringe separation decreases.

1A

Therefore, the error in measuring x will increase.

1A

15Ch.6 第6章Long Q 长题目Difficult 难10 marksPHY13063115

A beam of white light is reflected normally from a soap film of refractive index 1.33. It is then directed to a diffraction grating at normal incidence. A dark band is observed in the first order spectrum.

(a) Explain why a dark band is observed in the first order spectrum. (4 marks)

(b) It is known that, for a soup film of thickness p, a reflected ray with wavelength λ cannot be observed in the spectrum if λ satisfies the relation 2np = mλ, where n is the refractive index of the soap film and m = 1, 2, 3…

If the dark band appears at 18° to the normal and the grating has 500 lines per mm,

(i) calculate the wavelength of this dark band, (2 marks)

(ii) find the minimum thickness of soap film. (2 marks)

(iii) find the next possible thickness of the soap film. (2 marks)

Chinese translation · 中文对照

一束白光垂直射到折射率（refractive index）为 1.33 的肥皂薄膜上被反射，然后垂直射到衍射光栅上。在光的第 1 级光谱中观察到一条暗带（dark band）。

（见图）

(a) 解释为什么在第 1 级光谱中会观察到暗带。（4 分）

(b) 已知对于厚度为 p 的肥皂膜，如果波长 λ 满足关系 2np = mλ（n 是肥皂膜的折射率，m = 1, 2, 3…），则这波长的反射光在光谱中观察不到。

如果暗带出现在与法线成 18° 的方向，而光栅每毫米有 500 条刻线，

(i) 计算这条暗带的波长，（2 分）

(ii) 求肥皂膜的最小厚度，（2 分）

(iii) 求肥皂膜下一个可能的厚度。（2 分）

Approach · 思路提示

  • Explain the film with thin-film interference: light reflected from the top surface and from the bottom surface of the film recombine. Because the film is optically denser than air, one reflection gives a phase reversal of half a wavelength.
  • For that wavelength the two reflected waves cancel (destructive interference, condition 2np = mλ), so that colour is missing from the reflected beam.
  • The grating then separates the light by angle; at the angle where that missing wavelength's first-order maximum would be, nothing arrives → a dark band in the 1st order spectrum.
  • Grating spacing d = 1 mm / 500 = 2 × 10−6 m; first order: λ = d sin 18° = 2 × 10−6 × 0.309 ≈ 6.2 × 10−7 m (about 618 nm).
  • Minimum thickness uses m = 1: p = λ/(2n) = 618 × 10−9 / (2 × 1.33) ≈ 2.3 × 10−7 m; the next possible thickness uses m = 2, which is exactly double: ≈ 4.6 × 10−7 m.
  • 先用薄膜干涉（thin-film interference）解释：由薄膜上表面和下表面分别反射的两束光重新叠加。因为肥皂膜比空气光密，其中一次反射会有半波长的相位反转（phase reversal）。
  • 对这个波长而言，两束反射光互相抵消（相消干涉，条件 2np = mλ），于是这个颜色在反射光中完全缺失。
  • 光栅把光按角度分开，在「这个缺失波长本应出现第 1 级极大」的方向上收不到光，于是第 1 级光谱里就出现一条暗带。
  • 光栅常数 d = 1 mm ÷ 500 = 2 × 10−6 m；取第 1 级：λ = d sin 18° = 2 × 10−6 × 0.309 ≈ 6.2 × 10−7 m（约 618 nm）。
  • 最小厚度取 m = 1：p = λ ÷ (2n) = 618 × 10−9 ÷ (2 × 1.33) ≈ 2.3 × 10−7 m；下一个可能厚度取 m = 2，正好是它的两倍 ≈ 4.6 × 10−7 m。
refractive index 折射率soap film 肥皂薄膜normal incidence 垂直入射dark band 暗带
Suggested answer 参考答案

Solutions

Marks

(a) When light falls on the film, reflection occurs at both surfaces of the soup film.

1A

The two reflected rays interfere with each other.

1A

For a certain wavelength, destructive interference occurs.

1A

Hence, such a wavelength will be absent in the spectrum and a dark band will be observed.

1A

(b) (i) Slit separation d == 2 × 10–6 m

For the first order spectrum,

by d sin θ = nλ,

1M

λ =

= 6.18 × 10–7 m

1A

\ The wavelength is 6.18 × 10–7 m.

(ii) When the thickness is the minimum,

2np = λ

1M

p =

= 2.32 × 10–7 m

1A

\ The minimum thickness is 2.32 × 10–7 m.

(iii) The next possible thickness correspond to m = 2.

2np = 2λ

1M

p =

= 4.65 × 10–7 m

1A

\ The thickness is 4.65 × 10–7 m.

1Ch.6 第6章Short Q 短题目Average 中等6 marksPHY13062101

In the figure below, monochromatic light from a laser pointer is directed normally onto a double slit with slit separation 0.8 mm. Bright and dark fringes are observed on a screen at a distance of 5 m from the double slit. On the screen, the distance between the 1st bright fringe and the 4th bright fringe on the same side is 7.5 mm.

(a) Find the wavelength of the light and name its colour. (4 marks)

(b) Estimate the number of bright fringes within a width of 0.1 m at the centre of the screen.
(1 mark)

(c) A laser pointer emitting light of a longer wavelength is then used. Describe the change in the pattern on the screen. (1 mark)

Chinese translation · 中文对照

（见图）图中，激光笔发出的单色光（monochromatic light）垂直照射在一双缝（double slit）上，双缝的缝距为 0.8 mm。在离双缝 5 m 远的屏幕上可以观察到明、暗相间的条纹（fringe）。屏幕上，同一侧的中央第 1 条亮纹（即第 1 级亮纹）与第 4 条亮纹之间相距 7.5 mm。

(a) 求该光的波长，并说出它的颜色。(4 分)

(b) 估计屏幕中央宽度 0.1 m 的范围内有多少条亮纹。(1 分)

(c) 然后换用发出较长波长光的激光笔，描述屏幕上图案的变化。(1 分)

Approach · 思路提示

  • Use the double-slit relation d sin θ = nλ; for small angles it becomes Δy = λD / a, where a = slit separation and D = slit-to-screen distance.
  • The 1st and the 4th bright fringe are 3 fringe spacings apart, so Δy = 7.5 mm / 3 = 2.5 mm.
  • Substitute a = 0.8 mm = 8 × 10−4 m, D = 5 m, Δy = 2.5 × 10−3 m: λ = Δy a / D = 4 × 10−7 m = 400 nm, which corresponds to violet colour.
  • (b) Number of spacings in 0.1 m = 0.1 / (2.5 × 10−3) = 40, so about 40 bright fringes (about 41 if the central one is counted).
  • (c) From Δy = λD / a, a longer λ gives a larger Δy: the fringes become wider apart and fewer of them fit in the same width, while the central fringe stays at the centre.
  • 用双缝公式 d sin θ = nλ；小角度下可写成条纹间距 Δy = λD / a，其中 a 是缝距，D 是缝到屏幕的距离。
  • 第 1 条和第 4 条亮纹之间相隔 3 个条纹间距，所以 Δy = 7.5 mm ÷ 3 = 2.5 mm。
  • 代入 a = 0.8 mm = 8 × 10−4 m、D = 5 m、Δy = 2.5 × 10−3 m：λ = Δy a ÷ D = 4 × 10−7 m = 400 nm，属紫色。
  • (b) 0.1 m 内有几个条纹间距：0.1 ÷ (2.5 × 10−3) = 40，所以约有 40 条亮纹（若连中央亮纹一起数，约 41 条）。
  • (c) 由 Δy = λD / a 可知，λ 变长则 Δy 变大：条纹变宽、同样宽度内条纹数目变少，中央亮纹位置不变。
monochromatic 单色的slit separation 缝距fringe 条纹on the same side 在同一侧
Suggested answer 参考答案

Solutions

Marks

(a) Δy == 2.5 × 10−3 m

1M

By Δy =,

1M

λ === 4 × 10−7 m

1A

The wavelength of the light is 4 × 10−7 m

and it is a violet light.

1A

(b) Number of bright fringes ≈= 40

1A

(Accept 41)

(c) The fringes are more widely spaced.

1A

2Ch.6 第6章Short Q 短题目Difficult 难3 marksPHY13062102

In the set-up below, the transmitter emits microwave. A partial reflector is placed in front of the transmitter. It partially reflects about half of the microwave incident on it and allows the remaining microwave to pass through. The metal plate behind it reflects all the microwave incident on it. The receiver just next to the transmitter detects the intensity of the reflected microwave.

When the partial reflector is moved towards the transmitter/receiver slowly, the intensity detected by the receiver reaches maximum and minimum values alternately. Explain this phenomenon. (3 marks)

Chinese translation · 中文对照

（见图）在下面的装置中，发射器发出微波（microwave）。在发射器前方放置一块半反射片（partial reflector），它把照射到片上约一半的微波反射回去，并让其余微波通过。它后面的金属板则把照射到板上的微波全部反射。紧靠发射器的接收器用来探测反射回来的微波的强度（intensity）。

当半反射片缓慢地向发射器／接收器移动时，接收器探测到的强度交替出现最大值和最小值。解释这个现象。(3 分)

Approach · 思路提示

  • There are two reflected waves arriving at the receiver: one reflected by the partial reflector, and one that passed through it, was reflected by the metal plate and then came back through it.
  • The two waves come from the same source, so they are coherent and can interfere.
  • When the partial reflector moves a distance Δd, the path difference between the two reflected waves changes by 2Δd.
  • Path difference = nλ gives constructive interference → maximum intensity; path difference = (n + 1/2)λ gives destructive interference → minimum intensity.
  • As the reflector moves continuously the path difference changes continuously, so maxima and minima appear alternately; the separation between successive maxima is λ/2.
  • 到达接收器的反射波有两束：一束由半反射片反射，另一束穿过半反射片、被金属板反射后再穿回来。
  • 两束波来自同一个波源，所以它们是相干（coherent）的，可以产生干涉（interference）。
  • 半反射片移动 Δd 时，两束反射波的路程差（path difference）改变 2Δd。
  • 路程差 = nλ 时是相长干涉（constructive interference），强度最大；路程差 = (n + 1/2)λ 时是相消干涉（destructive interference），强度最小。
  • 半反射片持续移动，路程差就连续改变，所以极大值与极小值交替出现；相邻两个极大值对应半反射片移动 λ/2。
partial reflector 半反射片intensity 强度reflect 反射alternately 交替地
Suggested answer 参考答案

Solutions

Marks

The reflected wave from the partial reflector and the reflected wave from the metal plate interfere with each other.

1A

The reflected waves become in phase when the separation between the partial reflector and the metal plate equals to some values, and become in antiphase when the separation equals to some other values.

1A

Constructive interference occurs in the former case and destructive interference occurs in the latter case, giving maximum and minimum values of intensity.

1A

3Ch.6 第6章Short Q 短题目Average 中等5 marksPHY13062103

A beam of monochromatic light with wavelength 540 nm is directed normally onto a double slit and forms a pattern on the screen at 3 m behind the double slit. The separation between the first and the fifth bright fringes on the screen is 9 mm.

(a) Estimate the slit separation of the double slit. (3 marks)

(b) If the double slit is replaced by one with smaller slit separation, how will the fringe pattern change? (2 marks)

Chinese translation · 中文对照

一束波长为 540 nm 的单色光垂直照射到双缝上，在双缝后方 3 m 的屏幕上形成条纹图案。屏幕上第一条亮纹与第五条亮纹之间的距离是 9 mm。

(a) 估计双缝的缝距。(3 分)

(b) 如果把双缝换成缝距更小的双缝，条纹图案会怎样改变？(2 分)

Approach · 思路提示

  • The 1st and the 5th bright fringes are separated by 4 fringe spacings, so Δy = 9 mm / 4 = 2.25 mm.
  • Use Δy = λD / a and rearrange: a = λD / Δy.
  • Substitute λ = 540 nm = 5.4 × 10−7 m, D = 3 m, Δy = 2.25 × 10−3 m: a = 7.2 × 10−4 m = 0.72 mm.
  • (b) A smaller a makes Δy = λD / a larger: the bright fringes move farther apart and fewer fringes are seen, while their brightness and colour do not change.
  • 第 1 条到第 5 条亮纹之间相隔 4 个条纹间距，所以 Δy = 9 mm ÷ 4 = 2.25 mm。
  • 用 Δy = λD / a，变形得 a = λD ÷ Δy。
  • 代入 λ = 540 nm = 5.4 × 10−7 m、D = 3 m、Δy = 2.25 × 10−3 m：a = 7.2 × 10−4 m = 0.72 mm。
  • (b) a 变小，Δy = λD / a 就变大：亮纹间距变大、看到的条纹数目变少，但亮度、颜色不变，中央亮纹仍在中央。
estimate 估计fringe pattern 条纹图案slit separation 缝距
Suggested answer 参考答案

Solutions

Marks

(a) Δy == 2.25 × 10−3 m

1M

By Δy =,

1M

a === 7.2 × 10−4 m

1A

The slit separation of the double slit is 7.2 × 10−4 m.

(b) The fringe pattern will become dimmer.

1A

The fringes will become more widely spaced.

1A

4Ch.6 第6章Short Q 短题目Difficult 难6 marksPHY13062104

Monochromatic light is incident normally on a film with four narrow slits at equal spacing of 9.5 × 10−5 m. A screen is placed at 2.2 m behind this film.

(a) The figure below shows how the intensity of light varies along AC on the screen. Let x be the distance from the central point B and take x as positive at positions closer to A than C.

Estimate the wavelength of the light. (4 marks)

(b) The 1st and the 3rd slit counted from the top are then covered. Find the fringe separation in the pattern observed. (2 marks)

Chinese translation · 中文对照

单色光垂直照射在一张有 4 条窄缝（narrow slit）的胶片上，各缝的间距均为 9.5 × 10−5 m。在胶片后方 2.2 m 处放置一个屏幕。

(a)（见图）下图显示光强沿屏幕上 AC 方向的变化。设 x 为距中心点 B 的距离，靠近 A 的位置 x 取正值。

（见图）

估计该光的波长。(4 分)

(b) 然后遮住从上往下数的第 1 条和第 3 条缝。求所观察到的图案中的条纹间距（fringe separation）。(2 分)

Approach · 思路提示

  • Four slits behave like a grating: principal maxima obey d sin θ = nλ with d = 9.5 × 10−5 m.
  • Read from the graph the distance between two neighbouring principal maxima (the tallest peaks), which is the fringe separation Δx (see figure).
  • Between principal maxima there are weaker secondary maxima — do not mistake them for principal maxima when reading the graph.
  • For small angles, λ = Δx · d / D with D = 2.2 m.
  • (b) Covering the 1st and the 3rd slit leaves two slits (the 2nd and the 4th) separated by 2d = 1.9 × 10−4 m.
  • With a = 2d, the fringe separation Δy = λD / a becomes twice the value in (a).
  • 4 条缝相当于一个光栅（grating）：主极大（principal maximum）满足 d sin θ = nλ，其中 d = 9.5 × 10−5 m。
  • 从图中读出相邻两个主极大（最高峰）之间的距离，这就是条纹间距 Δx（见图）。
  • 主极大之间还有较弱的次极大（secondary maximum），读图时不要把次极大当成主极大。
  • 小角度下 λ = Δx · d ÷ D，其中 D = 2.2 m。
  • (b) 遮住第 1 和第 3 条缝后，剩下第 2、第 4 条缝，两缝相距 2d = 1.9 × 10−4 m。
  • 把 a = 2d 代入 Δy = λD / a，得到的条纹间距是 (a) 的两倍。
narrow slit 窄缝equal spacing 相等的间距intensity 强度principal maximum 主极大
Suggested answer 参考答案

Solutions

Marks

(a) The 1st order maximum is at x = 10 × 10−3 m.

1M

sin θ ≈ tan θ == 4.55 × 10−3

1M

By d sin θ = nλ,

1M

λ === 4.32 × 10−7 m

1A

The wavelength of the light is 4.32 × 10−7 m.

(b) Fringe separation Δy =

1M

=

= 5.00 × 10−3 m

1A

The fringe separation is 5.00 × 10−3 m.

5Ch.6 第6章Short Q 短题目Easy 易6 marksPHY13062105

In recent years, aerial photographing becomes very common. A video camera is installed at the bottom of a drone (unmanned aerial vehicle), which is lifted by at least four vertically oriented propellers. Radio signals are used to control the drone and transmit the image from the video camera to the observer on the ground. By using this, magnificent views could be recorded from the sky.

(Courtesy of Peter Linehan ½ Flickr.com)

The frequencies of the radio signals for controlling the drone and transmitting images are 2.4 GHz and 5.8 GHz respectively. (Given: 1 GHz = 109 Hz)

(a) Explain an advantage of using radio wave as control signals over other electromagnetic waves. (2 marks)

(b) Find the wavelengths of these radio signals. (3 marks)

(c) State one advantage of using radio signals of different frequencies. (1 mark)

Chinese translation · 中文对照

近年来，航拍变得十分普遍。无人驾驶飞行器（drone，unmanned aerial vehicle）的底部装有一台摄像机（video camera），无人机靠至少四个竖直方向的螺旋桨（propeller）升空。控制无人机以及把摄像机图像传送给地面观察者，都使用无线电信号。用这种方法可以从空中拍下壮丽的景色。

（图片由 Peter Linehan ½ Flickr.com 提供）

用于控制无人机的无线电信号频率是 2.4 GHz，用于传送图像的频率是 5.8 GHz。（已知：1 GHz = 109 Hz）

(a) 解释用无线电波（radio wave）作为控制信号，比其他电磁波（electromagnetic wave）有什么优点。(2 分)

(b) 求这两个无线电信号的波长。(3 分)

(c) 说出使用不同频率的无线电信号的一个优点。(1 分)

Approach · 思路提示

  • (a) Radio waves have a long wavelength, so they diffract strongly and can bend around buildings, trees and other obstacles; they are also barely absorbed by the atmosphere, so they can travel a long distance with little loss.
  • (b) Use λ = c / f with c = 3 × 108 m/s.
  • For 2.4 GHz: λ = 3 × 108 / 2.4 × 109 = 0.125 m.
  • For 5.8 GHz: λ = 3 × 108 / 5.8 × 109 = 0.0517 m ≈ 0.052 m.
  • (c) Two signals of different frequencies do not interfere with each other, so the control signal and the image signal can be transmitted at the same time without mutual interference (and more drones can be used at the same time).
  • (a) 无线电波波长很长，衍射（diffraction）明显，能绕过建筑物、树木等障碍物；而且大气对它的吸收很少，可以远距离传播、损耗小。
  • (b) 用 λ = c ÷ f，其中 c = 3 × 108 m/s。
  • 对 2.4 GHz：λ = 3 × 108 ÷ (2.4 × 109) = 0.125 m。
  • 对 5.8 GHz：λ = 3 × 108 ÷ (5.8 × 109) = 0.0517 m ≈ 0.052 m。
  • (c) 不同频率的信号互不干扰，所以控制信号和图像信号可以同时传送而不互相干扰（同时也允许多台无人机同时工作）。
propeller 螺旋桨unmanned aerial vehicle 无人驾驶飞行器transmit 传送advantage 优点
Suggested answer 参考答案

Solutions

Marks

(a) Radio wave has long wavelength.

1A

It can diffract around obstacles to maintain the transmission of signals.

1A

(b) Apply v = fλ ⇒ λ =.

1M

For 2.4 GHz signals,

wavelength == 0.125 m

1A

For 5.8 GHz signals,

wavelength == 0.0517 m

1A

(c) This can avoid interference of the two radio signals.

1A

6Ch.6 第6章Short Q 短题目Average 中等4 marksPHY13062106

The fringe separation of the interference pattern of light produced by a double slit is given by

(a) A beam of monochromatic light of wavelength 420 nm is incident normally on a double slit of slit separation 0.2 mm. An interference pattern is formed on a screen 2.4 m behind the double slit. Find the fringe separation of the interference pattern produced in mm.
(2 marks)

(b) The interference pattern of microwaves is studied using the following set-up.

The transmitter emits microwaves towards the slits. The receiver is moved along XY to locate the maxima. Ronald thinks that by applying the formula given above, the separation of the maxima obtained should match the experimental result. Is he correct? Explain your answer briefly. (2 marks)

Chinese translation · 中文对照

双缝产生的光的干涉（interference）图案，其条纹间距由下式给出：

（见图）

(a) 一束波长为 420 nm 的单色光垂直照射到缝距为 0.2 mm 的双缝上，在双缝后方 2.4 m 的屏幕上形成干涉图案。求所产生干涉图案的条纹间距，以 mm 为单位。(2 分)

(b) 用下面的装置研究微波（microwave）的干涉图案。

发射器向两条缝发出微波。接收器沿 XY 移动，以找出极大值（maxima）的位置。Ronald 认为：套用上面给出的公式，所得极大值的间距应该与实验结果相符。他是否正确？简要解释你的答案。(2 分)

Approach · 思路提示

  • (a) Use Δy = λD / a with λ = 420 nm = 4.2 × 10−7 m, a = 0.2 mm = 2 × 10−4 m, D = 2.4 m.
  • Δy = (4.2 × 10−7 × 2.4) / (2 × 10−4) = 5.04 × 10−3 m = 5.04 mm ≈ 5.0 mm.
  • (b) He is not correct.
  • The formula Δy = λD / a is only an approximation valid when the angle is small (λ << a and the fringes are near the centre).
  • For microwaves the wavelength is comparable with the slit separation, so the diffraction angle is large and sin θ is no longer equal to tan θ; the exact relation a sin θ = nλ must be used, so the measured separation does not match the formula.
  • (a) 用 Δy = λD / a，其中 λ = 420 nm = 4.2 × 10−7 m，a = 0.2 mm = 2 × 10−4 m，D = 2.4 m。
  • Δy = (4.2 × 10−7 × 2.4) ÷ (2 × 10−4) = 5.04 × 10−3 m = 5.04 mm ≈ 5.0 mm。
  • (b) 他不对。
  • 公式 Δy = λD / a 只是小角度下的近似（要求 λ 远小于 a、条纹靠近中央）。
  • 微波的波长与缝距差不多，衍射角很大，sin θ 不再约等于 tan θ，必须用严格的 a sin θ = nλ，所以实测的极大值间距与公式结果不符。
fringe separation 条纹间距maxima 极大值incident normally 垂直入射match 与……相符
Suggested answer 参考答案

Solutions

Marks

(a) Fringe separation Δy =

=

1M

= 0.005 04 m

= 5.04 mm

1A

(b) No.

1A

The formula is applicable only if D is much larger than a.

1A

7Ch.6 第6章Short Q 短题目Easy 易3 marksPHY13062107

In Figure a, a beam of red light falls onto a double slit and forms a fringe pattern on a screen as shown in Figure b.

Fig a

Fig b

Explain the formation of the fringe using the wave theory of light. (3 marks)

Chinese translation · 中文对照

在图 a 中，一束红光照射到双缝上，在屏幕上形成如图 b 所示的条纹图案。

（见图）

图 a

（见图）

图 b

用光的波动说（wave theory of light）解释这些条纹的形成。(3 分)

Approach · 思路提示

  • The light from the two slits comes from the same source, so it is coherent: the two waves have a constant phase difference.
  • The two waves superpose at each point on the screen; the result depends on their path difference.
  • Path difference = nλ → constructive interference → bright fringe; path difference = (n + 1/2)λ → destructive interference → dark fringe.
  • The path difference changes steadily as we move across the screen, so bright and dark fringes appear alternately at equal spacing, with a bright fringe at the centre where the path difference is zero.
  • 两条缝的光来自同一光源，所以是相干（coherent）的：两束波的相位差（phase difference）保持不变。
  • 两束波在屏幕上每一点叠加（superpose），结果取决于它们的光程差／路径差（path difference）。
  • 路程差 = nλ → 相长干涉 → 亮纹；路程差 = (n + 1/2)λ → 相消干涉 → 暗纹。
  • 沿屏幕移动时路程差连续变化，所以亮、暗纹等距交替出现；中央处路程差为零，是亮纹。
fringe 条纹wave theory 波动说formation 形成
Suggested answer 参考答案

Solutions

Marks

When a light wave falls onto the double slit,

diffraction occurs at each of the slit

1A

and the two diffracted waves interfere.

1A

Bright fringes are formed at places where constructive interference occur and dark fringes are formed at places where destructive interference occur.

1A

8Ch.6 第6章Short Q 短题目Easy 易5 marksPHY13062108

(a) Describe the Young’s double slit experiment and its results. (3 marks)

(b) Explain why this experiment provides evidence that light are waves instead of particles.
(2 marks)

Chinese translation · 中文对照

(a) 描述杨氏双缝实验（Young's double slit experiment）及其结果。(3 分)

(b) 解释为什么这个实验提供了「光是波，而不是粒子」的证据。(2 分)

Approach · 思路提示

  • (a) Monochromatic light passes first through a single slit (to make the light coherent) and then through two narrow, closely spaced slits.
  • The light emerging from the two slits falls on a screen placed some distance away.
  • (a) Result: a series of bright and dark fringes of equal spacing is seen on the screen, with a bright fringe at the centre.
  • (b) A bright fringe appears where the two waves arrive in phase (path difference = nλ) and a dark fringe where they are out of phase — this is constructive and destructive interference.
  • Particles travelling in straight lines could not cancel each other out to produce dark regions, so the fringes are evidence that light behaves as a wave.
  • (a) 单色光先经过一条单缝（使光相干），再经过两条很窄、靠得很近的缝。
  • 从两条缝出来的光射到放在一定距离外的屏幕上。
  • (a) 结果：屏幕上出现一系列等距的亮纹和暗纹，中央是一条亮纹。
  • (b) 两束波同相到达（路程差 = nλ）的地方是亮纹，反相到达的地方是暗纹——这就是相长干涉和相消干涉。
  • 沿直线运动的粒子不可能互相抵消而产生暗区，所以条纹证明光具有波动性。
evidence 证据particle 粒子result 结果
Suggested answer 参考答案

Solutions

Marks

(a) In the Young’s double slit experiment,

a beam of monochromatic light passes through a double slit

1A

and falls onto a screen behind the double slit.

1A

It is observed that alternative bright and dark fringes are formed on the screen.

1A

(b) The fringe pattern formed on the screen can be explained by the diffraction and interference of waves

1A

but cannot be explained by the particle theory of light.

1A

9Ch.6 第6章Short Q 短题目Easy 易3 marksPHY13062109

In a Young’s double slit experiment, a blue light is incident on a double slit and a fringe pattern is formed on a screen.

Describe what will be observed in the following situations.

(a) The blue light is replaced by a green light. (1 mark)

(b) The amplitude of the blue light is increased. (1 mark)

(c) The double slit is replaced by one with a larger slit separation. (1 mark)

Chinese translation · 中文对照

在杨氏双缝实验中，蓝光照射到双缝上，在屏幕上形成条纹图案。

（见图）

描述在下列情况下会观察到什么。

(a) 用绿光代替蓝光。(1 分)

(b) 增大蓝光的振幅（amplitude）。(1 分)

(c) 用缝距更大的双缝代替原来的双缝。(1 分)

Approach · 思路提示

  • (a) Green light has a longer wavelength than blue light.
  • From Δy = λD / a, a longer wavelength gives a larger fringe separation: the fringes become wider apart (the pattern spreads out), and the colour of the fringes becomes green.
  • (b) A larger amplitude means a brighter wave: the fringes become brighter, but their positions and spacing are unchanged.
  • (c) A larger slit separation a makes Δy smaller: the fringes become closer together (narrower spacing).
  • (a) 绿光的波长比蓝光长。
  • 由 Δy = λD / a，波长越长条纹间距越大：条纹变得更疏（图案变宽），条纹颜色变为绿色。
  • (b) 振幅变大表示光更亮：条纹变亮，但条纹的位置和间距都不变。
  • (c) 缝距 a 变大，Δy 变小：条纹变得更密（间距变窄）。
replaced by 用……代替amplitude 振幅observe 观察到
Suggested answer 参考答案

Solutions

Marks

(a) The fringe separation increases.

1A

(b) The fringe becomes brighter.

1A

(c) The fringe separation decreases.

1A

10Ch.6 第6章Short Q 短题目Easy 易4 marksPHY13062110

A beam of red light falls onto a plane transmission grating with 500 lines per mm and forms a fringe pattern.

(a) Calculate the grating spacing. (2 marks)

(b) Suggest two ways to increase the fringe separation. (2 marks)

Chinese translation · 中文对照

一束红光照射到每毫米有 500 条刻线的平面透射光栅（plane transmission grating）上，形成条纹图案。

(a) 计算光栅间距（grating spacing）。(2 分)

(b) 提出两种增大条纹间距的方法。(2 分)

Approach · 思路提示

  • (a) Grating spacing d = 1 / (number of lines per metre).
  • 500 lines per mm → d = 1 / 500 mm = 2 × 10−3 mm = 2 × 10−6 m = 2 μm.
  • (b) From d sin θ = nλ, a longer wavelength gives a larger angle: use light of longer wavelength (e.g. change red to a longer-wavelength light).
  • (b) Use a grating with fewer lines per mm (larger grating spacing d), or increase the distance between the grating and the screen (the fringe separation on the screen increases).
  • (a) 光栅间距 d = 1 ÷ 每米刻线数。
  • 每毫米 500 条 → d = 1 ÷ 500 mm = 2 × 10−3 mm = 2 × 10−6 m = 2 μm。
  • (b) 由 d sin θ = nλ，波长越长偏折角越大：改用波长更长的光。
  • (b) 改用每毫米刻线更少（光栅间距 d 更大）的光栅；或增大光栅与屏幕的距离（屏幕上的条纹间距会变大）。
plane transmission grating 平面透射光栅lines per mm 每毫米刻线数grating spacing 光栅间距suggest 提出
Suggested answer 参考答案

Solutions

Marks

(a) Grating spacing =

1M

= 2 × 10–6 m

1A

(b) Use a grating with more lines per mm

1A

Increase the distance between the grating and the screen

1A

11Ch.6 第6章Short Q 短题目Easy 易5 marksPHY13062111

(a) What are electromagnetic waves? (1 mark)

(b) State three common properties of electromagnetic waves. (3 marks)

(c) State the relationship between visible light and electromagnetic waves. (1 mark)

Chinese translation · 中文对照

(a) 什么是电磁波（electromagnetic wave）？(1 分)

(b) 说出电磁波的三个共同性质。(3 分)

(c) 说出可见光与电磁波之间的关系。(1 分)

Approach · 思路提示

  • (a) An electromagnetic wave consists of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of travel (it is a transverse wave).
  • (b) Common properties: they are transverse waves; they travel at the speed of light in a vacuum (3 × 108 m/s); they need no medium and can travel through a vacuum.
  • (b) Other acceptable properties: they carry energy, they can be reflected/refracted/diffracted, and they can be polarized.
  • (c) Visible light is only a small part of the electromagnetic spectrum — it is one member of the family of electromagnetic waves.
  • (a) 电磁波由互相垂直、且都垂直于传播方向的振荡电场和磁场组成（它是横波）。
  • (b) 共同性质：都是横波；在真空中都以光速 3 × 108 m/s 传播；不需要介质，可以在真空中传播。
  • (b) 也可答：能携带能量；能反射、折射、衍射；可以被偏振（polarization）。
  • (c) 可见光只是电磁波谱（electromagnetic spectrum）中很小的一部分，是电磁波家族中的一员。
electromagnetic waves 电磁波properties 性质relationship 关系
Suggested answer 参考答案

Solutions

Marks

(a) Electromagnetic waves are varying electric and magnetic fields that propagates through space.

1A

(b) Any three of the following:

3 × 1A

They are transverse waves.

They travel at the same speed (3 × 108 m s–1) in a vacuum.

They transmit energy from one place to another.

They obey the relation v = fλ.

They exhibit reflection, refraction, diffraction and interference.

(Or other reasonable answers)

(c) Visible light is a kind of electromagnetic wave.

1A

12Ch.6 第6章Short Q 短题目Easy 易4 marksPHY13062112

The frequency of red light is 4.6 × 1014 Hz.

(a) Find the wavelength of red light in a vacuum.
(Given: Speed of light in a vacuum = 3 × 108 m s–1) (2 marks)

(b) A student says that the electromagnetic spectrum is a part of the visible light spectrum. Is he correct? Explain briefly. (2 marks)

Chinese translation · 中文对照

红光的频率是 4.6 × 1014 Hz。

(a) 求红光在真空中的波长。（已知：真空中的光速 = 3 × 108 m s−1）(2 分)

(b) 有学生说「电磁波谱是可见光波谱的一部分」。他是否正确？简要解释。(2 分)

Approach · 思路提示

  • (a) Use v = fλ, so λ = c / f.
  • λ = (3 × 108) / (4.6 × 1014) = 6.52 × 10−7 m ≈ 6.5 × 10−7 m (about 652 nm).
  • (b) He is not correct.
  • It is the other way round: visible light occupies only a very small range of the electromagnetic spectrum, so visible light is part of the electromagnetic spectrum, not the reverse.
  • (a) 用 v = fλ，所以 λ = c ÷ f。
  • λ = (3 × 108) ÷ (4.6 × 1014) = 6.52 × 10−7 m ≈ 6.5 × 10−7 m（约 652 nm）。
  • (b) 他不对。
  • 正好相反：可见光只占电磁波谱中很小的一段，所以可见光是电磁波谱的一部分，而不是反过来。
vacuum 真空spectrum 波谱briefly 简要地
Suggested answer 参考答案

Solutions

Marks

(a) By v = fλ,

1M

wavelength of red light λ === 6.52 × 10−7 m

1A

(b) He is incorrect.

1A

The visible light spectrum is a part of the electromagnetic spectrum.

1A

13Ch.6 第6章Short Q 短题目Easy 易4 marksPHY13062113

Radio waves, visible light and X-rays belong to a family of waves that can travel at the speed of light in a vacuum.

(a) State the name of this family of waves. (1 mark)

(b) The following are some applications of the waves in this family.

Solar cells

Measuring body temperature

Radar

Boradcasting TV signal

(Courtesy of JVCAmerica | Photobucket)

Checking luggage for security

Sterilizing drinking water

Which of the above examples is a use of:

(i) radio waves, (1 mark)

(ii) visible light, (1 mark)

(iii) X-rays? (1 mark)

Chinese translation · 中文对照

无线电波、可见光和 X 射线都属于同一族波，它们在真空中都能以光速传播。

(a) 说出这一族波的名称。(1 分)

(b) 以下是这族波的一些应用。

表格：太阳能电池（solar cells） || 测量体温（measuring body temperature）

表格：雷达（radar） || 广播电视信号（broadcasting TV signal）

表格：检查行李作保安用途 || 为饮用水消毒（sterilizing drinking water）

（图片由 JVCAmerica | Photobucket 提供）

上面哪个例子是下列各项的用途：

(i) 无线电波，(1 分)

(ii) 可见光，(1 分)

(iii) X 射线？(1 分)

Approach · 思路提示

  • (a) They belong to the family of electromagnetic waves (the electromagnetic spectrum).
  • (b)(i) Radio waves are used in radar and in broadcasting TV signals.
  • (b)(ii) Visible light is used by solar cells (light energy is converted into electrical energy).
  • (b)(iii) X-rays are used for checking luggage for security (they pass through soft materials but are absorbed by denser objects).
  • Note: measuring body temperature uses infra-red radiation and sterilizing drinking water uses ultraviolet — they are not the answers here.
  • (a) 它们属于电磁波（electromagnetic waves），也就是电磁波谱这一族。
  • (b)(i) 无线电波用于雷达（radar）和广播电视信号。
  • (b)(ii) 可见光用于太阳能电池（光能转化为电能）。
  • (b)(iii) X 射线用于检查行李作保安用途（它能穿过软材料，但被较密实的物体吸收）。
  • 注意：测量体温用的是红外线，为饮用水消毒用的是紫外线，都不是本题答案。
solar cells 太阳能电池radar 雷达sterilizing 消毒checking luggage 检查行李
Suggested answer 参考答案

Solutions

Marks

(a) Electromagnetic waves

1A

(b) (i) Broadcasting TV signal

1A

(ii) Solar cells

1A

(iii) Checking luggage for security

1A

14Ch.6 第6章Short Q 短题目Easy 易3 marksPHY13062114

Electromagnetic waves have been widely used for different medical purposes. Name the electromagnetic wave that is used in the following cases:

(a) Medical imaging of the bones in a human body (1 mark)

(b) Radiotherapy for killing cancer cells (1 mark)

(c) Examining the duodenum (十二指腸) with an endoscope (1 mark)

Chinese translation · 中文对照

电磁波被广泛用于不同的医疗用途。说出在下列情况下所使用的电磁波：

(a) 对人体骨骼作医学成像（medical imaging）(1 分)

（见图）

(b) 用放射治疗（radiotherapy）杀死癌细胞 (1 分)

（见图）

(c) 用内窥镜（endoscope）检查十二指肠（duodenum）(1 分)

（见图）

Approach · 思路提示

  • (a) X-rays: they penetrate soft tissue but are absorbed more by bones, so bones show up on the image.
  • (b) Gamma rays (X-rays also accepted): they are highly penetrating and carry high energy, so they can kill cancer cells.
  • (c) Visible light: the endoscope carries light into the body through optical fibres and the doctor views the image directly.
  • Remember: infra-red is used for measuring temperature, not for these three medical uses.
  • (a) X 射线：它能穿透软组织，但较多被骨骼吸收，所以骨骼在影像中显出来。
  • (b) 伽马射线（答 X 射线也可接受）：穿透力强、能量高，能杀死癌细胞。
  • (c) 可见光：内窥镜用光纤把光送入体内，医生直接观察图像。
  • 记住：红外线是用来测温的，这三种医疗用途都不是红外线。
medical imaging 医学成像radiotherapy 放射治疗endoscope 内窥镜duodenum 十二指肠
Suggested answer 参考答案

Solutions

Marks

(a) X-rays

1A

(b) Gamma rays

1A

(c) Visible light

1A

15Ch.6 第6章Short Q 短题目Average 中等0 marksPHY13062115

lvl=easy><part=core><type=sq><mark=4><bk=3B><ch=06><content>

Name the electromagnetic wave(s) that is/are prevented from reaching us in the following cases.

(a) Applying suntan lotion on our body (1 mark)

(b) Wearing a face shield during welding (2 marks)

(c) Wearing silvery clothes while working near a blast furnace (1 mark)

Chinese translation · 中文对照

说出在下列情况下，哪些电磁波被阻止而不能到达我们身上。

(a) 在身体上涂抹防晒油（suntan lotion）(1 分)

（见图）

(b) 焊接时戴上面罩（face shield）(2 分)

（见图）

(c) 在高炉（blast furnace）附近工作时穿上银色衣服 (1 分)

（见图）

Approach · 思路提示

  • (a) Ultraviolet radiation (UV) — suntan lotion absorbs/blocks UV so that the skin is not burnt.
  • (b) Ultraviolet radiation (and also the very intense visible light) — welding arcs give out strong UV that can damage the eyes and skin.
  • (c) Infra-red radiation — the silvery surface reflects infra-red so less heat reaches the body.
  • Remember the order of the spectrum: radio → microwave → infra-red → visible → UV → X-ray → gamma, with wavelength decreasing in that order.
  • (a) 紫外线（ultraviolet, UV）——防晒油吸收／阻挡紫外线，使皮肤不被晒伤。
  • (b) 紫外线（也答很强烈的可见光）——焊接电弧发出很强的紫外光，会伤害眼睛和皮肤。
  • (c) 红外线（infra-red）——银色的表面把红外线反射掉，到达身体的热较少。
  • 记住波谱顺序：无线电波 → 微波 → 红外线 → 可见光 → 紫外线 → X 射线 → 伽马射线，波长依次变短。
suntan lotion 防晒油face shield 面罩blast furnace 高炉prevented from reaching 被阻挡而未能到达
Suggested answer 参考答案

Solutions

Marks

(a) Ultra-violet radiation

1A

(b) Ultra-violet radiation and visible light

1A + 1A

(c) Infra-red radiation

1A

16Ch.6 第6章Short Q 短题目Easy 易4 marksPHY13062116

In ancient times, armies use beacons (烽火) to alert defence. A fire would be lit at a well-known location on a high place to signal over land that enemy troops were approaching.

(a) What kind of electromagnetic wave was involved in the signal transmission using beacons? (1 mark)

(b) Nowadays, we have used other types of electromagnetic waves to send signals at a distance. Name the electromagnetic wave that is used in sending signals for each of the following purposes:

(i) Broadcasting TV signals (1 mark)

(ii) Satellite telecommunications (1 mark)

(iii) Controlling a TV set with a remote control (1 mark)

Chinese translation · 中文对照

古时军队用烽火（beacon）示警。他们会在高处的固定地点燃起火堆，用火光把「敌军来犯」的消息传过陆地。

（见图）

(a) 用烽火传递信号时涉及哪一种电磁波？(1 分)

(b) 今天我们已用其他种类的电磁波远距离传送信号。说出下列各项用途所用的电磁波：

(i) 广播电视信号 (1 分)

(ii) 卫星通讯（satellite telecommunications）(1 分)

(iii) 用遥控器控制电视机 (1 分)

Approach · 思路提示

  • (a) The light from the fire is visible light — it is an electromagnetic wave with wavelength about 4 × 10−7 to 7 × 10−7 m.
  • (b)(i) Radio waves are used for broadcasting TV signals.
  • (b)(ii) Microwaves are used for satellite telecommunications (they can pass through the atmosphere and be focused by a dish).
  • (b)(iii) Infra-red radiation is used in remote controls.
  • (a) 火光就是可见光，它是波长约 4 × 10−7 至 7 × 10−7 m 的电磁波。
  • (b)(i) 广播电视信号用的是无线电波。
  • (b)(ii) 卫星通讯用的是微波（microwave）：它能穿过大气层，并可用天线聚焦。
  • (b)(iii) 遥控器用的是红外线。
beacon 烽火；烽火台broadcast 广播satellite telecommunications 卫星通讯
Suggested answer 参考答案

Solutions

Marks

(a) Visible light

1A

(b) (i) Radio waves

1A

(ii) Microwaves

1A

(iii) Infra-red radiation

1A

17Ch.6 第6章Short Q 短题目Average 中等6 marksPHY13062117

A monochromatic light of wavelength 550 nm is incident on a plane transmission grating with 200 lines per mm.

(a) Find the maximum order of fringe. (3 marks)

(b) A screen is placed 0.6 m from the grating. Find the distance between the two 1st-order fringes on the screen. (3 marks)

Chinese translation · 中文对照

一束波长为 550 nm 的单色光照射到每毫米有 200 条刻线的平面透射光栅上。

(a) 求能出现的最高条纹级次。(3 分)

(b) 在离光栅 0.6 m 处放置一个屏幕。求屏幕上两条第 1 级条纹之间的距离。(3 分)

Approach · 思路提示

  • (a) Grating spacing d = 1 / 200 mm = 5 × 10−3 mm = 5 × 10−6 m.
  • (a) Use d sin θ = nλ. Since sin θ ≤ 1, n ≤ d / λ = 5 × 10−6 / 550 × 10−9 = 9.09, so the maximum order is n = 9.
  • (b) For the 1st order (n = 1): sin θ = λ / d = 550 × 10−9 / 5 × 10−6 = 0.11, so θ = 6.3°.
  • (b) Distance of each 1st-order fringe from the centre: y = D tan θ = 0.6 × tan 6.3° = 0.066 m.
  • (b) The two 1st-order fringes are on opposite sides of the centre, so the required distance is 2y = 0.13 m.
  • (a) 光栅间距 d = 1 ÷ 200 mm = 5 × 10−3 mm = 5 × 10−6 m。
  • (a) 用 d sin θ = nλ。因为 sin θ ≤ 1，所以 n ≤ d ÷ λ = 5 × 10−6 ÷ (550 × 10−9) = 9.09，最大级次 n = 9。
  • (b) 对第 1 级（n = 1）：sin θ = λ ÷ d = 550 × 10−9 ÷ (5 × 10−6) = 0.11，得 θ = 6.3°。
  • (b) 每条第 1 级条纹到中央的距离 y = D tan θ = 0.6 × tan 6.3° = 0.066 m。
  • (b) 两条第 1 级条纹在中央两侧，所以所求距离是 2y = 0.13 m。
maximum order 最高级次plane transmission grating 平面透射光栅lines per mm 每毫米刻线数
Suggested answer 参考答案

Solutions

Marks

(a) n =

≤

1M

=

= 9.10

1A

\The maximum order of fringe is 9.

1A

(b) d sinθ = nλ

1M

sin θ === 0.11

θ = 6.32°

distance between the two 1st-order fringes

= 2 × 0.6 × tan 6.32°

1M

= 0.133 m

1A

18Ch.6 第6章Short Q 短题目Average 中等6 marksPHY13062118

A beam of red light is travelling towards a narrow slit. The figure below shows the wavefronts of the incident light wave.

(a) Draw the diffraction pattern in the figure. (2 marks)

(b) If blue light is used instead of red light, describe and explain the difference of the diffraction patterns. (2 marks)

(c) Why a slit of width 5 cm cannot produce the same pattern as in (a)? (2 marks)

Chinese translation · 中文对照

一束红光射向一条窄缝（narrow slit）。下图显示入射光波的波前（wavefront）。

（见图）

(a) 在图中画出衍射图案（diffraction pattern）。(2 分)

(b) 如果改用蓝光代替红光，描述并解释衍射图案的差别。(2 分)

(c) 为什么宽 5 cm 的缝不能产生与 (a) 相同的图案？(2 分)

Approach · 思路提示

  • (a) Draw the wavefronts bending/spreading out after passing through the slit, with a broad central maximum brighter than the narrower, weaker side maxima on both sides.
  • (b) Blue light has a shorter wavelength than red light.
  • (b) From the diffraction formula, the angle of the first minimum satisfies sin θ ≈ λ / b, so a shorter wavelength gives a narrower pattern — the blue pattern is narrower (the central fringe is narrower).
  • (c) Diffraction is only noticeable when the slit width is comparable with, or not much larger than, the wavelength.
  • (c) Red light has λ ≈ 7 × 10−7 m, which is far smaller than 5 cm, so the light travels almost in straight lines — no observable diffraction pattern is formed.
  • (a) 画波前通过窄缝后向外弯曲、散开，中央是一块较宽较亮的主极大，两侧是较窄较暗的次极大。
  • (b) 蓝光的波长比红光短。
  • (b) 由衍射公式，第一暗纹满足 sin θ ≈ λ ÷ b，波长越短图案越窄：蓝光的衍射图案更窄（中央亮纹更窄）。
  • (c) 只有当缝宽与波长差不多（或不比波长大得多）时，衍射才明显。
  • (c) 红光的 λ ≈ 7 × 10−7 m，远小于 5 cm，光几乎沿直线传播，所以看不到明显的衍射图案。
wavefront 波前diffraction pattern 衍射图案narrow slit 窄缝
Suggested answer 参考答案

Solutions

Marks

(a)

(Wavelength is unchanged after diffraction)

1A

(Correct diffraction pattern)

1A

(b) The blue light spreads out less than the red light.

1A

This is because the wavelength of blue light is shorter than that of red light.

1A

(c) Diffraction is prominent only if the size of the gap is comparable to the wavelength of the wave.

1A

Since the wavelength of visible light is of the order of 10−7 m, the slit of width 5 cm is much larger than the wavelength and therefore cannot produce observable diffraction.

1A

19Ch.6 第6章Short Q 短题目Average 中等6 marksPHY13062119

A red laser beam is directed towards a single slit on a screen. A pattern is produced on the screen as shown below.

(a) Name this wave phenomenon. (1 mark)

(b) The wavelength of red light is 7 × 10−7 m. Find the frequency of the light.
(Given: speed of light in air = 3 × 108 m s−1) (2 marks)

(c) The width of the slit is 0.1 mm. Why does the slit have to be so narrow? (2 marks)

(d) How would the pattern be affected if a slit of width 1 cm is used? (1 mark)

Chinese translation · 中文对照

一束红色激光照射到屏幕上的单缝（single slit）上，在屏幕上产生如下图所示的图案。

（见图）

(a) 说出这种波的現象的名称。(1 分)

(b) 红光的波长是 7 × 10−7 m。求该光的频率。（已知：空气中的光速 = 3 × 108 m s−1）(2 分)

(c) 缝宽为 0.1 mm。为什么缝必须这么窄？(2 分)

(d) 如果改用宽 1 cm 的缝，图案会有什么变化？(1 分)

Approach · 思路提示

  • (a) The phenomenon is diffraction of light.
  • (b) Use v = fλ, so f = c / λ = 3 × 108 / 7 × 10−7 = 4.3 × 1014 Hz.
  • (c) Noticeable diffraction needs the slit width to be comparable with the wavelength of the light.
  • (c) 0.1 mm is still much larger than λ = 7 × 10−7 m, but it is narrow enough for the pattern to spread out and be wide enough to observe; a much wider slit would make the diffraction effect too small to see.
  • (d) With a 1 cm slit the diffraction pattern becomes much narrower, and the light behaves almost as if it travels in straight lines (a bright band of about the same width as the slit with sharp edges).
  • (a) 这个現象是光的衍射（diffraction）。
  • (b) 用 v = fλ，所以 f = c ÷ λ = 3 × 108 ÷ (7 × 10−7) = 4.3 × 1014 Hz。
  • (c) 要看到明显的衍射，缝宽必须与光的波长差不多。
  • (c) 0.1 mm 虽然仍远大于 λ = 7 × 10−7 m，但已足够窄，图案才能散开、宽到可以观察；缝再宽衍射效应就太小、看不见了。
  • (d) 缝宽 1 cm 时衍射图案变得很窄，光几乎沿直线传播（只看到一条宽度与缝差不多、边缘清晰的亮带）。
phenomenon 現象single slit 单缝narrow 窄的frequency 频率
Suggested answer 参考答案

Solutions

Marks

(a) Diffraction

1A

(b) By v = fλ,

1M

frequency === 4.29 × 1014 Hz

1A

(c) This is because the wavelength of light is very small (~10−7 m)

1A

and diffraction of light is observable only when the slit width is comparable to wavelength of the light.

1A

(d) No diffraction pattern will be observed. There will be a bright spot on the screen.

1A

20Ch.6 第6章Short Q 短题目Average 中等3 marksPHY13062120

(a) What is meant by interference of wave? (1 mark)

(b) State two necessary conditions for the production of an observable interference pattern.
(2 marks)

Chinese translation · 中文对照

(a) 什么是波的干涉（interference）？(1 分)

(b) 说出产生可观察的干涉图案的两个必要条件。(2 分)

Approach · 思路提示

  • (a) Interference is the superposition of two or more waves: where they arrive in phase they reinforce each other (constructive interference), and where they arrive out of phase they cancel (destructive interference).
  • (b) The two waves must have the same frequency (same wavelength) and a constant phase difference — the sources must be coherent.
  • (b) Their amplitudes should be nearly equal (otherwise the contrast is poor), and they must be polarized in the same direction / travel in overlapping regions.
  • The result is a stable pattern of alternating maxima and minima.
  • (a) 干涉是两个或多个波叠加：同相到达时互相加强（相长干涉），反相到达时互相抵消（相消干涉）。
  • (b) 两列波频率（波长）相同，且相位差保持不变——即波源必须是相干（coherent）的。
  • (b) 两列波的振幅最好接近（否则明暗对比很差），并且偏振方向相同、在重叠的区域相遇。
  • 结果是形成稳定的、明暗交替的图案。
interference 干涉observable 可观察的phase difference 相位差
Suggested answer 参考答案

Solutions

Marks

(a) Interference is the superposition of two or more waves when they meet.

1A

(b) Any two of the following:

2 × 1A

The waves must be produced by coherent sources.

The amplitudes of the waves must be the same or close to each other.

The distance(s) between sources should be comparable to the wavelength.

21Ch.6 第6章Short Q 短题目Average 中等5 marksPHY13062121

A beam of yellow light passes through a single slit and a double slit mounted on the end of a cardboard tube as shown in the Figure a.

Fig a

Fig b

An image is recorded on a piece of photographic film placed at the end of the tube. When the film is developed, an interference pattern can be seen as shown in Figure b.

(a) In an experiment, the separation between X and Y is found to be 7 mm. The slit separation of the double slit is 0.2 mm. Find the wavelength of the yellow light. (2 marks)

(b) State one method to reduce the error in the value of the wavelength found in (a). (1 mark)

(c) Describe how will the pattern on the film be different if one side of the double slit is covered. (2 marks)

Chinese translation · 中文对照

一束黄光通过装在硬纸筒（cardboard tube）一端的单缝和双缝，如图 a 所示。

表格：图 a || 图 b

在筒的另一端放一张摄影底片（photographic film），图像就记录在底片上。底片冲洗（developed）后，可以看到如图 b 所示的干涉图案。

(a) 实验中测得 X 与 Y 之间的距离为 7 mm，双缝的缝距为 0.2 mm。求黄光的波长。(2 分)

(b) 说出一种可减少 (a) 中所求波长误差的方法。(1 分)

(c) 如果把双缝的一边遮住，底片上的图案会怎样不同？描述一下。(2 分)

Approach · 思路提示

  • (a) Use Δy = λD / a, so λ = Δy · a / D, where Δy is the separation between X and Y on the film and D is the distance from the double slits to the film (see figure for D).
  • Convert to SI units: a = 0.2 mm = 2 × 10−4 m, Δy = 7 mm = 7 × 10−3 m, and use the D read from the figure.
  • (b) Measure the distance across several fringes and divide by the number of intervals (multi-fringe method) to average out the reading error; also measure the fringe positions with a travelling microscope and D more carefully.
  • (c) Covering one side of the double slit leaves only one slit, so the double-slit interference fringes disappear.
  • (c) A single-slit diffraction pattern is seen instead: a broad central bright band with narrower, dimmer bands on both sides.
  • (a) 用 Δy = λD / a，即 λ = Δy · a ÷ D；Δy 是底片上 X 与 Y 之间的距离（条纹间距），D 是双缝到底片的距离（D 见图）。
  • 换算成国际单位：a = 0.2 mm = 2 × 10−4 m，Δy = 7 mm = 7 × 10−3 m，D 从图中读出。
  • (b) 测量相隔多个条纹的总距离再除以间隔数（多条纹法），以减小读数误差；也可用读数显微镜测量条纹位置，并把 D 量得更准。
  • (c) 遮住双缝的一边后只剩一条缝，双缝干涉条纹消失。
  • (c) 改为看到单缝衍射图案：中间一条较宽的亮带，两侧是较窄、较暗的条纹（见图）。
cardboard tube 硬纸筒photographic film 摄影底片developed 冲洗（底片）separation 间距
Suggested answer 参考答案

Solutions

Marks

(a) By Δy =,

1M

wavelength === 5.83 × 10−7 m

1A

(b) Measure the distance across several fringes and calculate the average fringe spacing Δy.

1A

(Or other reasonable answers)

(c) A diffraction pattern will be formed

1A

instead of an interference pattern.

1A

22Ch.6 第6章Short Q 短题目Average 中等5 marksPHY13062122

(a) State three precautions for generating observable interference fringes in a Young’s double slit experiment. (3 marks)

(b) A double slit with slit separation 0.8 mm is 2.5 m away from a screen as shown below.

When a coherent light of wavelength 589 nm is directed to the double slit, fringes are observed on the screen. Find the fringe separation. (2 marks)

Chinese translation · 中文对照

(a) 说出在杨氏双缝实验中产生可观察干涉条纹的三项注意事项。(3 分)

(b) 缝距为 0.8 mm 的双缝与屏幕相距 2.5 m，如下图所示。

（见图）

当波长为 589 nm 的相干光（coherent light）射向双缝时，屏幕上可观察到条纹。求条纹间距。(2 分)

Approach · 思路提示

  • (a) Use monochromatic light (a single wavelength), otherwise the fringes of different colours overlap and blur.
  • (a) Place a single slit in front of the source so the light reaching the two slits is coherent (same phase).
  • (a) Use narrow slits with small slit separation, place the screen far away, and work in a dark room to make the fringes clear and bright.
  • (b) Use Δy = λD / a with λ = 589 × 10−9 m, a = 0.8 × 10−3 m, D = 2.5 m.
  • (b) Δy = (589 × 10−9 × 2.5) / (0.8 × 10−3) = 1.84 × 10−3 m = 1.84 mm.
  • (a) 要用单色光（单一波长），否则不同颜色的条纹重叠、图案模糊。
  • (a) 在光源前放一条单缝，使到达双缝的光相干（相位相同）。
  • (a) 缝要窄、缝距要小，屏幕要放得远，并在暗室中做实验，条纹才清晰明亮。
  • (b) 用 Δy = λD / a，其中 λ = 589 × 10−9 m，a = 0.8 × 10−3 m，D = 2.5 m。
  • (b) Δy = (589 × 10−9 × 2.5) ÷ (0.8 × 10−3) = 1.84 × 10−3 m = 1.84 mm。
precaution 注意事项coherent light 相干光fringe separation 条纹间距
Suggested answer 参考答案

Solutions

Marks

(a) Monochromatic light should be used.

1A

The slits should be narrow.

1A

The slit separation should be small.

1A

(Or other reasonable answers)

(b) By Δy =,

1M

fringe separation == 1.84 mm

1A

23Ch.6 第6章Short Q 短题目Average 中等6 marksPHY13062123

The following figure (not drawn to scale) shows the apparatus used to measure the wavelength of a light.

(a) In order to form measurable fringes on the screen, the slit separation a must be small and the distance D between the slits and the screen must be large. Explain briefly. (3 marks)

(b) The double slit used in the experiment has a slit separation of 0.25 mm. The distance between the double slit and the translucent screen is 5 m. The separation between the central bright fringe and the 8th bright fringe on the screen is 88 mm. Find the wavelength of the light used. (3 marks)

Chinese translation · 中文对照

下图（不按比例）显示用来测量某光波长的实验装置。

（见图）

(a) 为了在屏幕上形成可测量的条纹，缝距 a 必须很小，而缝与屏幕的距离 D 必须很大。简要解释。(3 分)

(b) 实验所用双缝的缝距是 0.25 mm，双缝与半透明屏幕（translucent screen）的距离是 5 m。屏幕上中央亮纹与第 8 条亮纹相距 88 mm。求所用光的波长。(3 分)

Approach · 思路提示

  • (a) The fringe separation is given by Δy = λD / a.
  • (a) A small a makes Δy larger, and a large D also makes Δy larger; a large Δy is easier to measure accurately (a small Δy would be too small to read on a ruler).
  • (a) So a must be small and D must be large so that the fringes are far apart and measurable.
  • (b) The separation between the central bright fringe and the 8th bright fringe is 8 fringe spacings: Δy = 88 mm / 8 = 11 mm = 1.1 × 10−2 m.
  • (b) λ = Δy · a / D = (1.1 × 10−2 × 0.25 × 10−3) / 5 = 5.5 × 10−7 m = 550 nm.
  • (a) 条纹间距由 Δy = λD / a 决定。
  • (a) a 小会使 Δy 变大，D 大也会使 Δy 变大；Δy 大就容易量得准（Δy 太小时用米尺根本读不出来）。
  • (a) 所以缝距要小、缝到屏幕的距离要大，条纹才会分得开、量得到。
  • (b) 中央亮纹到第 8 条亮纹之间相隔 8 个条纹间距：Δy = 88 mm ÷ 8 = 11 mm = 1.1 × 10−2 m。
  • (b) λ = Δy · a ÷ D = (1.1 × 10−2 × 0.25 × 10−3) ÷ 5 = 5.5 × 10−7 m = 550 nm。
apparatus 仪器装置measurable 可测量的translucent screen 半透明屏幕
Suggested answer 参考答案

Solutions

Marks

(a) From a sinθ = nλ, it can be deduced that sin θ = for n = 1.

1M

Besides, the fringe separation is given by Δy = D tan θ.

1M

Since the wavelength λ is very small,

1A

a must be small so that θ will not be too small, and D must be large so that Δy will not be too small.

(b) Fringe separation == 11 mm

1A

By Δy =,

1M

Wavelength λ === 5.5 × 10−7 m

1A

24Ch.6 第6章Short Q 短题目Average 中等6 marksPHY13062124

In a Young’s double slit experiment, the slit separation is 1.5 mm. The distance between the double slit and the screen is 6.0 m. The monochromatic light used has a wavelength of 589 nm.

(a) Find the separation between the central bright fringe and the first order bright fringe.
(2 marks)

(b) Find the separation between the central bright fringe and the third order dark fringe.
(2 marks)

(c) What would happen to the interference pattern if a slit with larger slit separation is used?
(2 marks)

Chinese translation · 中文对照

在杨氏双缝实验中，缝距是 1.5 mm，双缝与屏幕的距离是 6.0 m，所用单色光的波长是 589 nm。

(a) 求中央亮纹与第 1 级亮纹之间的距离。(2 分)

(b) 求中央亮纹与第 3 级暗纹（third order dark fringe）之间的距离。(2 分)

(c) 如果改用缝距更大的缝，干涉图案会有什么变化？(2 分)

Approach · 思路提示

  • (a) Use Δy = λD / a = (589 × 10−9 × 6.0) / (1.5 × 10−3) = 2.36 × 10−3 m ≈ 2.4 mm.
  • (b) The nth dark fringe occurs at path difference (n − 1/2)λ, so the 3rd dark fringe is at path difference 2.5λ.
  • (b) Its distance from the centre = 2.5 Δy = 2.5 × 2.36 mm = 5.9 mm.
  • (c) A larger slit separation gives a smaller Δy, so the fringes become closer together (narrower spacing), while the central fringe stays bright and at the same place.
  • (a) 用 Δy = λD / a = (589 × 10−9 × 6.0) ÷ (1.5 × 10−3) = 2.36 × 10−3 m ≈ 2.4 mm。
  • (b) 第 n 条暗纹对应的路程差是 (n − 1/2)λ，所以第 3 级暗纹的路程差是 2.5λ。
  • (b) 它离中央的距离 = 2.5Δy = 2.5 × 2.36 mm = 5.9 mm。
  • (c) 缝距变大，Δy 变小，条纹变得更密（间距变窄），但中央亮纹位置不变、仍然明亮。
central bright fringe 中央亮纹third order dark fringe 第 3 级暗纹separation 间距
Suggested answer 参考答案

Solutions

Marks

(a) By Δy =,

1M

the separation between the fringes == 2.36 mm

1A

(b) The separation between the fringes = 2.5Δy

1M

=

= 5.89 mm

1A

(c) By Δy =, if a increases, Δy decreases.

1A

Therefore, the fringes become closer together.

1A

25Ch.6 第6章Short Q 短题目Average 中等4 marksPHY13062125

In a Young’s double slit experiment, the slit separation is 0.07 cm and the distance between the double slit and the screen is 200 cm. When a blue light is used, the distance of the 1st-order bright fringe from the centre of the interference pattern is 0.13 cm.

(a) Find the wavelength of the blue light used in the experiment. (2 marks)

(b) Find the distance of the 4th-order dark fringe from the centre of the interference pattern.
(2 marks)

Chinese translation · 中文对照

在杨氏双缝实验中，缝距是 0.07 cm，双缝与屏幕的距离是 200 cm。当使用蓝光时，第 1 级亮纹离干涉图案中心的距离是 0.13 cm。

(a) 求实验所用蓝光的波长。(2 分)

(b) 求第 4 级暗纹离干涉图案中心的距离。(2 分)

Approach · 思路提示

  • (a) The 1st-order bright fringe is at a distance equal to one fringe spacing: Δy = 0.13 cm.
  • (a) Use λ = Δy · a / D with a = 0.07 cm and D = 200 cm (the same unit cm can be used throughout for Δy, a and D).
  • (a) λ = (0.13 × 0.07) / 200 cm = 4.55 × 10−5 cm = 4.55 × 10−7 m = 455 nm.
  • (b) The 4th dark fringe is at path difference (4 − 1/2)λ = 3.5λ, so its distance from the centre is 3.5 Δy.
  • (b) Distance = 3.5 × 0.13 cm = 0.455 cm ≈ 0.46 cm.
  • (a) 第 1 级亮纹离中央的距离就是一个条纹间距：Δy = 0.13 cm。
  • (a) 用 λ = Δy · a ÷ D，其中 a = 0.07 cm、D = 200 cm（Δy、a、D 都用 cm 就可以，单位会约掉）。
  • (a) λ = (0.13 × 0.07) ÷ 200 cm = 4.55 × 10−5 cm = 4.55 × 10−7 m = 455 nm。
  • (b) 第 4 级暗纹对应的路程差是 (4 − 1/2)λ = 3.5λ，所以离中央的距离是 3.5Δy。
  • (b) 距离 = 3.5 × 0.13 cm = 0.455 cm ≈ 0.46 cm。
centre of the interference pattern 干涉图案的中心slit separation 缝距order 级次
Suggested answer 参考答案

Solutions

Marks

(a) By Δy =,

1M

wavelength === 4.55 × 10−7 m

1A

(b) The distance between the 4th-order dark fringe and the central fringe

= 3.5 Δy

1M

= 3.5 × 0.13 × 10−2

= 4.55 × 10−3 m

1A

26Ch.6 第6章Short Q 短题目Average 中等5 marksPHY13062126

In a Young’s double slit experiment, a laser of wavelength 650 nm is used. The slit separation of the double slit is 0.05 mm. 11 bright fringes are observed on the screen as shown.

(a) Find θ. (3 marks)

(b) How does θ change in the following cases?

(i) Increasing the distance between the screen and the double slit. (1 mark)

(ii) Replacing the double slit with another one of wider slit separation. (1 mark)

Chinese translation · 中文对照

在杨氏双缝实验中，使用波长为 650 nm 的激光。双缝的缝距是 0.05 mm。屏幕上观察到 11 条亮纹，如图所示。

（见图）

(a) 求 θ。(3 分)

(b) 在下列情况下 θ 会怎样变化？

(i) 增大屏幕与双缝之间的距离。(1 分)

(ii) 用缝距更宽的双缝代替原来的双缝。(1 分)

Approach · 思路提示

  • (a) First see from the figure which fringes the angle θ is measured between (see figure). There are 11 bright fringes, i.e. 5 on each side of the central one.
  • (a) If θ is measured from the central fringe to the outermost (5th) bright fringe, use a sin θ = nλ with n = 5: sin θ = (5 × 650 × 10−9) / (0.05 × 10−3) = 0.065, so θ = 3.7°.
  • (a) If the θ marked in the figure is the total angle between the two outermost fringes, then θ = 2 × 3.7° = 7.5°.
  • (b)(i) The angle of each order depends only on λ and a, so θ is unchanged — but the fringes on the screen are farther apart because the screen is farther away.
  • (b)(ii) A wider slit separation a makes sin θ = nλ / a smaller, so θ becomes smaller (the fringes are closer together).
  • (a) 先看图确认 θ 是哪两条亮纹之间的角（见图）。共有 11 条亮纹，即中央两侧各有 5 条。
  • (a) 如果 θ 是由中央亮纹量到最外侧（第 5 条）亮纹的角，用 a sin θ = nλ，n = 5：sin θ = (5 × 650 × 10−9) ÷ (0.05 × 10−3) = 0.065，得 θ = 3.7°。
  • (a) 如果图中标出的 θ 是左右两条最外侧亮纹之间的总角，则 θ = 2 × 3.7° = 7.5°。
  • (b)(i) 每一级条纹的角度只由 λ 和 a 决定，所以 θ 不变——但屏幕离得更远，屏幕上的条纹间距会变大。
  • (b)(ii) 缝距 a 变宽，sin θ = nλ / a 变小，所以 θ 变小（条纹靠得更近）。
fringe 条纹wider 更宽的slit separation 缝距
Suggested answer 参考答案

Solutions

Marks

(a) Let D be the distance between the double slit and the screen.

Fringe separation === 0.013D

1M

tan =

1M

θ = 7.44°

1A

(b) (i) Remains unchanged

1A

(ii) Decreases

1A

27Ch.6 第6章Short Q 短题目Average 中等5 marksPHY13062127

The figure below shows the interference pattern of red light formed in a Young’s double slit experiment. Points A and B are two points on the screen 4.0 cm apart.

The distance between the double slit and the screen is 1.5 m and the slit separation is 0.2 mm.

(a) Estimate the wavelength of the red light. (3 marks)

(b) Find the path difference at point A. (2 marks)

Chinese translation · 中文对照

下图显示杨氏双缝实验中形成的红光的干涉图案。A 和 B 是屏幕上相距 4.0 cm 的两点。

（见图）

双缝与屏幕的距离是 1.5 m，缝距是 0.2 mm。

(a) 估计红光的波长。(3 分)

(b) 求 A 点的路径差（path difference）。(2 分)

Approach · 思路提示

  • (a) From the figure, count how many fringe spacings lie between A and B, say n intervals (see figure).
  • (a) Then Δy = 4.0 cm / n, and λ = Δy · a / D with a = 0.2 mm = 2 × 10−4 m and D = 1.5 m.
  • (a) Take care that A and B are compared at the same kind of fringes (e.g. centre to centre of bright fringes) when counting.
  • (b) The path difference at a point a distance y from the centre is a·y / D.
  • (b) Equivalently: if A is the nth bright fringe the path difference is nλ; if A is the nth dark fringe it is (n − 1/2)λ. Use the fringe number read from the figure.
  • (a) 先从图中数出 A、B 之间相隔多少个条纹间距，设为 n 个（见图）。
  • (a) 则 Δy = 4.0 cm ÷ n，再用 λ = Δy · a ÷ D，其中 a = 0.2 mm = 2 × 10−4 m，D = 1.5 m。
  • (a) 数条纹时要注意 A、B 要取同一类条纹（例如都以亮纹中心为准）。
  • (b) 离中央距离为 y 的地方，路程差 = a·y ÷ D。
  • (b) 也可以这样算：若 A 是第 n 条亮纹，路程差 = nλ；若 A 是第 n 条暗纹，路程差 = (n − 1/2)λ。条纹编号从图中读出。
path difference 路径差estimate 估计pattern 图案
Suggested answer 参考答案

Solutions

Marks

(a) By Δy =,

1M

wavelength =

=

1M

= 6.67 × 10−7 m

1A

(b) A is at the 4th order bright fringe.

\ Path difference at A = 4 × λ

1M

= 4 (6.67 × 10−7)

= 2.67 × 10−6 m

1A

28Ch.6 第6章Short Q 短题目Average 中等6 marksPHY13062128

Red light is directed towards a fine grating and an interference pattern is formed on a screen.

(a) Explain how the interference pattern will change if the red light is replaced by blue light.
(3 marks)

(b) Explain how the interference pattern will change if the fine grating is replaced by a coarse one. (3 marks)

Chinese translation · 中文对照

红光照射到一块细光栅（fine grating）上，在屏幕上形成干涉图案。

(a) 解释如果改用蓝光代替红光，干涉图案会怎样改变。(3 分)

(b) 解释如果把细光栅换成粗光栅（coarse grating），干涉图案会怎样改变。(3 分)

Approach · 思路提示

  • Use the grating equation d sin θ = nλ (or Δy = λD / a for small angles).
  • (a) Blue light has a shorter wavelength, so sin θ = nλ / d is smaller: the fringes move closer to the centre and the angular separation between orders decreases.
  • (a) Because d / λ is larger, more orders can be seen (the maximum order n = d / λ increases).
  • (b) A coarse grating has fewer lines per unit length, i.e. a larger grating spacing d.
  • (b) A larger d makes sin θ = nλ / d smaller, so the fringes become closer together (smaller fringe separation).
  • (b) A larger d / λ also allows a higher maximum order, so more orders appear.
  • 用光栅公式 d sin θ = nλ（小角度下也可用 Δy = λD / a）。
  • (a) 蓝光波长更短，sin θ = nλ / d 更小：条纹更靠向中央，各级之间的角间距变小。
  • (a) 因为 d/λ 变大，能看到的级次更多（最高级次 n = d/λ 变大）。
  • (b) 粗光栅每单位长度的刻线更少，也就是光栅间距 d 更大。
  • (b) d 变大，sin θ = nλ / d 变小，所以条纹靠得更近（条纹间距变小）。
  • (b) d/λ 变大也允许更高的最高级次，所以出现的级次更多。
fine grating 细光栅coarse grating 粗光栅interference pattern 干涉图案
Suggested answer 参考答案

Solutions

Marks

(a) Since the wavelength of blue light is shorter than that of red light,

1A

the fringe separation will decrease

1A

and the fringes will increase in number.

1A

(b) Since the grating spacing of a coarse grating is larger than that of a fine grating,

1A

the fringe separation will decrease

1A

and the fringes will increase in number.

1A

29Ch.6 第6章Short Q 短题目Average 中等6 marksPHY13062129

Parallel monochromatic light is incident normally on a plane transmission grating of 3000 lines per cm. A metre rule is positioned 2 m from the grating and parallel to its plane. Bright lines are observed at the 9 cm, 45 cm and 81 cm marks of the ruler.

(a) What is the grating spacing? (1 mark)

(b) What is the order of each bright line? (2 marks)

(c) Find the wavelength of the light. (3 marks)

Chinese translation · 中文对照

平行单色光垂直照射到每厘米有 3000 条刻线的平面透射光栅上。一把米尺（metre rule）放在离光栅 2 m、并与光栅平面平行的位置。在米尺的 9 cm、45 cm 和 81 cm 刻度处观察到亮线。

（见图）

(a) 光栅间距是多少？(1 分)

(b) 每条亮线是第几级？(2 分)

(c) 求该光的波长。(3 分)

Approach · 思路提示

  • (a) d = 1 / 3000 cm = 3.33 × 10−4 cm = 3.33 × 10−6 m.
  • (b) The 45 cm mark is the centre (the central maximum, zero order), because the 9 cm and 81 cm lines are symmetric about it: 45 − 9 = 36 cm and 81 − 45 = 36 cm.
  • (b) So both outer lines are the 1st order: tan θ = 36 cm / 2 m = 0.18, giving θ = 10.2°; if they were 2nd order the wavelength would be about 3 × 10−7 m, which is not in the visible range.
  • (c) Use d sin θ = nλ with n = 1: sin 10.2° = 0.177.
  • (c) λ = d sin θ / n = (3.33 × 10−6 × 0.177) / 1 = 5.9 × 10−7 m.
  • (a) d = 1 ÷ 3000 cm = 3.33 × 10−4 cm = 3.33 × 10−6 m。
  • (b) 45 cm 处是中央（零级主极大），因为 9 cm 和 81 cm 的亮线关于它对称：45 − 9 = 36 cm，81 − 45 = 36 cm。
  • (b) 所以两侧的亮线都是第 1 级：tan θ = 36 cm ÷ 2 m = 0.18，得 θ = 10.2°；若是第 2 级，波长会约 3 × 10−7 m，不在可见光范围。
  • (c) 用 d sin θ = nλ，n = 1：sin 10.2° = 0.177。
  • (c) λ = d sin θ ÷ n = (3.33 × 10−6 × 0.177) ÷ 1 = 5.9 × 10−7 m。
metre rule 米尺grating spacing 光栅间距order 级次parallel 平行的
Suggested answer 参考答案

Solutions

Marks

(a) Grating spacing == 3.33 ×10−6 m

1A

(b) The bright line at the 45 cm mark is the 0th-order bright fringe.

1A

The bright lines at the 9 cm and 81 cm marks are the 1st-order bright fringes.

1A

(c) For the 1st-order fringe,

sin θ == 0.1772

1M

By d sin θ = nλ,

1M

wavelength === 590 nm

1A

30Ch.6 第6章Short Q 短题目Average 中等6 marksPHY13062130

A plane transmission grating with a grating spacing of 2.2 × 10–6 m is used to examine the light from a glowing gas. It is found that the first order indigo light emerges at an angle of 11.8° and the first order orange light at an angle of 15.8° as shown below.

(a) Find the wavelengths of lights with these two colours. (3 marks)

(b) Will orange and indigo fringes be observed at 54.8°? Explain your answer. (3 marks)

Chinese translation · 中文对照

用光栅间距为 2.2 × 10−6 m 的平面透射光栅观察发光气体发出的光。测得第 1 级靛色（indigo）光以 11.8° 的角度射出，第 1 级橙色光以 15.8° 的角度射出，如下图所示。

（见图）

(a) 求这两种颜色光的波长。(3 分)

(b) 在 54.8° 处会观察到橙色和靛色条纹吗？解释你的答案。(3 分)

Approach · 思路提示

  • (a) Use d sin θ = nλ with n = 1: λ = d sin θ.
  • (a) Indigo: λ = 2.2 × 10−6 × sin 11.8° = 2.2 × 10−6 × 0.2045 = 4.5 × 10−7 m.
  • (a) Orange: λ = 2.2 × 10−6 × sin 15.8° = 2.2 × 10−6 × 0.2723 = 6.0 × 10−7 m.
  • (b) d sin 54.8° = 2.2 × 10−6 × 0.8171 = 1.80 × 10−6 m.
  • (b) 1.80 × 10−6 / (6.0 × 10−7) = 3 for orange, and 1.80 × 10−6 / (4.5 × 10−7) ≈ 4 for indigo.
  • (b) Both are whole numbers, so yes: the 3rd-order orange and the 4th-order indigo appear at 54.8° at the same time (they overlap).
  • (a) 用 d sin θ = nλ，n = 1：λ = d sin θ。
  • (a) 靛色：λ = 2.2 × 10−6 × sin 11.8° = 2.2 × 10−6 × 0.2045 = 4.5 × 10−7 m。
  • (a) 橙色：λ = 2.2 × 10−6 × sin 15.8° = 2.2 × 10−6 × 0.2723 = 6.0 × 10−7 m。
  • (b) d sin 54.8° = 2.2 × 10−6 × 0.8171 = 1.80 × 10−6 m。
  • (b) 1.80 × 10−6 ÷ (6.0 × 10−7) = 3（橙色），1.80 × 10−6 ÷ (4.5 × 10−7) ≈ 4（靛色）。
  • (b) 两者都是整数，所以会：第 3 级橙纹和第 4 级靛纹同时出现在 54.8°（两纹重叠）。
indigo 靛蓝色emerge 射出；出现order 级次
Suggested answer 参考答案

Solutions

Marks

(a) By d sin θ = nλ,

1M

wavelength of indigo light

=== 4.499 m ≈ 4.50 ×10−7 m

1A

wavelength of orange light

=== 5.99 × 10−7 m

1A

(b) At θ = 54.8°,

for indigo light, n === 4.00

1M

for orange light, n === 3.00

1M

Since n is an integer in both cases, orange and indigo fringes can be observed.

1A

31Ch.6 第6章Short Q 短题目Average 中等4 marksPHY13062131

(a) State three properties common to all electromagnetic waves. (3 marks)

(b) State one property that varies among electromagnetic waves. (1 mark)

Chinese translation · 中文对照

(a) 说出所有电磁波共有的三个性质。(3 分)

(b) 说出一种在不同电磁波之间有差异的性质。(1 分)

Approach · 思路提示

  • (a) All electromagnetic waves are transverse waves.
  • (a) They all travel at the same speed in a vacuum (3 × 108 m/s) and need no medium.
  • (a) They all consist of oscillating electric and magnetic fields, carry energy, and show reflection, refraction, diffraction and interference.
  • (b) Wavelength, frequency and photon energy differ from one part of the spectrum to another (e.g. wavelength decreases from radio waves to gamma rays).
  • (a) 所有电磁波都是横波。
  • (a) 它们在真空中的速度都相同（3 × 108 m/s），都不需要介质。
  • (a) 它们都由振荡的电场和磁场组成，都能携带能量，都能反射、折射、衍射和干涉。
  • (b) 波长、频率和光子能量在不同波段的电磁波之间是不同的（例如从无线电波到伽马射线波长依次变短）。
properties 性质vary 有差异；变化common 共同的
Suggested answer 参考答案

Solutions

Marks

(a) Any three of the following:

3 × 1A

They all show reflection.

They all show refraction.

They all show diffraction.

They all show interference.

They travel through a vacuum at the same speed (3 × 108 m).

They obey the equation v = fλ.

(Or other reasonable answers)

(b) They have different wavelengths / frequencies.

1A

(Or other reasonable answers)

32Ch.6 第6章Short Q 短题目Average 中等6 marksPHY13062132

The electromagnetic spectrum is shown in the figure.

radio waves

X

infra-red radiations

visible light

Y

X-rays

Z

(a) Name the parts X, Y and Z. (3 marks)

(b) (i) Which part of the spectrum has the shortest wavelength? (1 mark)

(ii) Which part of the spectrum has the lowest frequency? (1 mark)

(c) State one application of Z. (1 mark)

Chinese translation · 中文对照

下图显示电磁波谱（electromagnetic spectrum）。

表格：无线电波（radio waves） || X || 红外辐射（infra-red radiations） || 可见光（visible light） || Y || X 射线（X-rays） || Z

(a) 写出 X、Y、Z 各部分的名称。(3 分)

(b) (i) 波谱中哪一部分的波长最短？(1 分)

(ii) 波谱中哪一部分的频率最低？(1 分)

(c) 说出 Z 的一种应用。(1 分)

Approach · 思路提示

  • (a) The order of the spectrum by decreasing wavelength is: radio waves → microwaves → infra-red → visible light → ultraviolet → X-rays → gamma rays.
  • (a) So X = microwaves, Y = ultraviolet (UV), Z = gamma rays.
  • (b)(i) The shortest wavelength belongs to Z, the gamma rays (at the far right of the spectrum).
  • (b)(ii) The lowest frequency belongs to radio waves (longest wavelength, far left).
  • (c) Gamma rays: killing cancer cells in radiotherapy / sterilizing medical equipment and food (X-rays are also accepted for some of these).
  • (a) 按波长由长到短的顺序是：无线电波 → 微波 → 红外线 → 可见光 → 紫外线 → X 射线 → 伽马射线。
  • (a) 所以 X = 微波，Y = 紫外线，Z = 伽马射线。
  • (b)(i) 波长最短的是 Z，即伽马射线（在波谱最右端）。
  • (b)(ii) 频率最低的是无线电波（波长最长，在最左端）。
  • (c) 伽马射线的应用：放射治疗杀死癌细胞／为医疗器材和食品消毒（部分也可答 X 射线）。
infra-red radiations 红外辐射shortest wavelength 最短波长application 应用
Suggested answer 参考答案

Solutions

Marks

(a) X : microwaves

1A

Y : ultra-violet radiation

1A

Z : gamma rays

1A

(b) (i) Gamma rays

1A

(ii) Radio waves

1A

(c) Any one of the following:

1A

Radiotherapy

Sterilization

Medical imaging

(Or other reasonable answer)

33Ch.6 第6章Short Q 短题目Average 中等4 marksPHY13062133

Radio signals are broadcast in a form of wave, called radio waves. As shown below, a house is situated on one side of a hill, while a radio station is located on the other side of the hill. The station transmits radio waves with wavelengths ranging from a few metres to a few kilometres.

(a) Explain, with the help of a diagram, how the antenna of the house receives the signals transmitted by the radio station. (2 marks)

(b) Of which wavelength, long or short, would the antenna have a better reception? Explain briefly. (2 marks)

Chinese translation · 中文对照

无线电信号以波的形式广播，这种波称为无线电波。如下图所示，一座房子位于山的一侧，而广播电台位于山的另一侧。电台发出的无线电波波长由几米到几公里不等。

（见图）

(a) 借助示意图，解释房子的天线（antenna）如何接收到电台发出的信号。(2 分)

(b) 对哪一种波长（长波还是短波），天线会有较好的接收效果？简要解释。(2 分)

Approach · 思路提示

  • (a) The radio waves do not only travel in straight lines: they diffract at the top of the hill and spread out behind it, so part of the wave reaches the house on the other side.
  • (a) Diagram: draw the waves bending round the hill and spreading into the region behind it (the shadow region).
  • (b) Long waves give better reception.
  • (b) Diffraction is more significant when the wavelength is comparable with, or larger than, the size of the obstacle; long radio waves bend round the hill much more easily, so a stronger signal reaches the antenna.
  • (a) 无线电波不只沿直线传播：它会在山顶处衍射（diffraction），并绕到山后散开，所以有一部分波能到达山的另一侧的房子。
  • (a) 画图：画出波绕过山、向山后的阴影区散开。
  • (b) 长波长的接收效果较好。
  • (b) 当波长与障碍物尺寸差不多或更大时，衍射更明显；长波的无线电波更容易绕过山，所以到达天线的信号更强。
antenna 天线diffraction 衍射reception 接收comparable 可相比的
Suggested answer 参考答案

Solutions

Marks

(a)

(Diffraction shown)

1A

The waves diffract round the hill and reach the antenna.

1A

(b) The antenna would have a better reception of radio waves of long wavelengths.

1A

This is because radio waves of long wavelengths diffract more than those with short wavelengths.

1A

34Ch.6 第6章Short Q 短题目Average 中等5 marksPHY13062134

(a) The TV signals received by the house shown below is disturbed when an aeroplane flies over the house.

(i) Which part of the electromagnetic spectrum do TV signals belong to? (1 mark)

(ii) Explain briefly how the TV signals received at the house are affected by the aeroplane. (2 marks)

(b) Explain why mobile phones should be turned off in an aeroplane. (2 marks)

Chinese translation · 中文对照

(a) 下图所示的房子所接收的电视信号，在有飞机飞过房子上方时会受到干扰。

（见图）

(i) 电视信号属于电磁波谱的哪一部分？(1 分)

(ii) 简要解释飞机如何影响房子接收到的电视信号。(2 分)

(b) 解释为什么在飞机上要把手机关掉。(2 分)

Approach · 思路提示

  • (a)(i) TV signals are radio waves (UHF radio waves), part of the radio-wave region of the electromagnetic spectrum.
  • (a)(ii) The aeroplane reflects the TV signals; the reflected wave reaches the house together with the direct signal.
  • (a)(ii) The two waves have different path lengths, so they interfere — sometimes constructively (stronger) and sometimes destructively (weaker), which disturbs the picture and sound.
  • (b) A mobile phone communicates with base stations by emitting microwaves (radio waves).
  • (b) These signals can be picked up by the aeroplane's electronic and navigation systems and interfere with them, which may be dangerous — so phones must be switched off.
  • (a)(i) 电视信号属于无线电波（UHF 无线电波），是电磁波谱中无线电波这一段。
  • (a)(ii) 飞机会反射电视信号，反射波和直接射来的信号一起到达房子。
  • (a)(ii) 两束波走的路程不同，会发生干涉：有时相长（信号变强），有时相消（信号变弱），于是画面和声音受干扰。
  • (b) 手机靠发射微波（无线电波）与基站通讯。
  • (b) 这些信号可能被飞机的电子和导航系统接收到并造成干扰，可能很危险，所以必须关掉手机。
disturbed 受干扰aeroplane 飞机switch off 关掉
Suggested answer 参考答案

Solutions

Marks

(a) (i) Radio waves

1A

(ii) The signal reflected from the aeroplane

1A

interferes with the signal emitted directly from the transmitter.

1A

(b) Mobile phones emit radio waves

1A

which may interfere with signals in electronic devices of the plane.

1A

35Ch.6 第6章Short Q 短题目Average 中等5 marksPHY13062135

A student uses an auto-focus camera to take a photo of an object. The camera emits a electromagnetic pulse to the object and detects a pulse reflected from it.

(a) Which part of the electromagnetic spectrum does the pulse belong to? (1 marks)

(b) If the time difference between emitting and receiving the pulse is 15 ns, what is the distance between the object and the camera? (Given: 1 ns = 10–9 s) (2 marks)

(c) The auto-focus function makes focusing easy, but it does not work well when the object is dull black in colour. Explain why. (2 marks)

Chinese translation · 中文对照

某学生用自动对焦（auto-focus）相机给一个物体拍照。相机会向物体发出一个电磁脉冲（pulse），并探测由物体反射回来的脉冲。

(a) 这个脉冲属于电磁波谱的哪一部分？(1 分)

(b) 如果发出和接收到脉冲的时间差是 15 ns，物体与相机的距离是多少？（已知：1 ns = 10−9 s）(2 分)

(c) 自动对焦功能使对焦变得容易，但当物体是暗黑色时效果不好。解释原因。(2 分)

Approach · 思路提示

  • (a) The pulse is infra-red radiation.
  • (b) The pulse travels to the object and back, so the total distance 2s = c t.
  • (b) s = c t / 2 = (3 × 108 × 15 × 10−9) / 2 = 4.5 / 2 = 2.25 m.
  • (c) A dull black surface absorbs most of the infra-red radiation and reflects very little.
  • (c) The reflected pulse is then too weak to be detected, so the camera cannot measure the distance and the auto-focus fails.
  • (a) 这个脉冲是红外线（infra-red）。
  • (b) 脉冲要走到物体再返回，所以总路程 2s = c t。
  • (b) s = c t ÷ 2 = (3 × 108 × 15 × 10−9) ÷ 2 = 4.5 ÷ 2 = 2.25 m。
  • (c) 暗黑色的表面会吸收大部分红外线，只反射很少。
  • (c) 反射回来的脉冲太弱，探测不到，相机就无法测出距离，自动对焦失效。
auto-focus 自动对焦pulse 脉冲dull black 暗黑色的detect 探测
Suggested answer 参考答案

Solutions

Marks

(a) Infra-red radiation.

1A

(b) Distance = speed × time

1M

= 3 × 108 ×× 15 × 10−9

= 2.25 m

1A

(c) The dull black object is a good absorber of radiation. It can absorb the infra-red pulses

1A

and so the camera cannot detect a reflected pulse.

1A

36Ch.6 第6章Short Q 短题目Average 中等4 marksPHY13062136

A man stands 4 m from a plane mirror and takes a photo of himself in the mirror.

(a) What is the distance between him and his image? (2 marks)

(b) The auto-focus function of the camera uses infra-red radiation to measure its distance from an object. Can the man obtain a sharp image of a candle fire using the auto-focus function? (2 marks)

Chinese translation · 中文对照

一个人站在离平面镜（plane mirror）4 m 的地方，在镜中给自己拍照。

（见图）

(a) 他与自己的像之间的距离是多少？(2 分)

(b) 相机的自动对焦功能利用红外线来测量它与物体之间的距离。他能否用自动对焦功能拍到清晰的烛火（candle fire）图像？(2 分)

Approach · 思路提示

  • (a) For a plane mirror the image is as far behind the mirror as the object is in front of it.
  • (a) So the image is 4 m behind the mirror, and the distance between the man and his image is 4 m + 4 m = 8 m.
  • (b) Yes, he can.
  • (b) A candle flame is a strong emitter of infra-red radiation, so even though it reflects little, enough infra-red from the flame reaches the camera for it to focus. (Contrast with a dull black object, which absorbs infra-red and reflects almost none, so focusing fails.)
  • (a) 平面镜成的像，在镜后的距离等于物在镜前的距离。
  • (a) 所以像在镜后 4 m，人与像的距离是 4 m + 4 m = 8 m。
  • (b) 可以。
  • (b) 烛火本身是红外线的强发射源，虽然反射很少，但从火焰发出的红外线足以让相机对焦。（相比暗黑色物体，它吸收红外线、几乎不反射，所以对焦会失败。）
plane mirror 平面镜image 像sharp 清晰的candle fire 烛火
Suggested answer 参考答案

Solutions

Marks

(a) Distance between the man and his image

= 4 m + 4 m

1M

= 8 m

1A

(b) No

1A

The camera measures the distance of an object by emitting an infra-red pulse and detecting the reflected pulse. The infra-red radiation from the open flame can confuse the infra-red sensor of the camera.

1A

37Ch.6 第6章Short Q 短题目Average 中等6 marksPHY13062137

(a) You are given the following equipment: a double slit , a single slit, a red laser pointer, a screen and two ray-box lamps. Select suitable equipments and describe a method to demonstrate the interference of light. State the expected results. (5 marks)

(b) State one measure to ensure a clear observable result in the demonstration. (1 mark)

Chinese translation · 中文对照

(a) 给你以下器材：一个双缝、一个单缝、一个红色激光笔、一个屏幕和两个光线盒（ray-box lamp）。选取合适的器材，并描述一个演示光的干涉（interference）的方法，说出预期结果。(5 分)

(b) 说出一个可确保演示结果清晰可观察的措施。(1 分)

Approach · 思路提示

  • (a) Choose the double slit, the red laser pointer and the screen (the single slit is not needed, because laser light is already coherent and monochromatic).
  • (a) Method: place the double slit in front of the laser pointer so the beam passes through both slits, and put the screen a few metres away behind the double slit.
  • (a) Expected result: a series of equally spaced bright and dark fringes appears on the screen, with a bright fringe at the centre; the fringes are red because the light is red.
  • (b) Do the demonstration in a darkened room (and keep the screen far from the slits) so that the fringes are bright and clear against a dark background.
  • (a) 选用双缝、红色激光笔和屏幕（不需要单缝，因为激光本身已经是相干、单色的）。
  • (a) 方法：把双缝放在激光笔前面，让光束同时穿过两条缝；在双缝后方几米处放屏幕。
  • (a) 预期结果：屏幕上出现一系列等距的亮纹和暗纹，中央是一条亮纹；因为用的是红光，条纹是红色的。
  • (b) 在暗室中做演示（并把屏幕放得远一些），条纹在黑暗背景上才明亮清晰、容易观察。
demonstrate 演示ray-box lamp 光线盒expected results 预期结果observable 可观察的
Suggested answer 参考答案

Solutions

Marks

(a) Use a laser pointer, a double slit and a screen.

1A

Direct the laser pointer towards the double slit

1A

and let the light fall on the screen after passing through the double slit.

1A

A pattern consisting of bright and dark fringes

1A

which are evenly spaced can be observed on the screen.

1A

(b) Any one of the following:

1A

Black out the laboratory and shield the screen from stray lights.

The slit widths should be comparable to the wavelength of the light.

The slit separation should be very small (~ 0.5 mm).

The distance between the double slit and the screen should be large
(~ 1 m, or above).

1Ch.6 第6章Reading 阅读题Average 中等8 marksPHY13064101

Anti-reflection coatings

When you look at the reflections of lights by your glasses, you may find that the images appear to be purple in colour. This is a result of an anti-reflection coating on the surface of the lens.

Fig a

Refer to Figure b below. When a monochromatic light falls onto the anti-reflection coating, it will be partly reflected. When the transmitted light hits the coating–lens interface, the same process will happen again. If destructive interference occurs for the reflected rays, the reflected light will not be noticeable, thus resulting in an anti-reflection effect. However, for the same coating, destructive interference will only occur for light of one wavelength. Most coatings are designed such that destructive interference occurs for green light. For any incident light, the closer is its wavelength to green light, the weaker the reflected light will be.

Fig b

Other coatings such as anti-UV coating, anti-fog coating and scratch-resistance coating may also be applied on the lens. They affect the behavior of the lens in other different ways.

(a) Explain why purple images can be seen in Figure a. (2 marks)

(b) Suppose a green light hits the coating normally. It is given that the wavelength of the green light in vacuum is 550 nm and the reflective index of the coating is 1.2.

(i) Find the wavelength of the green light in the coating. (2 marks)

(ii) What is the minimum thickness d of the coating for destructive interference to occur? (2 marks)

(iii) A red light has wavelength λ′ = 583 nm in the coating. If it hits the coating normally, find the path difference of the two reflected rays in terms of λ′. (2 marks)

Chinese translation · 中文对照

抗反射涂层（anti-reflection coating）

当你看眼镜镜片上灯光的反射时，可能会发现反射出来的像呈紫色。这是镜片表面的抗反射涂层造成的。

（见图 a）

请参考下面的图 b。当单色（monochromatic）光射到抗反射涂层上时，会有一部分被反射。透射光到达涂层–镜片界面时，又会发生同样的过程。如果两束反射光线发生相消干涉（destructive interference），反射光就不会很明显，从而产生抗反射的效果。不过，对同一层涂层来说，相消干涉只会对某一个波长的光发生。大多数涂层都设计成对绿光发生相消干涉。对任何入射光来说，它的波长越接近绿光，反射光就越弱。

（见图 b）

镜片上还可能加上其他涂层，例如防紫外线涂层、防雾涂层和抗刮涂层。它们会以其他不同的方式影响镜片的特性。

(a) 解释为什么在图 a 中可以看到紫色的像。（2 分）

(b) 假设有一束绿光垂直射到涂层上。已知绿光在真空中的波长为 550 nm，涂层的折射率（refractive index）为 1.2。

(i) 求绿光在涂层中的波长。（2 分）

(ii) 要使相消干涉发生，涂层的最小厚度 d 是多少？（2 分）

(iii) 有一束红光，它在涂层中的波长为 λ′ = 583 nm。如果它垂直射到涂层上，求两束反射光线的光程差（path difference），并用 λ′ 表示答案。（2 分）

Approach · 思路提示

  • Part (a): the coating is designed so that green light (about 550 nm) is cancelled by destructive interference, so green is hardly reflected.
  • The farther a wavelength is from green, the more strongly it is reflected. Red and blue/violet are reflected most strongly, and red + violet light mixed together looks purple - that is the purple image.
  • Part (b)(i): in the coating the wavelength is shorter: lambda_coating = lambda_vacuum / n = 550 / 1.2 = 458 nm (about 4.6 x 10-7 m).
  • Both reflected rays are reflected at a boundary with a higher refractive index (air to coating, then coating to lens), so each gains a phase change of pi; the two pi shifts cancel and the relative phase change is zero.
  • Part (b)(ii): so destructive interference needs 2d = (m + 1/2)lambda_coating. The minimum thickness (m = 0) is d = lambda_coating / 4 = 458 / 4 = 115 nm (about 1.2 x 10-7 m).
  • Part (b)(iii): the path difference stays 2d = 2 x 115 = 229 nm, so 2d / lambda' = 229 / 583 = 0.39, i.e. the path difference is 0.39 lambda'. It is not 0.5 lambda', so red light is not cancelled and is still reflected.
  • (a)：涂层被设计成让绿光（约 550 nm）发生相消干涉（destructive interference），所以绿光几乎不会被反射出去。
  • 波长离绿光越远，反射越强。红光和蓝紫光反射得最多，红光和紫光混合起来看上去就是紫色，所以看到紫色的像。
  • (b)(i)：光进入介质后波长变短，用 λ_涂层 = λ_真空 / n = 550 / 1.2 = 458 nm（约 4.6 × 10-7 m）。
  • 两束反射光都是在「光疏到光密」的界面上反射（空气→涂层，涂层→镜片），各多出 π 的相位变化；两个 π 互相抵消，所以相对相位差为 0。
  • (b)(ii)：因此相消干涉的条件是 2d = (m + 1/2)λ_涂层。取 m = 0 得到最小厚度 d = λ_涂层 / 4 = 458 / 4 = 115 nm（约 1.2 × 10-7 m）。
  • (b)(iii)：光程差（path difference）仍然是 2d = 2 × 115 = 229 nm，所以 2d / λ′ = 229 / 583 = 0.39，即光程差是 0.39λ′。它不是 0.5λ′，所以红光不会相消，仍会被反射（这也解释了(a)中的紫色）。
anti-reflection coating 抗反射涂层（防反光镀膜）monochromatic 单色的destructive interference 相消干涉refractive index 折射率（原文写作 reflective index）
Suggested answer 参考答案

Solutions

Marks

(a) When white light fall on the anti-reflection coating, the green component of the reflected light diminishes

1A

while the other components form purple images.

1A

(b) (i) Let λ1 be the wavelength of the green light in the coating.

n ==

1M

λ1 === 458 nm

1A

(ii) 2d =

1M

d == 114.5 ≈ 115 nm

1A

(iii) Path difference = 2d

1M

= 2 × 114.5

= 229 nm

= 0.393λ1

1A

1Ch.7 第7章MC 选择题Easy 易2 marksPHY13071101

A musical note is played in air. Which of the following will remain unchanged when it enters water?

ASpeed of the sound wave
BLoudness of the note
CPitch of the note
DWavelength of the sound wave
Chinese · 中文对照

在空气中演奏一个音符（musical note）。当这个声音传入水中时，下列哪一项保持不变？

musical note 音符remain unchanged 保持不变refraction 折射amplitude 振幅
Correct answer 正确答案C

During refraction, the frequency of the sound wave will remain unchanged while its speed and wavelength will change.

∴ A and D are incorrect.

The pitch of the note remains unchanged as the frequency of the sound wave remains unchanged.

∴ C is correct.

The amplitude of the sound wave will decrease when travelling in air and water, as well as during refraction at the boundary. The loudness of the note decreases as the amplitude of the sound wave decreases.

∴ B is incorrect.

中文解析

在折射（refraction）过程中，声波的频率（frequency）保持不变，而它的速度和波长都会改变。

∴ A 和 D 不正确。

因为声波的频率不变，所以这个音的音调（pitch）保持不变。

∴ C 正确。

声波在空气中和水中传播时，以及在边界处发生折射时，它的振幅（amplitude）都会减小。声波的振幅减小，这个音的响度（loudness）就减小。

∴ B 不正确。

2Ch.7 第7章MC 选择题Easy 易2 marksPHY13071102

The sonar system of a ship emits an ultrasound pulse. It receives the echo reflected by the seabed after 0.6 s. The speed of ultrasound in sea water is 1500 m s−1. What is the distance between the ship and the seabed?

A300 m
B400 m
C450 m
D500 m
Chinese · 中文对照

某艘船的声呐（sonar）系统发出一个超声脉冲（ultrasound pulse），0.6 s 后接收到由海底反射回来的回声（echo）。超声波在海水中的速度是 1500 m s−1。船与海底之间的距离是多少？

sonar 声呐ultrasound 超声波echo 回声seabed 海底
Correct answer 正确答案C

Distance between the ship and the seabed

= vt = 1500 × 0.62 = 450 m

中文解析

船与海底之间的距离

= vt = 1500 × (0.6)/(2) = 450 m

（提示：回声走的是「去 + 回来」两段路，所以时间要除以 2。）

3Ch.7 第7章MC 选择题Average 中等2 marksPHY13071103

Figure a shows the equilibrium positions of particles a to k in a medium. The particles are separated by 2 cm from each other. A longitudinal wave is travelling from left to right. At a certain instant, the positions of the particles are shown in Figure b. Which of the following is/are correct?

(1)The amplitude of the wave is 2.4 cm.
(2)The wavelength of the wave is 8 cm.
(3)Particle f is moving to the right.
A(1) only
B(2) only
C(1) and (2) only
D(1) and (3) only
Chinese · 中文对照

（见图 a、图 b）图 a 显示介质中粒子 a 到 k 的平衡位置（equilibrium position），相邻粒子之间相距 2 cm。一个纵波（longitudinal wave）从左向右传播。在某一时刻，粒子的位置如图 b 所示。下列哪项／哪些正确？

（1）这个波的振幅是 2.4 cm。

（2）这个波的波长是 8 cm。

（3）粒子 f 正在向右移动。

equilibrium position 平衡位置longitudinal wave 纵波amplitude 振幅wavelength 波长
Correct answer 正确答案D

Particles b and f are at their equilibrium positions. The separation between them is equal to 0.5λ.

0.5λ = 8

λ = 16 cm

∴ (2) is incorrect.

Particle d is midway between particles b and f. The magnitude of its displacement is equal to the amplitude at the instant shown.

Amplitude = 2.4 cm

∴ (1) is correct.

Take rightwards as positive. When the wave moves to the right, the displacement of particle f changes from 0 to positive value. Therefore, particle f is moving to the right.

∴ (3) is correct.

中文解析

粒子 b 和 f 都位于平衡位置，它们之间的距离等于 0.5λ。

0.5λ = 8

λ = 16 cm

∴（2）不正确。

粒子 d 位于粒子 b 和 f 的中点，在图示这一刻，它的位移大小就等于振幅。

振幅 = 2.4 cm

∴（1）正确。

取向右为正方向。波向右传播时，粒子 f 的位移由 0 变成正值，所以粒子 f 正在向右移动。

∴（3）正确。

4Ch.7 第7章MC 选择题Easy 易2 marksPHY13071104

Sound wave travels the fastest in

Aa vacuum.
Bgas.
Cliquid.
Dsolid.
Chinese · 中文对照

声波在下列哪种物质中传播得最快？

vacuum 真空liquid 液体solid 固体
Correct answer 正确答案D

Sound wave cannot travel through a vacuum. In general, sound wave travels faster in solids than in liquids, and in turn, faster in liquids than in gases.

中文解析

声波不能在真空中传播。一般来说，声波在固体中比在液体中传播得快，在液体中又比在气体中快。所以固体中最快。

5Ch.7 第7章MC 选择题Average 中等2 marksPHY13071105

Which of the following statements about a longitudinal wave are correct?

(1)A longitudinal wave is one in which the oscillation of the transmitting medium is parallel to the direction of travel of the wave.
(2)When a particle is at the centre of compression or the centre of rarefaction at an instant, it has the largest magnitude of displacement from its equilibrium position at that instant.
(3)When a particle is at the centre of compression or the centre of rarefaction at an instant, it has the highest instantaneous speed.
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

下列关于纵波（longitudinal wave）的说法，哪些正确？

（1）纵波是指传播介质（transmitting medium）的振动方向与波的传播方向平行的波。

（2）当某粒子在某一时刻位于密部中心（centre of compression）或疏部中心（centre of rarefaction）时，它在那一刻离平衡位置的位移最大。

（3）当某粒子在某一时刻位于密部中心或疏部中心时，它的瞬时速率最大。

longitudinal wave 纵波oscillation 振动parallel to 与⋯⋯平行compression 密部（压缩区）
Correct answer 正确答案B

(1) is the definition of longitudinal wave.

When a particle is at the centre of compression or the centre of rarefaction at an instant, it is at its equilibrium position and has the highest instantaneous speed.

∴ (2) is incorrect. (3) is correct.

中文解析

（1）正是纵波的定义。

当粒子在某一时刻位于密部中心或疏部中心时，它正处在平衡位置上，瞬时速率最大。

∴（2）不正确，（3）正确。

6Ch.7 第7章MC 选择题Easy 易2 marksPHY13071106

The following figures show two CRO displays of notes produced by musical instruments. The setting of the CRO remains unchanged.

Which of the following statements about the two notes is/are correct?

(1)They are produced by different musical instruments.
(2)They have the same pitch.
(3)They have the same loudness.
A(1) only
B(3) only
C(1) and (2) only
D(2) and (3) only
Chinese · 中文对照

（见图）下图显示两台乐器所发出声音在示波器（CRO）上的波形。示波器的设置保持不变。关于这两个音，下列哪些说法正确？

（1）它们由不同的乐器产生。

（2）它们的音调（pitch）相同。

（3）它们的响度（loudness）相同。

CRO 示波器（阴极射线示波器）waveform 波形pitch 音调loudness 响度
Correct answer 正确答案C

The two waveforms are different.

∴ (1) is correct.

The two sound waves have the same frequency, so the two notes have the same pitch.

∴ (2) is correct.

The two sound waves have different amplitudes, so the two notes have different loudness.

∴ (3) is incorrect.

中文解析

两个波形（waveform）的形状不同。

∴（1）正确。

两个声波的频率相同，所以两个音的音调相同。

∴（2）正确。

两个声波的振幅不同，所以两个音的响度不同。

∴（3）不正确。

7Ch.7 第7章MC 选择题Easy 易2 marksPHY13071107

Which of the following phenomena is observed in sound?

(1)Reflection
(2)Refraction
(3)Diffraction
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

下列哪些现象可以在声音中观察到？

（1）反射（reflection）

（2）折射（refraction）

（3）衍射（diffraction）

reflection 反射refraction 折射diffraction 衍射interference 干涉
Correct answer 正确答案D

Like other waves, sound waves exhibit reflection, refraction, diffraction and interference.

中文解析

和其他波一样，声波也会表现出反射、折射、衍射和干涉（interference）。

8Ch.7 第7章MC 选择题Easy 易2 marksPHY13071108

Arrange the following media according to the speed of sound in them.

Aair < water < iron
Bair < iron < water
Cwater < iron < air
Diron < water < air
Chinese · 中文对照

按声音在下列介质中传播的速度，由小到大排列。

arrange 排列media 介质（medium 的复数）according to 按照
Correct answer 正确答案A

In general, sound wave travels faster in solids than in liquids, and in turn, faster in liquids than in gases.

中文解析

一般来说，声波在固体中比在液体中传播得快，在液体中又比在气体中快。铁是固体、水是液体、空气是气体，所以速度由小到大是：空气 < 水 < 铁。

9Ch.7 第7章MC 选择题Easy 易2 marksPHY13071109

Which of the following statements about noise is correct?

ANoise is caused by regular vibrations.
BExposure to noise causes stress.
CHearing damage by noise can always be healed.
DTraffic noise can be completely avoided by reflective barriers.
Chinese · 中文对照

下列关于噪音（noise）的说法，哪一项正确？

noise 噪音exposure to 暴露于hearing damage 听力损伤reflective barrier 反射屏障
Correct answer 正确答案B
中文解析

噪音是由不规则的振动产生的，所以 A 错。

长期处在噪音环境中，人会出现烦躁、失眠、精神紧张等压力反应，所以 B 正确。

噪音造成的听力损伤（例如听觉细胞受损）通常是永久性的，不能治愈，所以 C 错。

反射屏障只能减弱噪音，不能完全避免，而且低频噪音还能绕过屏障，所以 D 错。

10Ch.7 第7章MC 选择题Easy 易2 marksPHY13071110

Which of the following noises are least effectively shielded by reflective barriers?

ANoises with high amplitudes
BNoises with low frequencies
CNoises produced by multiple objects
DNoises produced by large objects
Chinese · 中文对照

下列哪些噪音最难被反射屏障（reflective barrier）屏蔽？

shield 屏蔽、遮挡reflective barrier 反射屏障diffraction 衍射
Correct answer 正确答案B

Noises with low frequencies have long wavelengths. They have a large degree of diffraction around reflective barriers.

中文解析

低频率的噪音波长大。波长越长，绕过反射屏障时衍射（diffraction）的程度就越大，所以最难被屏住。

11Ch.7 第7章MC 选择题Easy 易2 marksPHY13071111

Which of the following are major sources of noises in Hong Kong?

(1)Factories
(2)Road traffic
(3)Construction sites
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

下列哪些是香港主要的噪音来源？

（1）工厂

（2）道路交通

（3）建筑工地

major 主要的source 来源construction site 建筑工地
Correct answer 正确答案C

Traffic and construction work are two major sources of noise in Hong Kong.

中文解析

交通和建筑工程是香港两个主要的噪音来源。

12Ch.7 第7章MC 选择题Easy 易2 marksPHY13071112

A longitudinal wave passes through a medium. The following figure shows the equilibrium positions of some particles in the medium and their positions at a certain instant.

Which of the following pair of particles vibrate in antiphase?

AP and V
BS and Y
CR and T
DV and Y
Chinese · 中文对照

一个纵波通过某种介质。（见图）下图显示介质中一些粒子的平衡位置，以及它们在某一时刻的位置。下列哪一对粒子是反相（antiphase）振动的？

antiphase 反相（相差半个周期）vibrate 振动separation 距离、间隔
Correct answer 正确答案D

λ = separation between P and V = 6 cm

Two particles whose equilibrium positions are separated by (n + 0.5)λ vibrate in antiphase.

P and V: 6 cm = 1λ

S and Y: 6 cm = 1λ

R and T: 2 cm = 13λ

V and Y: 3 cm = 0.5λ

中文解析

λ = P 与 V 之间的距离 = 6 cm

平衡位置相隔 (n + 0.5)λ 的两个粒子，是反相振动的。

P 和 V：6 cm = 1λ

S 和 Y：6 cm = 1λ

R 和 T：2 cm = (1)/(3)λ

V 和 Y：3 cm = 0.5λ

只有 V 和 Y 相隔半个波长，所以它们反相振动。

13Ch.7 第7章MC 选择题Easy 易2 marksPHY13071113

The following figure shows a longitudinal wave travelling in a medium.

At t = 0.1 s, which particle(s) move(s) at the highest speed?

AR
BQ and S
CQ, R and S
DP, R and T
Chinese · 中文对照

（见图）下图显示一个在介质中传播的纵波。在 t = 0.1 s 时，哪个／哪些粒子运动得最快？

at the highest speed 以最大速率（运动）particle 粒子
Correct answer 正确答案D

Particles at their equilibrium positions (P, R and T) are moving at the highest speed.

中文解析

位于平衡位置上的粒子（P、R 和 T）运动得最快。

14Ch.7 第7章MC 选择题Easy 易2 marksPHY13071114

The figure below shows a longitudinal wave travelling through some particles.

What is the amplitude of the wave?

A2 cm
B3 cm
C4 cm
D5 cm
Chinese · 中文对照

（见图）下图显示一个纵波通过一些粒子传播。这个波的振幅（amplitude）是多少？

amplitude 振幅displacement 位移equilibrium position 平衡位置
Correct answer 正确答案C

Particles C and G are at their equilibrium positions and particle E is midway between them.

Amplitude = magnitude of displacement of particle E = 4 cm

中文解析

粒子 C 和 G 位于平衡位置，粒子 E 位于它们的中点。

振幅 = 粒子 E 的位移大小 = 4 cm

15Ch.7 第7章MC 选择题Easy 易2 marksPHY13071115

Which of the following statements about sound are correct?

(1)Sound is a kind of longitudinal wave.
(2)The speed of sound in air is higher than that in water.
(3)Sound is produced by vibrations.
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

下列关于声音的说法，哪些正确？

（1）声音是一种纵波。

（2）声音在空气中的速度比在水中高。

（3）声音是由振动产生的。

longitudinal wave 纵波vibration 振动speed of sound 声速
Correct answer 正确答案B

In general, sound wave travels faster in solids than in liquids, and in turn, faster in liquids than in gases.

∴ (2) is incorrect.

中文解析

一般来说，声波在固体中比在液体中传播得快，在液体中又比在气体中快。

∴（2）不正确。

（1）声音是纵波，正确；（3）声音由振动产生，正确。

16Ch.7 第7章MC 选择题Easy 易2 marksPHY13071116

Which of the following statements about sound waves is correct?

AThey are transverse waves.
BThey are electromagnetic waves.
CThey travel at a speed of 3 × 108 m s–1 in air.
DThey cannot travel through a vacuum.
Chinese · 中文对照

下列关于声波的说法，哪一项正确？

transverse wave 横波electromagnetic wave 电磁波mechanical wave 机械波vacuum 真空
Correct answer 正确答案D

Sound waves are longitudinal waves but not transverse waves.

∴ A is incorrect.

Sound waves are mechanical waves but not electromagnetic waves. Mechanical waves cannot travel through a vacuum.

∴ B is incorrect. D is correct.

Sound waves travel at 340 m s−1 in air.

∴ C is incorrect.

中文解析

声波是纵波，不是横波。

∴ A 不正确。

声波是机械波（mechanical wave），不是电磁波。机械波不能在真空中传播。

∴ B 不正确，D 正确。

声波在空气中的速度是 340 m s−1。

∴ C 不正确。

17Ch.7 第7章MC 选择题Easy 易2 marksPHY13071117

Which of the following statements about sound is incorrect?

ASound waves are longitudinal waves.
BParticles in a medium vibrate as sound waves travel through them.
CSound waves have higher speeds in gases than in solids.
DSound waves may change the direction of travel when transmitting from one medium to another.
Chinese · 中文对照

下列关于声音的说法，哪一项不正确？

transmit through 通过⋯⋯传播medium 介质direction of travel 传播方向
Correct answer 正确答案C

In general, sound wave travels faster in solids than in liquids, and in turn, faster in liquids than in gases.

∴ C is incorrect.

中文解析

一般来说，声波在固体中比在液体中传播得快，在液体中又比在气体中快。

∴ C 不正确。

18Ch.7 第7章MC 选择题Easy 易2 marksPHY13071118

Which of the following statements about sound waves is/are correct?

(1)Sound travels at about 340 m s–1 in a vacuum.
(2)Sound travels at about 340 m s−1 in air.
(3)Air particles travels at about 340 m s−1 in a sound wave.
A(1) only
B(2) only
C(1) and (2) only
D(2) and (3) only
Chinese · 中文对照

下列关于声波的说法，哪项／哪些正确？

（1）声音在真空中以约 340 m s−1 传播。

（2）声音在空气中以约 340 m s−1 传播。

（3）在声波中，空气粒子以约 340 m s−1 移动。

vacuum 真空amplitude 振幅period 周期varies with time 随时间变化
Correct answer 正确答案B

Sound cannot travel in a vacuum.

∴ (1) is incorrect.

When a sound wave passes through some air particles, each air particle vibrates about its equilibrium position along the direction of travel of the sound wave. The speed of each air particle varies with time and depends on the amplitude and period of vibration, and is not the same as the speed of the sound wave.

∴ (3) is incorrect.

中文解析

声音不能在真空中传播。

∴（1）不正确。

当声波通过一些空气粒子时，每个空气粒子都沿声波的传播方向、在各自的平衡位置附近振动。每个空气粒子的速度随时间变化，取决于振动的振幅和周期，与声波的速度并不相同。

∴（3）不正确。

19Ch.7 第7章MC 选择题Easy 易2 marksPHY13071119

A wave can travel through a vacuum if

Ait is transverse.
Bit has a very high speed.
Cit has a very large amplitude.
DNone of above
Chinese · 中文对照

一个波能满足下列哪一项，就可以在真空中传播？

transverse 横的amplitude 振幅None of above 以上都不对
Correct answer 正确答案D

A transverse mechanical wave with a very high speed and a very large amplitude cannot travel through a vacuum.

中文解析

即使是一个速度非常高、振幅非常大的横的机械波，也不能在真空中传播。能在真空中传播的只有电磁波（例如光和微波）。

∴ 答案是 D。

20Ch.7 第7章MC 选择题Easy 易2 marksPHY13071120

A student shouts towards a wall and hears the echo after 1.5 s. If the speed of sound is
340 m s–1, what is the distance between the student and the wall?

A255 m
B510 m
C1020 m
D2040 m
Chinese · 中文对照

一个学生朝墙壁喊叫，1.5 s 后听到回声。若声速是 340 m s−1，这个学生与墙壁之间的距离是多少？

echo 回声distance 距离speed of sound 声速
Correct answer 正确答案A

Distance between the student and the wall

= vt = 340 × 1.52 = 255 m

中文解析

学生与墙壁之间的距离

= vt = 340 × (1.5)/(2) = 255 m

（回声要走「去 + 回来」两段，所以时间要除以 2。）

21Ch.7 第7章MC 选择题Easy 易2 marksPHY13071121

A man facing a wall claps his hands and hears the echo after 2.5 s.

Given that the speed of sound is 340 m s−1, what is the distance between the man and the wall?

A68 m
B136 m
C425 m
D850 m
Chinese · 中文对照

一个男人面对着墙壁拍手，2.5 s 后听到回声。（见图）已知声速是 340 m s−1，他和墙壁之间的距离是多少？

facing 面对、朝向clap his hands 拍手echo 回声
Correct answer 正确答案C

Distance between the student and the wall

= vt = 340 × 2.52 = 425 m

中文解析

学生与墙壁之间的距离

= vt = 340 × (2.5)/(2) = 425 m

22Ch.7 第7章MC 选择题Easy 易2 marksPHY13071122

Leo is singing inside a house. Joey is standing outside the house and cannot see Leo, as shown in the following figure, but she can still hear his singing.

Joey can hear Leo because of the

Areflection of sound waves.
Brefraction of sound waves.
Cinterference of sound waves.
Ddiffraction of sound waves.
Chinese · 中文对照

Leo 在屋内唱歌，Joey 站在屋外，看不见 Leo（见图），但仍然听得见他的歌声。Joey 能听见 Leo 的歌声，是因为声波的

diffraction 衍射noticeable 显著的、明显的comparable to 与⋯⋯差不多wavelength 波长
Correct answer 正确答案D

Joey can still hear Leo singing because sound undergoes diffraction after passing through the door. The diffraction is noticeable as the typical wavelength of a human voice is comparable to the width of a doorway.

中文解析

Joey 仍听得见 Leo 唱歌，是因为声音经过门口后发生了衍射。人声的典型波长与门口宽度差不多，所以衍射很显著。

23Ch.7 第7章MC 选择题Easy 易2 marksPHY13071123

A signal of frequency 1.4 kHz is fed into two loudspeakers. How far apart should the two loudspeakers be placed to make their separation equal to 2 wavelengths? The speed of sound in air is 340 m s–1.

A0.243 m
B0.486 m
C0.729 m
D0.971 m
Chinese · 中文对照

把频率为 1.4 kHz 的信号输入两个扬声器。要使两个扬声器之间的距离等于 2 个波长，它们应该相隔多远？空气中的声速是 340 m s−1。

loudspeaker 扬声器separation 间距signal 信号
Correct answer 正确答案B

Wavelength λ = vf = 3401400 = 0.243 m

Separation of the two loudspeakers = 2 × 0.243 = 0.486 m

中文解析

波长 λ = (v)/(f) = (340)/(1400) = 0.243 m

两个扬声器之间的距离 = 2 × 0.243 = 0.486 m

24Ch.7 第7章MC 选择题Easy 易2 marksPHY13071124

Which of the following about the audible frequency range is correct?

(1)The audible frequency range is the same for all people.
(2)The audible frequency range narrows with age.
(3)The audible frequency range is from about 20 Hz to 20 000 Hz.
A(1) only
B(3) only
C(1) and (3) only
D(2) and (3) only
Chinese · 中文对照

下列关于可听频率范围（audible frequency range）的说法，哪项／哪些正确？

（1）所有人的可听频率范围都一样。

（2）可听频率范围会随年龄增长而变窄。

（3）可听频率范围大约是 20 Hz 到 20 000 Hz。

audible 可听见的frequency range 频率范围narrow 变窄vary from person to person 因人而异
Correct answer 正确答案D

The audible frequency range varies from person to person.

∴ (1) is incorrect.

中文解析

可听频率范围因人而异。

∴（1）不正确。

（2）和（3）正确：人的可听频率范围大约由 20 Hz 到 20 000 Hz，并会随年龄增长而变窄。

25Ch.7 第7章MC 选择题Easy 易2 marksPHY13071125

Which of the following about ultrasound is correct?

(1)The frequency of ultrasound is above 20 000 Hz.
(2)All animals, including human beings, cannot hear ultrasound.
(3)Ultrasound is commonly used in scanning of foetuses.
A(1) only
B(3) only
C(1) and (3) only
D(2) and (3) only
Chinese · 中文对照

下列关于超声波（ultrasound）的说法，哪项／哪些正确？

（1）超声波的频率高于 20 000 Hz。

（2）所有动物（包括人类）都听不到超声波。

（3）超声波常用于扫描胎儿。

ultrasound 超声波foetus 胎儿scanning 扫描
Correct answer 正确答案C

Human beings cannot hear ultrasound but many animals can.

∴ (2) is incorrect.

中文解析

人类听不到超声波，但很多动物（例如蝙蝠、海豚）可以听到。

∴（2）不正确。

（1）和（3）正确：超声波指频率高于 20 000 Hz 的声波，常用于胎儿扫描（以及声呐、工业探伤等）。

26Ch.7 第7章MC 选择题Easy 易2 marksPHY13071126

Which of the following is not an application of ultrasonic waves?

ACooking food
BSonar system
CMedical check-up of foetus
DCleaning small and delicate objects
Chinese · 中文对照

下列哪一项不是超声波的应用？

application 应用sonar system 声呐系统delicate 精致的、易损的cleaning 清洁
Correct answer 正确答案A
中文解析

超声波可以用来做声呐探测、胎儿医学检查，以及清洗细小精致的物件（超声波清洗机）。煮食食物靠的是加热，不是超声波，所以不是超声波的应用。

27Ch.7 第7章MC 选择题Easy 易2 marksPHY13071127

Which of the following statements about light and sound is correct?

ABoth light and sound are transverse waves.
BBoth light and sound can travel in a vacuum.
CBoth light and sound travel faster in air than in water.
DBoth light and sound undergo refraction when travelling from one medium to anther medium at an angle.
Chinese · 中文对照

下列关于光和声音的说法，哪一项正确？

transverse wave 横波refraction 折射at an angle 斜着（以一定角度）
Correct answer 正确答案D

Sound is a longitudinal wave.

∴ A is incorrect.

Sound cannot travel in a vacuum.

∴ B is incorrect.

In general, sound travels faster in solids than in liquids, and in turn, faster in liquids than in gases.

∴ C is incorrect.

中文解析

声音是纵波。

∴ A 不正确。

声音不能在真空中传播。

∴ B 不正确。

一般来说，声音在固体中比在液体中快，在液体中又比在气体中快，也就是说声音在水中比在空气中快，所以 C 不正确。

D 正确：光和声音斜射入另一种介质时都会折射。

28Ch.7 第7章MC 选择题Easy 易2 marksPHY13071128

Which of the following comparisons between ultrasound and microwaves are correct?

(1)Microwaves are electromagnetic waves while ultrasound is a mechanical wave.
(2)Microwaves can travel through a vacuum while ultrasound cannot.
(3)Microwaves are transverse waves while ultrasound is a longitudinal wave.
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

下列关于超声波与微波（microwaves）的比较，哪些正确？

（1）微波是电磁波，而超声波是机械波。

（2）微波能在真空中传播，而超声波不能。

（3）微波是横波，而超声波是纵波。

microwave 微波electromagnetic wave 电磁波mechanical wave 机械波vacuum 真空
Correct answer 正确答案D
中文解析

微波是电磁波，属于横波，可以在真空中传播；超声波是机械波（纵波），必须靠介质传播，不能在真空中传播。

所以（1）、（2）、（3）都正确。

29Ch.7 第7章MC 选择题Easy 易2 marksPHY13071129

A boy sees a lightning at the sky and hears the sound of thunder 3 s later. It is given that the speed of sound in the air is 340 m s−1, how far is he from the thunderstorm?

A510 m
B1020 m
C2040 m
D2550 m
Chinese · 中文对照

一个男孩看见天空中的闪电，3 s 后才听到雷声。已知空气中的声速是 340 m s−1，他与雷暴（thunderstorm）相距多远？

lightning 闪电thunder 雷声negligible 可忽略的thunderstorm 雷暴
Correct answer 正确答案B

The speed of light in air is very high, so the time taken for light to reach the boy from the thunderstorm is negligible.

Distance between the boy and the thunderstorm

= vt = 340 × 3 = 1020 m

中文解析

光在空气中的速度非常高，所以光从雷暴传到男孩所需的时间可以忽略。

男孩与雷暴之间的距离

= vt = 340 × 3 = 1020 m

30Ch.7 第7章MC 选择题Easy 易2 marksPHY13071130

The figures below show the traces on a CRO of three notes produced by different musical instruments. The graphs are drawn in the same scale.

(1)

(2)

(3)

Which pair of notes has the same pitch?

A(1) and (2)
B(1) and (3)
C(2) and (3)
DNone of the above
Chinese · 中文对照

（见图）下图显示三台不同乐器发出的三个音在示波器（CRO）上的波形，三个图用同一比例画出。哪一对音的音调（pitch）相同？

trace （示波器上的）波形轨迹same scale 同一比例period 周期pitch 音调
Correct answer 正确答案D

The three waveforms have different periods and hence different frequencies. Therefore, the three notes have different pitches.

中文解析

三个波形的周期（period）都不同，所以频率都不同，因此三个音的音调都不同。

∴ 没有哪一对音的音调相同。

31Ch.7 第7章MC 选择题Easy 易2 marksPHY13071131

The following figures show the waveforms of the four different musical notes. The graphs are drawn in the same scale.

(1)

(2)

(3)

(4)

Which pair of notes has the same loudness?

A(1) and (3)
B(2) and (3)
C(2) and (4)
D(1) and (4)
Chinese · 中文对照

（见图）下图显示四个不同音符的波形，各图用同一比例画出。哪一对音符的响度（loudness）相同？

amplitude 振幅loudness 响度waveform 波形
Correct answer 正确答案B

The amplitudes of waveforms in (2) and (3) are the same, hence the notes in (2) and (3) have the same loudness.

中文解析

（2）和（3）两个波形的振幅相同，所以这两个音符的响度相同。

32Ch.7 第7章MC 选择题Easy 易2 marksPHY13071132

The following figures show the waveforms of four different musical notes shown on a CRO. The settings of the CRO remain unchanged. Which note is the loudest?

A
B
C
D
Chinese · 中文对照

（见图）下图显示四个不同音符在示波器（CRO）上的波形。示波器的设置保持不变。哪个音符最响？

loudest 最响的amplitude 振幅CRO settings 示波器设置
Correct answer 正确答案A

The amplitude of waveform in A is the largest, hence the note in A is the loudest.

中文解析

A 中波形的振幅最大，所以 A 中的音符最响。

33Ch.7 第7章MC 选择题Easy 易2 marksPHY13071133

The following figures show the waveforms of four different musical notes shown on a CRO. The settings of the CRO remain unchanged. Which note has the lowest pitch?

A
B
C
D
Chinese · 中文对照

（见图）下图显示四个不同音符在示波器（CRO）上的波形。示波器的设置保持不变。哪个音符的音调（pitch）最低？

lowest pitch 最低音调period 周期frequency 频率
Correct answer 正确答案D

The waveform in D has the longest period and hence the lowest frequency. Therefore, the note in D has the lowest pitch.

中文解析

D 中的波形周期最长，所以频率最低，因此 D 中的音符音调最低。

34Ch.7 第7章MC 选择题Easy 易2 marksPHY13071134

A student uses a microphone and a CRO to detect notes from two musical instruments X and Y in turn. The following figures show the traces obtained with the same CRO settings.

note from X

note from Y

Which of the following statements is correct?

AThe note form X has a different quality from that from Y.
BThe note from X is of a lower pitch than that from Y.
CThe note from X is much louder than that from Y.
DAll of the above
Chinese · 中文对照

一个学生用麦克风和示波器（CRO）依次检测乐器 X 和 Y 发出的音。（见图）下图是在相同示波器设置下得到的波形。下列哪一项正确？

quality 音品（音色）pitch 音调period 周期amplitude 振幅
Correct answer 正确答案A

The notes from X and Y have different waveforms, hence they have different qualities.

∴ A is correct.

The waveform of the note from X has a shorter period and hence a higher frequency. Therefore, the note from X has a higher pitch.

∴ B is incorrect.

Both waveforms have similar amplitudes, so both notes have similar loudness.

∴ C is incorrect.

中文解析

X 和 Y 的音波形不同，所以它们的音品不同。

∴ A 正确。

X 的音波形周期较短，因此频率较高，所以 X 的音调较高。

∴ B 不正确。

两个波形的振幅相近，所以两个音的响度也相近。

∴ C 不正确。

35Ch.7 第7章MC 选择题Easy 易2 marksPHY13071135

The following figures show the CRO displays of two musical notes X and Y. The settings of the CRO are identical in both cases.

note X

note Y

What is the ratio of the frequency of the waveform of X to that of Y?

A1 : 1
B2 : 1
C4 : 1
D5 : 4
Chinese · 中文对照

（见图）下图显示两个音符 X 和 Y 在示波器上的波形，两次的示波器设置完全相同。X 的波形频率与 Y 的波形频率之比是多少？

ratio 比率period 周期identical settings 相同的设置
Correct answer 正确答案A

The periods of the waveforms of notes X and Y are the same, so their frequencies are the same.

中文解析

音符 X 和 Y 的波形周期相同，所以它们的频率相同，频率之比是 1 : 1。

36Ch.7 第7章MC 选择题Easy 易2 marksPHY13071136

Which of the following statements is incorrect?

ANoises above 140 dB causes permanent damage to the ear.
BNoises at 130 dB can cause ear pain.
CThere is no vibration in the air at 0 dB.
DA person living in a noisy place may feel and stressed.
Chinese · 中文对照

下列哪一项不正确？

permanent damage 永久损伤ear pain 耳朵疼痛sound intensity level 声强级stressed 有压力的
Correct answer 正确答案C

Humans cannot hear a sound of a sound intensity level lower than 0 dB. However, it does not mean that there is no vibration in the air at 0 dB.

中文解析

人听不到声强级（sound intensity level）低于 0 dB 的声音。但这并不表示 0 dB 时空气中没有振动——0 dB 只是人耳刚好能听到的最小声强级而已。

∴ C 不正确。

37Ch.7 第7章MC 选择题Average 中等2 marksPHY13071137

The following shows the displacement–distance graph of a longitudinal wave travelling to the right, taking the direction to the right as positive.

Which of the following statements about particle X is incorrect?

AX is at a centre of rarefaction.
BX is at its equilibrium position.
CX is moving to the right.
DX is moving with its maximum speed.
Chinese · 中文对照

（见图）下图显示一个向右传播的纵波的位移–距离图（displacement–distance graph），取向右为正方向。关于粒子 X，下列哪一项不正确？

displacement–distance graph 位移–距离图rarefaction 疏部（稀疏区）maximum speed 最大速率positive direction 正方向
Correct answer 正确答案C

The displacement of X is zero.

∴ B is correct.

Particles at their equilibrium positions are moving with their maximum speed.

∴ D is correct.

Particles on the immediate left of X are displaced to the left from their equilibrium positions while particles on the immediate right of X are displaced to the right from their equilibrium positions. Therefore, X is at a centre of rarefaction.

∴ A is correct.

The figure below shows the displacement−distance graph after a very short time.

X is moving to the negative direction, i.e. to the left.

∴ C is incorrect.

中文解析

X 的位移为零。

∴ B 正确。

位于平衡位置的粒子正以最大速率运动。

∴ D 正确。

X 左边紧邻的粒子位移偏左，X 右边紧邻的粒子位移偏右，所以 X 位于疏部中心。

∴ A 正确。

下图显示经过很短时间后的位移−距离图（见图）。

X 正向负方向移动，也就是向左。

∴ C 不正确。

38Ch.7 第7章MC 选择题Average 中等2 marksPHY13071138

A longitudinal wave is travelling through a medium at speed 3000 m s–1. The following shows the displacement–time graph of a particle in the medium, taking the direction of travel of the wave as positive.

At what time is the particle at a centre of rarefaction?

At = 4 ms
Bt = 6 ms
Ct = 8 ms
Dt = 10 ms
Chinese · 中文对照

一个纵波以 3000 m s−1 的速度在介质中传播。（见图）下图显示介质中某个粒子的位移–时间图（displacement–time graph），取波的传播方向为正方向。这个粒子在哪个时刻位于疏部中心（centre of rarefaction）？

displacement–time graph 位移–时间图rarefaction 疏部（稀疏区）compression 密部（压缩区）
Correct answer 正确答案C

Particles at the centres of rarefaction are at their equilibrium positions.

∴ B and D are incorrect.

From the graph, we can see that the particle is moving to the positive direction at t = 4 ms. Therefore, it is at the centre of compression at this moment.

On the other hand, the particle is moving to the negative direction at t = 8 ms. Therefore, it is at the centre of rarefaction at this moment.

∴ A is incorrect. C is correct.

中文解析

位于疏部中心的粒子都在平衡位置上。

∴ B 和 D 不正确。

从图中可见，粒子在 t = 4 ms 时向正方向运动，所以这一刻它位于密部中心（centre of compression）。

另一方面，粒子在 t = 8 ms 时向负方向运动，所以这一刻它位于疏部中心。

∴ A 不正确，C 正确。

39Ch.7 第7章MC 选择题Average 中等2 marksPHY13071139

As shown in the figure below, a tuning fork emits sound waves after being struck. It is given that the speed of sound in air is 330 m s–1.

What is the separation between two adjacent compressions?

A0.25 m
B0.50 m
C0.75 m
D1.00 m
Chinese · 中文对照

（见图）如下图所示，音叉（tuning fork）被敲击后发出声波。已知空气中的声速是 330 m s−1。相邻两个密部之间的距离是多少？

tuning fork 音叉struck 被敲击adjacent 相邻的compression 密部（压缩区）
Correct answer 正确答案B

Separation between two adjacent compressions

= wavelength of the sound wave

= vf

= 330659

= 0.50 m

中文解析

相邻两个密部之间的距离

= 声波的波长

= (v)/(f)

= (330)/(659)

= 0.50 m

40Ch.7 第7章MC 选择题Average 中等2 marksPHY13071140

At t = 0, a car is at a distance 10 m from a motion sensor and is moving away from it at constant velocity. The motion sensor emits an ultrasound pulse at t = 0 and receives the reflected pulse from the car at t = 0.06 s. Find the speed of the car. It is known that the speed of sound in air is 340 m s–1.

A6.67 m s–1
B13.3 m s–1
C26.7 m s–1
D173 m s–1
Chinese · 中文对照

在 t = 0 时，一辆车离运动传感器（motion sensor）10 m，并以恒定速度远离传感器。传感器在 t = 0 时发出一个超声脉冲，并在 t = 0.06 s 接收到从车反射回来的脉冲。求车的速度。已知空气中的声速是 340 m s−1。

motion sensor 运动传感器pulse 脉冲reflected 反射回来的constant velocity 恒定速度
Correct answer 正确答案A

The ultrasound pulse reaches the car at t = 0.03 s.

Distance between the car and the motion sensor at t = 0.03 s

= vt = 340 × 0.03 = 10.2 m

Speed of the car = dt = 10.2-100.03 = 6.67 m s−1

中文解析

超声脉冲在 t = 0.03 s 时到达车。

在 t = 0.03 s 时，车与运动传感器之间的距离

= vt = 340 × 0.03 = 10.2 m

车的速度 = (d)/(t) = (10.2-10)/(0.03) = 6.67 m s−1

41Ch.7 第7章MC 选择题Average 中等2 marksPHY13071141

A loudspeaker contains two cones of different diameters as shown below. The larger cone is called the woofer and the smaller one is called the tweeter. They are designed to emit sound waves of different wavelengths so that the audiences do not miss out sound of some wavelengths.

Which of the following statements gives the correct description about the sound waves emitted by these cones?

AThe tweeter emits sounds of larger wavelengths so that sounds of different wavelengths have similar amplitudes.
BThe tweeter emits sounds of larger wavelengths so that sounds of different wavelengths have similar qualities.
CThe woofer emits sounds of larger wavelengths so that sounds of different wavelengths have similar degrees of diffraction.
DThe woofer emits sounds of larger wavelengths so that sounds of different wavelengths form similar interference patterns.
Chinese · 中文对照

（见图）如下图所示，一个扬声器内有两个直径不同的锥体。较大的锥体叫低音单元（woofer），较小的锥体叫高音单元（tweeter）。它们的用途是发出不同波长的声波，使听众不会漏掉某些波长的声音。下列哪一项正确描述这些锥体发出的声波？

cone 锥体（喇叭单元）woofer 低音单元tweeter 高音单元diffraction 衍射
Correct answer 正确答案C

If sounds of different wavelengths are emitted from the same cone, the sounds of larger wavelengths will have a larger degree of diffraction. Sounds of larger wavelengths should be emitted from a larger cone to reduce the degree of diffraction while sounds of shorter wavelengths should be emitted from a smaller cone to increase the degree of diffraction.

中文解析

如果不同波长的声音都由同一个锥体发出，波长较大的声音衍射程度会较大。因此，波长较大的声音应由较大的锥体发出，以减小衍射程度；波长较短的声音应由较小的锥体发出，以增大衍射程度。这样，不同波长的声音就有相近的衍射程度。

42Ch.7 第7章MC 选择题Average 中等2 marksPHY13071142

A longitudinal wave passes through a medium. The following figure shows the equilibrium positions of some particles in the medium and their positions at a certain instant.

What are the directions of motions of Q and Y?

AQ · towards the left
Y · towards the left
BQ · towards the left
Y · towards the right
CQ · towards the right
Y · towards the left
DQ · towards the right
Y · towards the right
Chinese · 中文对照

一个纵波通过某种介质。（见图）下图显示介质中一些粒子的平衡位置，以及它们在某一时刻的位置。Q 和 Y 的运动方向分别是什么？

direction of motion 运动方向equilibrium position 平衡位置momentarily at rest 瞬间静止
Correct answer 正确答案B

Take rightwards as positive.

P and S are at their equilibrium positions. Q is not midway between P and S, so Q is not momentarily at rest.

When the wave moves to the right, the displacement of Q changes from positive value to zero. Therefore, Q is moving to the left.

When the wave moves to the right, the displacement of Y changes from zero to positive value. Therefore, Y is moving to the right.

中文解析

取向右为正方向。

P 和 S 位于平衡位置。Q 并不在 P 和 S 的中点，所以 Q 这一刻并不是静止的。

当波向右传播时，Q 的位移由正值变成零，所以 Q 正在向左运动。

当波向右传播时，Y 的位移由零变成正值，所以 Y 正在向右运动。

43Ch.7 第7章MC 选择题Average 中等2 marksPHY13071143

Which of the following statements about travelling longitudinal waves is incorrect?

AAll sound waves are longitudinal waves.
BA particle at the centre of compression is momentarily at rest.
CAll particles carry energy.
DParticles vibrate along the travelling direction of the wave.
Chinese · 中文对照

关于行进的纵波（travelling longitudinal wave），下列哪一项不正确？

travelling wave 行进波centre of compression 密部中心momentarily at rest 瞬间静止carry energy 携带能量
Correct answer 正确答案B

Particles at the centres of compression are moving at the highest speed.

∴ B is incorrect.

中文解析

位于密部中心的粒子正以最大速率运动，并不是静止的。

∴ B 不正确。

44Ch.7 第7章MC 选择题Average 中等2 marksPHY13071144

Which of the following are the evidences showing that sound is a kind of wave?

(1)When a struck tuning fork touches a water surface, disturbance is caused on the surface.
(2)We can hear people talking around a corner of a building.
(3)Alternating loud and soft sounds are heard when we walk across two separate loudspeakers playing the same note.
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

下列哪些是「声音是一种波」的证据？

（1）被敲击的音叉接触水面时，水面会产生扰动。

（2）我们可以听见建筑物拐角后的人说话。

（3）走过两个播放同一音符的独立扬声器时，会听到忽大忽小的声音。

evidence 证据disturbance 扰动interference 干涉tuning fork 音叉
Correct answer 正确答案C

Only waves exhibit diffraction and interference. Therefore, these two criteria can be used to determine whether something is a wave or not.

中文解析

只有波才会表现出衍射和干涉，所以这两种现象可以用来判断某样东西是不是波。

（2）是衍射（绕过拐角），（3）是干涉（忽大忽小），所以（2）和（3）正确。

（1）只显示音叉在振动、把能量传给水，不能证明声音是波。

45Ch.7 第7章MC 选择题Average 中等2 marksPHY13071145

Sound waves cannot travel through a vacuum because

Athey are transmitted by vibrations of particles.
Bthey are longitudinal.
Ctheir speeds are not high enough.
Dthey have different speeds in different media.
Chinese · 中文对照

声波不能穿过真空，因为

vacuum 真空vibration of particles 粒子的振动transmitted 被传播
Correct answer 正确答案A
中文解析

声波是机械波，必须靠介质粒子的振动才能传播；真空中没有粒子，所以声波不能穿过真空。

46Ch.7 第7章MC 选择题Average 中等2 marksPHY13071146

The direction of travel of a sound wave may change when

(1)it travels from a medium to another.
(2)it passes through a small gap.
(3)it interferes with another sound wave.
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

声波的传播方向可能会改变，当

（1）它从一种介质传到另一种介质；

（2）它通过一个小缝隙；

（3）它与另一个声波发生干涉。

medium 介质gap 缝隙interfere 干涉refracted （被）折射
Correct answer 正确答案A

Sound wave is refracted when it travels from a medium to another.

Sound wave is diffracted when it passes through a small gap.

In both cases, the direction of travel of the sound wave may change.

中文解析

声波从一种介质传到另一种介质时会发生折射（refraction）。

声波通过小缝隙时会发生衍射（diffraction）。

这两种情况下，声波的传播方向都可能改变。

而干涉只是两个波叠加，不会改变波的传播方向，所以（3）不对。

47Ch.7 第7章MC 选择题Average 中等2 marksPHY13071147

A sound wave travels towards a house. Part of the wave does not hit the house and bends around the corners of the house. Which of the following statements are correct?

(1)The phenomenon is called diffraction.
(2)The speed of the waves remains unchanged as they bend around the corners.
(3)The degree of bending of the waves depends on the size of the house relative to the wavelength of the waves.
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

一个声波传向一栋房子。部分声波没有打中房子，而是绕过房子的墙角弯过去。下列哪些说法正确？

（1）这个现象叫衍射（diffraction）。

（2）声波绕过墙角时，波速保持不变。

（3）声波弯曲的程度取决于房子的大小相对于波长的比例。

bend around 绕过⋯⋯弯曲corner 角落diffraction 衍射wavelength 波长
Correct answer 正确答案A

The degree of bending of the waves depends on the wavelength of the waves but not the size of the house.

∴ (3) is incorrect.

中文解析

声波弯曲的程度取决于声波的波长，而与房子的大小无关。

∴（3）不正确。

48Ch.7 第7章MC 选择题Average 中等2 marksPHY13071148

In the following figure, two loudspeakers X and Y are connected to a signal generator, and a microphone is connected to a CRO.

When the microphone moves from P to Q, the amplitude of the waveform displayed in the CRO varies as shown in the following figure.

If the frequency of the signal generator is set at 850 Hz, what is the path difference at Q? It is given that the speed of sound in air is 340 m s–1.

A0.5 m
B0.75 m
C1 m
D1.5 m
Chinese · 中文对照

（见图）在下图中，两个扬声器 X 和 Y 接在信号发生器（signal generator）上，一个麦克风接在示波器（CRO）上。当麦克风由 P 移到 Q 时，示波器上显示的波形振幅变化如下图（见图）。若信号发生器的频率设为 850 Hz，Q 处的路程差（path difference）是多少？已知空气中的声速是 340 m s−1。

signal generator 信号发生器microphone 麦克风path difference 路程差amplitude 振幅
Correct answer 正确答案C

Wavelength λ = vf = 340850 = 0.4 m

Path difference at Q = 2.5λ = 2.5 × 0.4 = 1 m

中文解析

波长 λ = (v)/(f) = (340)/(850) = 0.4 m

Q 处的路程差 = 2.5λ = 2.5 × 0.4 = 1 m

49Ch.7 第7章MC 选择题Average 中等2 marksPHY13071149

In the following figure, two loudspeakers X and Y are connected to a signal generator, and a microphone is connected to a CRO.

When the microphone moves from P to Q, the amplitude of the waveform displayed in the CRO varies as shown in the following figure.

Which of the following figures best shows how the CRO reading varies when the frequency is doubled?

A
B
C
D
Chinese · 中文对照

（见图）在下图中，两个扬声器 X 和 Y 接在信号发生器上，一个麦克风接在示波器（CRO）上。当麦克风由 P 移到 Q 时，示波器上显示的波形振幅变化如下图（见图）。下列哪一幅图最能表示频率加倍后示波器读数如何变化？

doubled 加倍path difference 路程差wavelength 波长CRO reading 示波器读数
Correct answer 正确答案D

Before the frequency is doubled, the path difference at Q is 2.5λ.

When the frequency is doubled, by v = fλ, the wavelength is halved. The path difference at Q becomes 5λ.

中文解析

频率加倍之前，Q 处的路程差是 2.5λ。

当频率加倍时，由 v = fλ，波长会减半。于是 Q 处的路程差变成 5λ。

50Ch.7 第7章MC 选择题Average 中等2 marksPHY13071150

A factory uses ultrasonic waves to detect flaws in a steel rod, as shown in the following figure.

The figure below shows the transmitted signal and the reflected signal displayed on the screen.

Given that the speed of sound in steel is 6000 m s−1, what is the distance between the flaw and the receiver?

A0.15 m
B0.4 m
C0.6 m
D1.2 m
Chinese · 中文对照

（见图）下图显示一家工厂用超声波检测钢棒内部的裂缝（flaw）。下方图显示屏幕上发射信号与反射信号的波形（见图）。已知声音在钢中的速度是 6000 m s−1，裂缝与接收器之间的距离是多少？

flaw 裂缝、缺陷steel rod 钢棒transmitted signal 发射信号receiver 接收器
Correct answer 正确答案C

Distance between the flaw and the receiver

= vt = 6000 × (50×10−6)×42 = 0.6 m

中文解析

裂缝与接收器之间的距离

= vt = 6000 × ((50×10−6)×4)/(2) = 0.6 m

51Ch.7 第7章MC 选择题Average 中等2 marksPHY13071151

Which of the following is the reason why light can travel through a vacuum but sound cannot?

AThe wavelength of sound is much larger than that of light.
BThe speed of light is much higher than that of sound.
CLight is a transverse wave and sound is a longitudinal wave.
DSound is a mechanical wave and require a medium to transmit.
Chinese · 中文对照

下列哪一项是「光能穿过真空，而声音不能」的原因？

vacuum 真空mechanical wave 机械波medium 介质transmit 传播
Correct answer 正确答案D
中文解析

光（电磁波）不需要介质也能在真空中传播；声音是机械波，必须靠介质粒子的振动来传播，所以真空中无法传播声音。

波长、速度大小、横波还是纵波，都不是能否在真空中传播的原因。

52Ch.7 第7章MC 选择题Average 中等2 marksPHY13071152

If two students sing the same note ‘middle C’, they will not sound the same. Which of the following cannot explain this phenomenon?

AThe two notes have different qualities.
BThe two notes have different waveforms when displayed on a CRO screen.
CThe two notes have different frequencies.
DThe two notes have different amplitudes.
Chinese · 中文对照

如果两个学生唱同一个音「中央 C」（middle C），听起来还是不一样。下列哪一项不能解释这个现象？

explain 解释quality 音品（音色）waveform 波形amplitude 振幅
Correct answer 正确答案C

The frequency of the note ‘middle C’ is fixed.

中文解析

「中央 C」这个音的频率是固定的。

所以两个学生唱这个音时，频率应该相同，不能用「频率不同」来解释听起来不一样。

∴ 答案是 C。

53Ch.7 第7章MC 选择题Average 中等2 marksPHY13071153

When two different musical instruments each plays a note with the same pitch, their waveforms displayed on a CRO screen may have different

(1)amplitudes.
(2)shapes.
(3)frequencies.
A(2) only
B(3) only
C(1) and (2) only
D(1), (2) and (3)
Chinese · 中文对照

当两台不同的乐器各奏出一个音调（pitch）相同的音时，它们在示波器屏幕上显示的波形可能会有不同的

（1）振幅（amplitude）

（2）形状（shape）

（3）频率（frequency）

pitch 音调amplitude 振幅shape 形状musical instrument 乐器
Correct answer 正确答案C

The frequency of a note with a certain pitch is fixed.

中文解析

音调一定的音，它的频率是固定的。所以两个音调相同的音，频率一定相同，（3）不可能不同。

不同乐器发声的振幅和波形形状则可以不同，所以（1）和（2）正确。

54Ch.7 第7章MC 选择题Average 中等2 marksPHY13071154

Different notes sung by the same person may have different

(1)loudness.
(2)pitches.
(3)qualities.
A(1) and (2) only
B(1) and (3) only
C(2) and (3) only
D(1), (2) and (3)
Chinese · 中文对照

同一个人唱出的不同音，可能有不同的

（1）响度（loudness）

（2）音调（pitch）

（3）音品（quality）

loudness 响度pitch 音调quality 音品（音色）the same person 同一个人
Correct answer 正确答案D
中文解析

同一个人唱不同的音时，可以唱得大声或小声，所以响度可以不同（1 正确）；

可以唱高音或低音，所以音调可以不同（2 正确）；

不同音的泛音（overtone）成分不同，波形形状就不同，所以音品也可以不同（3 正确）。

∴ 三项都正确。

1Ch.7 第7章Long Q 长题目Average 中等10 marksPHY13073101

A longitudinal wave is travelling from right to left in a medium. Figure a shows the equilibrium positions of particles A to L in the medium. Figure b shows the positions of the particles at a certain time t = 0.

(a) State one particle which is

(i) at the centre of rarefaction,

(ii) moving to the left

at time t = 0. (2 marks)

(b) Suppose each particle takes 2 s to make one complete vibration. Find

(i) the frequency of the wave,

(ii) the wave speed.

(4 marks)

(c) Sketch the displacement−distance graph of the wave at t = 0. Take the displacement to the left as positive. (2 marks)

(d) Draw the positions of particles D and G at t = 0.5 s in Figure c. (2 marks)

Chinese translation · 中文对照

一列纵波（longitudinal wave）在介质中从右向左传播。图 a 显示介质中粒子 A 到 L 的平衡位置（equilibrium positions）。图 b 显示某一时刻 t = 0 时这些粒子的位置。

(a) 说出在时刻 t = 0 时符合下列情况的一个粒子：

(i) 位于疏部（rarefaction）的中心，

(ii) 正在向左运动。

（2 分）

(b) 假设每个粒子完成一次完整振动需要 2 s。求

(i) 这列波的频率（frequency），

(ii) 波速（wave speed）。

（4 分）

(c) 画出 t = 0 时这列波的位移–距离图（displacement–distance graph）。取向左的位移为正。（见图）

（2 分）

(d) 在图 c 中画出 t = 0.5 s 时粒子 D 和 G 的位置。（见图）

（2 分）

Approach · 思路提示

  • In a longitudinal wave, particles vibrate along the direction of travel; a compression is where particles are closest, a rarefaction is where they are most spread out.
  • Read one wavelength λ from Figure b: the distance between the centres of two neighbouring compressions (or rarefactions).
  • The wave travels right to left, so a particle always follows the displacement of the particle on its right side. A particle at zero displacement has maximum speed, not zero speed.
  • The period is given: T = 2 s, so f = 1/T = 0.5 Hz.
  • Then use v = fλ with the λ measured from the figure.
  • 0.5 s is one quarter of a period (T = 2 s), so each particle moves to its 90-degree (quarter-cycle) position.
  • 纵波里粒子的振动方向和波的传播方向平行。压部（compression）是粒子最密集处，疏部是粒子最稀疏处。
  • 从图 b 量出波长 λ：相邻两个压部中心（或相邻两个疏部中心）之间的距离就是一个 λ。
  • 波从右向左传，所以粒子的位移总是跟着它右边（上游）那个粒子的位移走。位移为零的粒子速率最大，不是静止。
  • 题目已给周期 T = 2 s，所以 f = 1/T = 0.5 Hz。
  • 再用 v = fλ，把从图中量到的 λ 代入。
  • 0.5 s 正好是四分之一个周期（T = 2 s），所以每个粒子都走到相差 90° 的位置。
longitudinal wave 纵波equilibrium position 平衡位置rarefaction 疏部（粒子最稀处）sketch 画出（草图）
Suggested answer 参考答案

Solutions

Marks

(a) (i) Particle A

1A

(ii) Particle E (or F, G, H, I)

1A

(b) (i) Frequency =

1M

=

= 0.5 Hz

1A

(ii) Wave speed = fλ

1M

= 0.5 × 2.4

= 1.2 m s−1

1A

(c)

(Correct amplitude and wavelength)

1A

(Correct graph)

1A

(d)

(Correct position of particle D)

1A

(Correct position of particle G)

1A

2Ch.7 第7章Long Q 长题目Difficult 难10 marksPHY13073102

In the figure below, two loudspeakers, X and Y, are connected to a signal generator so that they produce coherent sound waves.

(a) (i) Give the definition of longitudinal wave. (1 mark)

(ii) Suggest one practical way to prove that sound waves produced by the loudspeakers are longitudinal waves. (2 marks)

(b) O is a point on line MN. It is equidistant from the two loudspeakers. A microphone, which is connected to a CRO, is moved slowly from O towards M. It detects the first minimum at point A. When the microphone stays at A, the CRO display is as follows.

(i) Explain why a minimum is detected at A. (2 marks)

(ii) Suggest why the amplitude is not zero when the microphone stays at A. (1 mark)

(iii) If XA = 2.4 m, YA = 2.8 m, estimate the wavelength of the sound waves. (3 marks)

(iv) The frequency of the sound waves is increased and the first minimum is then detected at B. Is B closer to or farther away from O than A is? (1 mark)

Chinese translation · 中文对照

下图中，两个扬声器（loudspeaker）X 和 Y 连接到一个信号发生器（signal generator），因此它们发出相干（coherent）的声波。（见图）

(a) (i) 写出纵波的定义。（1 分）

(ii) 提出一个实际可行的方法，证明扬声器发出的声波是纵波。（2 分）

(b) O 是直线 MN 上的一点，它到两个扬声器的距离相等。一个连接着示波器（CRO）的麦克风（microphone）从 O 慢慢向 M 移动，在 A 点第一次测到极小值（minimum）。当麦克风停在 A 处时，示波器的显示如下。（见图）

(i) 解释为什么在 A 点会测到极小值。（2 分）

(ii) 提出为什么麦克风停在 A 处时振幅（amplitude）不是零。（1 分）

(iii) 若 XA = 2.4 m、YA = 2.8 m，估计该声波的波长（wavelength）。（3 分）

(iv) 把声波的频率调高，这时第一次在 B 点测到极小值。B 比 A 更接近 O，还是更远离 O？（1 分）

Approach · 思路提示

  • Define a longitudinal wave: particles of the medium vibrate parallel to the direction in which the wave travels.
  • To show sound is longitudinal, show that air particles move to and fro along the direction of travel (a candle flame in front of the loudspeaker wobbles back and forth, or dust on the sound path moves along it).
  • A minimum is heard where the two waves arrive in anti-phase, i.e. path difference = (n + 1/2)λ; at O the path difference is 0 (loud), and A is the FIRST minimum, so at A it is exactly λ/2.
  • Path difference = YA − XA = 2.8 − 2.4 = 0.4 m = λ/2, so λ = 0.8 m.
  • If the frequency increases, λ becomes smaller, so the distance from O needed for a path difference of λ/2 is smaller: B is closer to O than A.
  • The amplitude at a minimum is not zero because the two waves are not exactly equal in amplitude (they travel different distances) and there is background noise.
  • 纵波的定义：介质粒子的振动方向和波的传播方向平行。
  • 要证明声音是纵波，就要显示空气粒子沿传播方向前后来回移动（例如扬声器前的烛焰前后摆动，或声波路径上的纸屑沿该方向来回移动）。
  • 极小值出现在两列波反相到达的地方，即路程差 = (n + ½)λ；O 处路程差为 0（响亮），而 A 是「第一次」极小值，所以 A 处路程差正好是 λ/2。
  • 路程差 = YA − XA = 2.8 − 2.4 = 0.4 m = λ/2，所以 λ = 0.8 m。
  • 频率调高后 λ 变小，要产生 λ/2 的路程差所需离 O 的距离也变小，所以 B 比 A 更接近 O。
  • 极小值处振幅不是零，因为两列波到达时振幅不完全相等（走过不同距离），而且环境还有噪声。
coherent 相干的path difference 路程差loudspeaker 扬声器microphone 麦克风
Suggested answer 参考答案

Solutions

Marks

(a) (i) A longitudinal wave is one in which the particle vibrations are along the direction of travel of the wave.

1A

(ii) Put a candle flame in front of the loudspeaker.

1A

When the loudspeaker is turned on, the candle flame moves back and forth.

1A

(b) (i) The sound waves from the two loudspeakers arrive at A in antiphase.

1A

Destructive interference occurs at A and a minimum is detected.

1A

(ii) The amplitudes of the sound waves from the two loudspeakers are not the same at A.

1A

(iii) Path difference at A = 2.8 − 2.4 = 0.4 m

1M

For the 1st minimum,

path difference = 0.5λ = 0.4

1M

λ = 0.8 m

1A

The wavelength of the sound waves is 0.8 m.

(iv) B is closer to O than A is.

1A

3Ch.7 第7章Long Q 长题目Average 中等11 marksPHY13073103

The figure below shows the floor plan of a room. A loudspeaker is placed at X and Richard is standing at Y in front of a corner.

A loud sound of frequency 800 Hz is produced by the loudspeaker at time t = 0. Assume the reflection of sound by the roof is negligible and the surfaces of the walls are flat and smooth. The speed of sound in air is 340 m s−1.

(a) Find the wavelength of the sound produced. (2 marks)

(b) (i) In the figure, draw the three different paths for the sound to reach Richard. (3 marks)

(ii) Hence, estimate the time instants when Richard hears the sound. (4 marks)

(c) A tall barrier is added as shown below and a loud sound is produced again. Can Richard still hear the sound? Explain your answer. (2 marks)

Chinese translation · 中文对照

下图显示一个房间的平面图。扬声器放在 X 处，Richard 站在一个角落前方的 Y 处。（见图）

在时刻 t = 0，扬声器发出频率 800 Hz 的响亮声音。假设天花板（roof）对声音的反射可以忽略，而各墙壁的表面又平又滑。空气中的声速为 340 m s−1。

(a) 求所发出声音的波长。（2 分）

(b) (i) 在图中画出声音传到 Richard 处的三条不同路径。（3 分）

(ii) 由此，估计 Richard 听到声音的各个时刻。（4 分）

(c) 如下图加上一道很高的屏障（barrier），再发出一次响亮声音。Richard 还能听到声音吗？解释你的答案。（2 分）

Approach · 思路提示

  • Use λ = v/f = 340/800 = 0.425 m in (a).
  • Three paths: (1) the direct path X to Y, (2) one reflection off one wall, (3) one reflection off the other wall (the corner has two walls).
  • Draw reflected paths with the image method: mirror Y (or X) in the wall, join the straight line, and the crossing point on the wall is the reflection point.
  • Find each path length from the distances in the floor plan (add up two straight pieces, or use Pythagoras).
  • Each arrival time is t = distance ÷ 340; the direct sound arrives first, reflections arrive later.
  • For (c) remember diffraction: sound is a wave and can bend around the edge of a barrier. A very large barrier diffracts very little, so Richard would hear almost nothing; decide using the size of the barrier in the figure.
  • (a) 用 λ = v/f = 340/800 = 0.425 m。
  • (b) 三条路径：(1) 由 X 直接到 Y；(2) 经其中一面墙反射一次；(3) 经另一面墙反射一次（角落有左右两面墙）。
  • 画反射路径可用镜像法：把 Y（或 X）沿墙面作对称点，连接成直线，直线与墙的交点就是反射点。
  • 用平面图上的距离求出每条路径的长度（把两段直线相加，或用勾股定理）。
  • 每个时刻用 t = 路程 ÷ 340 计算；直接传来的声音最早到达，反射的声音较晚。
  • (c) 要想到绕射（diffraction）：声音是波，可以绕过屏障边缘传到 Y 处。若屏障很大，绕射很弱，就几乎听不到；要按图中屏障的大小来判断。
floor plan 平面图barrier 屏障/障碍物negligible 可忽略的reflection 反射
Suggested answer 参考答案

Solutions

Marks

(a) By v = fλ,

1M

λ === 0.425 m

1A

The wavelength of the sound produced is 0.425 m.

(b) (i)

(Correct paths)

3 × 1A

(ii) Apply v =⇒ t =.

1M

For path I,

time == 0.158 s

1A

For path II,

time == 0.106 s

1A

For path III,

time == 0.197 s

1A

Richard hears the sound at t = 0.106 s, 0.158 s and 0.197 s.

(c) Yes.

1A

The sound will diffract around the edges of the barrier.

1A

4Ch.7 第7章Long Q 长题目Average 中等11 marksPHY13073104

The following figure shows the positions of particles on a travelling longitudinal wave at time t1 and their equilibrium positions. The displacement–distance graph of the wave at time t1 is plotted, taking the direction to the right as positive.

(a) From the figure, what can you say about the particles having zero displacement at a certain time? (2 marks)

(b) State one pair of particles in the figure that are

(i) in phase. (1 mark)

(ii) out of phase. (1 mark)

(iii) in anti-phase. (1 mark)

(c) (i) State the wavelength of the wave. (1 mark)

(ii) State the amplitude of the wave. (1 mark)

(d) The frequency of the wave is 3 Hz. Find

(i) the period of vibration of a particle. (2 marks)

(ii) the speed of the wave. (2 marks)

Chinese translation · 中文对照

下图显示在时刻 t_1 时，一列行进中的纵波（travelling longitudinal wave）上各粒子的位置，以及它们的平衡位置。图中也画出了时刻 t_1 时的位移–距离图（displacement–distance graph），取向右的方向为正。（见图）

(a) 从图中，关于在某一时刻位移为零的粒子，你可以得到什么结论？（2 分）

(b) 说出图中符合下列条件的一对粒子：

(i) 同相（in phase），（1 分）

(ii) 不同相（out of phase），（1 分）

(iii) 反相（in anti-phase）。（1 分）

(c) (i) 说出这列波的波长。（1 分）

(ii) 说出这列波的振幅。（1 分）

(d) 这列波的频率为 3 Hz。求

(i) 一个粒子振动的周期（period），（2 分）

(ii) 波速。（2 分）

Approach · 思路提示

  • A particle with zero displacement is passing through its equilibrium position; at that instant its speed is maximum (it is NOT at rest).
  • In phase: two particles separated by a whole number of wavelengths (nλ) — they always move the same way.
  • In anti-phase: two particles separated by (n + 1/2)λ — they always move in opposite directions.
  • Out of phase simply means the phases are different (any pair that is not in phase qualifies).
  • Wavelength = compression centre to compression centre, i.e. between two neighbouring peaks of the graph; amplitude = the maximum displacement from equilibrium, read from the scale.
  • T = 1/f = 1/3 ≈ 0.33 s, then v = fλ = 3 × λ (λ from the figure).
  • 位移为零的粒子正通过平衡位置，这一瞬间它的速率最大，并非静止。
  • 同相：两个粒子相距整数个波长（nλ），它们的运动状态永远一样。
  • 反相：两个粒子相距 (n + ½)λ，它们的运动方向永远相反。
  • 不同相就是相位不同，只要不是同相的一对都符合。
  • 波长 = 由一个压部中心到下一个压部中心，也就是图上相邻两波峰之间的距离；振幅 = 离开平衡位置的最大距离，按刻度读出。
  • 先算 T = 1/f = 1/3 ≈ 0.33 s；再用 v = fλ = 3 × λ（λ 从图中量出）。
in phase 同相anti-phase 反相out of phase 不同相displacement 位移
Suggested answer 参考答案

Solutions

Marks

(a) They are either at a centre of compression

1A

or a centre of rarefaction.

1A

(b) (i) a and i / b and j / c and k / d and l / e and m

1A

(ii) (Any pair of particles except the above)

1A

(iii) a and e / b and f / c and g / d and h / e and i /

f and j / g and k / h and l / i and m / a and m

1A

(c) (i) 8 cm

1A

(ii) 1 cm

1A

(d) (i) Period =

1M

== 0.333 s

1A

(ii) Wave speed = fλ

1M

= 3 × 8 = 24 cm s–1

1A

5Ch.7 第7章Long Q 长题目Average 中等10 marksPHY13073105

A small plasticine is stuck on a slinky spring which is pushed and pulled from the left end continuously, generating a wave.

(a) Name

(i) the kind of wave generated. (1 mark)

(ii) the regions X and Y. (2 marks)

(b) State the direction of motion of the plasticine at the instant shown. (1 mark)

(c) (i) State the wavelength of the wave. (1 mark)

(ii) Calculate the separation between the centres of X and Y. (2 marks)

(d) (i) How is the frequency of wave motion related to the frequency of the plasticine oscillation? (1 mark)

(ii) Calculate the wave speed if the frequency of the plasticine oscillation is 0.5 Hz.
(2 marks)

Chinese translation · 中文对照

一小块橡皮泥（plasticine）粘在一根 slinky 弹簧上，弹簧被从它的左端不断推拉，因而产生一列波。（见图）

(a) 说出

(i) 所产生的波的种类，（1 分）

(ii) 区域 X 和 Y 的名称。（2 分）

(b) 说出在所示的一瞬间橡皮泥的运动方向。（1 分）

(c) (i) 说出这列波的波长。（1 分）

(ii) 计算 X 与 Y 中心之间的距离。（2 分）

(d) (i) 波动的频率与橡皮泥振动的频率有什么关系？（1 分）

(ii) 若橡皮泥振动的频率为 0.5 Hz，计算波速。（2 分）

Approach · 思路提示

  • Pushing and pulling the spring makes a longitudinal wave (a compression–rarefaction wave).
  • X is a compression (coils crowded together), Y is a rarefaction (coils spread apart).
  • In a longitudinal wave the plasticine only vibrates about its equilibrium position; it does not travel along the spring.
  • Direction of motion of the plasticine: it takes on the displacement of the part of the spring on the side of the wave source (upstream). If the coil next to it on the source side is being pulled left, it moves left.
  • Wavelength = compression centre to compression centre; the separation between the centres of X (compression) and Y (rarefaction) is λ/2.
  • Frequency of the wave equals the frequency of the plasticine's oscillation, and v = fλ = 0.5 × λ.
  • 推拉弹簧产生的是纵波（压缩–稀疏波）。
  • X 是压部（线圈挤在一起），Y 是疏部（线圈散开）。
  • 在纵波中橡皮泥只在平衡位置附近振动，不会随波沿弹簧跑。
  • 判断橡皮泥的运动方向：它总是跟着靠近波源那一侧（上游）的弹簧做同样的位移。若波源那一侧的线圈正被向左拉，它就向左动。
  • 波长 = 由一个压部中心到下一个压部中心；而 X（压部）与 Y（疏部）中心之间的距离等于 λ/2。
  • 波动的频率等于橡皮泥振动的频率；再代入 v = fλ = 0.5 × λ。
slinky spring （玩具）弹簧compression 压部/压缩区separation （两者的）间距oscillation 振动
Suggested answer 参考答案

Solutions

Marks

(a) (i) Longitudinal wave

1A

(ii) Region X: compression

1A

Region Y: rarefaction

1A

(b) Towards the left

1A

(c) (i) 20 cm

1A

(ii) Separation =

1M

=

= 30 cm

1A

(d) (i) They are equal.

1A

(ii) v = fλ

1M

= 0.5 × 0.2

= 0.1 m s–1

1A

6Ch.7 第7章Long Q 长题目Average 中等10 marksPHY13073106

During daytime, the air near the ground is usually warmer than the air at high altitudes. The case during the night is usually the opposite. When sound waves are emitted from the ground, they travel in different pathways during daytime and the night, as shown in Figures a and b.

Fig a Fig b

(a) State the conditions for total internal reflection to occur. (2 marks)

(b) Explain why the sound waves travel in pathways as shown in the two figures. (4 marks)

(c) It is given that the speeds of sound in carbon dioxide and air under room conditions are 259 m s–1 and 343 m s–1 respectively.

(i) Find the critical angle at an air–carbon dioxide interface. (2 marks)

(ii) Suppose a thick layer of carbon dioxide is floating near the ground. Sketch the pathway of the sound in the following figure. (2 marks)

Chinese translation · 中文对照

日间，靠近地面的空气通常比高空的空气暖；夜间的情况通常相反。当声波从地面发出时，日间和夜间会走不同的路径，如图 a 和图 b 所示。（见图）

(a) 说出发生全反射（total internal reflection）的条件。（2 分）

(b) 解释为什么声波会像两幅图所示那样走这种路径。（4 分）

(c) 已知在室温条件下，声音在二氧化碳和空气中的速度分别为 259 m s^–1 和 343 m s^–1。

(i) 求空气–二氧化碳交界面处的临界角（critical angle）。（2 分）

(ii) 假设有一层厚的二氧化碳浮在地面附近，在下面的图中画出声音的路径。（2 分）（见图）

Approach · 思路提示

  • Conditions for total internal reflection: the wave must travel from a medium where it is slower into one where it is faster (optically 'dense' to 'less dense'), AND the angle of incidence must be greater than the critical angle.
  • The speed of sound depends on air temperature: sound travels faster in warmer air, slower in cooler air.
  • Daytime: air near the ground is warm (fast) and higher air is cool (slow). Sound going upwards bends towards the normal — it curves UPWARDS and never comes back; sound does not travel far.
  • Night: air near the ground is cool (slow) and the air above is warm (fast). Sound going upwards bends away from the normal, the angle grows, and when it exceeds the critical angle total internal reflection turns the sound back DOWN — so it is heard far away.
  • For the critical angle use sin C = v_slower / v_faster = 259 / 343, which gives C ≈ 49°.
  • In (c)(ii) note that total internal reflection can only happen for sound going from the SLOWER medium (CO2) into the FASTER one (air); draw the bending of the ray whenever the speed changes as the ray crosses the layer.
  • 发生全反射需要两个条件：波必须由传播较慢的介质射向传播较快的介质，而且入射角（angle of incidence）大于临界角。
  • 声速跟气温有关：空气越暖，声速越大；空气越冷，声速越小。
  • 日间：近地面空气暖（快）、高空空气冷（慢），声音向上升时向法线（normal）偏折，不断向上弯，不会折返，所以传不远。
  • 夜间：近地面空气冷（慢）、高空空气暖（快），声音向上升时偏离法线，入射角越来越大；超过临界角后发生全反射，声音被折回地面，所以能传得很远。
  • 临界角用 sin C = 慢的一方声速 ÷ 快的一方声速 = 259 / 343，得 C ≈ 49°。
  • (c)(ii) 要留意：全反射只会发生在声音由较慢的介质（二氧化碳）射向较快的介质（空气）时；画出声音穿过这层气体时因速度改变而偏折的路径。
total internal reflection 全反射critical angle 临界角altitude 高度/海拔interface 交界面
Suggested answer 参考答案

Solutions

Marks

(a) Total internal reflection occurs when a wave travels from a medium with a lower wave speed to a medium with a high wave speed

1A

at an angle of incidence greater than the critical angle.

1A

(b) Sound waves travel faster if the air is warmer.

1A

Therefore, the direction of travel of sound waves bends towards the normal when they travel from warm air to cold air in Figure a.

1A

On the other hand, the direction of travel of sound waves bends away from the normal when they travel from cold air to warm air in Figure b.

1A

At a certain height, the angle of incidence exceeds the critical angle. Total internal reflection occurs and the sound waves travel back towards the ground.

1A

(c) (i) Critical angle =

1M

= 49.0°

1A

(ii)

(Total internal reflection occurs at the air-carbon dioxide interface)

1A

(Correct shape)

1A

7Ch.7 第7章Long Q 长题目Average 中等11 marksPHY13073107

Two loudspeakers X and Y separated by 1 m are connected to the same signal generator and emit coherent sound waves at frequency 5 kHz. A student hears alternate loud and soft sounds as he walks on the straight line as shown in the following figure. He finds that the distance x between two points P and O, where adjacent loud sounds are heard, is 0.54 m.

(a) Explain what coherent sources are. (2 marks)

(b) (i) Find the path difference at point P. (3 marks)

(ii) Find the wavelength of the wave. (2 marks)

(iii) Hence, calculate the speed of sound in air. (2 marks)

(c) State and explain how x changes if

(i) the frequency of the sound decreases. (1 mark)

(ii) the two sources are in antiphase. (1 mark)

Chinese translation · 中文对照

两个相距 1 m 的扬声器 X 和 Y 连接到同一个信号发生器，发出频率为 5 kHz 的相干声波。一名学生沿如下图所示的直线行走时，听到响亮与轻微的声音交替出现。他发现相邻两个听到响亮声音的点 P 和 O 之间的距离 x 是 0.54 m。（见图）

(a) 解释什么是相干波源（coherent sources）。（2 分）

(b) (i) 求在 P 点的路程差。（3 分）

(ii) 求该波的波长。（2 分）

(iii) 由此，计算空气中的声速。（2 分）

(c) 说明并解释在下列情况下 x 会怎样改变：

(i) 声音的频率降低。（1 分）

(ii) 两个波源的相位相反（antiphase）。（1 分）

Approach · 思路提示

  • Coherent sources: two sources producing waves of the same frequency with a constant phase difference.
  • P is a loud point, so the path difference there is XP − YP = nλ; get XP and YP from the figure by Pythagoras and subtract them.
  • Since P is the next loud point after O, the path difference at P is exactly one wavelength larger than at O; if O has zero path difference then λ = XP − YP.
  • Wave speed: v = fλ = 5000 × λ.
  • If the frequency decreases, λ increases, so adjacent loud points are further apart and x increases.
  • If the two sources are in anti-phase, every maximum becomes a minimum and every minimum a maximum, but the spacing between adjacent maxima is unchanged, so x is the same.
  • 相干波源：两个波源发出频率相同、相位差恒定的波。
  • P 是听到响亮声音的点，所以 P 处路程差 = XP − YP = nλ；用图中给出的距离和勾股定理（Pythagoras' theorem）求出 XP 与 YP，再相减。
  • P 是紧邻 O 的下一个响亮点，所以 P 处路程差比 O 处多正好一个波长；若 O 处路程差为 0，则 λ = XP − YP。
  • 声速 v = fλ = 5000 × λ。
  • 频率降低 → λ 变大 → 相邻两个响亮点之间的距离变大，所以 x 增加。
  • 两个波源反相时，原来的极大值变成极小值、极小值变成极大值，但相邻极大值之间的距离不变，所以 x 不变。
coherent sources 相干波源path difference 路程差antiphase 反相adjacent 相邻的
Suggested answer 参考答案

Solutions

Marks

(a) Coherent sources are sources with the same frequency

1A

and constant phase difference.

1A

(b) (i) PX == 8.0001 m

1M

PY == 8.0673 m

Path difference = PY – PX

1M

= 8.0673 – 8.0001

= 0.0672 m

1A

(ii) Wavelength = path difference at P

1M

= 0.0672 m

1A

(iii) Speed of sound = fλ

1M

= 5000 × 0.0672

= 336 m s–1

1A

(c) (i) Increase

1A

(ii) Approximately unchanged

1A

8Ch.7 第7章Long Q 长题目Average 中等11 marksPHY13073108

A fishing boat is equipped with a radar and a sonar. The detector and transmitter of one of the devices are shown in the figure below.

(a) (i) Name the types of waves used by the radar and the sonar. (2 marks)

(ii) State two differences between these two kinds of waves. (2 marks)

(b) State one function for each of the devices. (2 marks)

(c) (i) An object is 600 m below the boat. A signal is emitted from the boat and detected after 0.8 s. Calculate the speed of the emitted waves. (2 marks)

(ii) What kind of wave is used? State and explain one advantage of this kind of wave.
(3 marks)

Chinese translation · 中文对照

一艘渔船装有一部雷达（radar）和一部声纳（sonar）。下图显示其中一个装置的探测器和发射器。（见图）

(a) (i) 说出雷达和声纳所用的波的类型。（2 分）

(ii) 说出这两种波的两个差异。（2 分）

(b) 说出这两个装置各自的其中一个功用。（2 分）

(c) (i) 一个物体在船下方 600 m 处。船发出一个讯号，0.8 s 后收到。计算所发出波的速率。（2 分）

(ii) 所用的波是哪一种？说出并解释这种波的一个优点。（3 分）

Approach · 思路提示

  • The radar uses electromagnetic waves (microwaves). The sonar uses ultrasonic sound waves.
  • Two differences you can quote: electromagnetic waves can travel through a vacuum while sound waves cannot; their speeds are very different (3×108 m/s for radio/microwaves, about 1500 m/s for sound in water); electromagnetic waves are transverse while sound waves are longitudinal.
  • Function of the radar: detect aircraft/other ships, measure distance or speed. Function of the sonar: detect shoals of fish/submarines, measure the depth of the sea, navigation.
  • In (c)(i) the signal travels there and back, so the total distance is 2 × 600 = 1200 m.
  • Speed = distance ÷ time = 1200 / 0.8 = 1500 m/s.
  • For (c)(ii) the wave is ultrasonic: high frequency means a short wavelength, so it diffracts very little and forms a narrow, well-directed beam; it also carries a lot of energy and is inaudible to humans (no disturbance to fish/people).
  • 雷达用电磁波（微波 microwave）；声纳用超声波（ultrasonic wave）。
  • 可以写出的差异：电磁波能在真空中传播，声波不能；速度差别很大（电磁波约 3×108 m/s，声音在水中约 1500 m/s）；电磁波是横波，声波是纵波。
  • 雷达的功用：探测飞机、其他船只，测量距离或速度。声纳的功用：探测鱼群或潜艇、测量水深、导航。
  • (c)(i) 讯号要来回走一趟，所以总路程 = 2 × 600 = 1200 m。
  • 速率 = 路程 ÷ 时间 = 1200 / 0.8 = 1500 m/s。
  • (c)(ii) 用的是超声波：频率高、波长短，所以绕射很小，能形成狭窄、方向性好的波束；它还能携带较多能量，而且人耳听不到，不会干扰。
radar 雷达sonar 声纳shoal of fish 鱼群detect 探测/侦测
Suggested answer 参考答案

Solutions

Marks

(a) (i) The radar uses radio waves

1A

while the sonar uses ultrasound waves.

1A

(ii) Radio waves are transverse while ultrasound waves are longitudinal.

1A

Radio waves can travel in a vacuum while ultrasound waves cannot.

1A

(Or other reasonable answers)

(b) Radar can be used to detect objects above sea-level,

1A

such as other boats on the sea.

Sonar can be used to detect a shoal of fish below sea-level.

1A

(Or other reasonable answers)

(c) (i) Speed =

1M

=

= 1500 m s–1

1A

(ii) Ultrasound

1A

It has a very short wavelength and does not diffract greatly,

1A

so it can be detected easily.

1A

9Ch.7 第7章Long Q 长题目Average 中等11 marksPHY13073109

A longitudinal wave of frequency 2 Hz travels to the right along a slinky spring. The figure below shows the positions of some particles on the spring at instant t. The equilibrium positions of the particles are also shown in the figure. Particle B is at its maximum displacement at instant t.

(a) At the instant shown,

(i) name a particle which is at the centre of compression. (1 mark)

(ii) name a particle which is at the centre of rarefaction. (1 mark)

(iii) name a particle which is at rest. (1 mark)

(b) Plot a displacement−distance graph of the particles at the instant shown. Take the displacement towards the right as positive. (4 marks)

(c) Find the amplitude of the longitudinal wave. (1 mark)

(d) Find the wavelength of the longitudinal wave. (1 mark)

(e) Find the speed of the longitudinal wave. (2 marks)

Chinese translation · 中文对照

一列频率为 2 Hz 的纵波沿着一根 slinky 弹簧向右传播。下图显示在某一瞬间 t，弹簧上一些粒子的位置。图中也画出了这些粒子的平衡位置。在瞬间 t，粒子 B 位于最大位移处。（见图）

(a) 在所示的一瞬间，

(i) 说出一个位于压部中心的粒子，（1 分）

(ii) 说出一个位于疏部中心的粒子，（1 分）

(iii) 说出一个静止的粒子。（1 分）

(b) 画出所示瞬间各粒子的位移–距离图。取向右的位移为正。（4 分）

(c) 求这列纵波的振幅。（1 分）

(d) 求这列纵波的波长。（1 分）

(e) 求这列纵波的速度。（2 分）

Approach · 思路提示

  • Compare each particle with its equilibrium (dotted) position: the compression centre is where particles are most crowded, the rarefaction centre where they are most spread out.
  • A particle is at rest only at maximum displacement (the turning points) — its velocity is momentarily zero there.
  • For the graph the horizontal axis is the equilibrium position (distance) and the vertical axis is displacement, positive to the right.
  • A compression centre has the largest positive or negative displacement; the graph is the usual sine curve.
  • Amplitude = maximum displacement from equilibrium read from the figure's scale; wavelength = compression centre to the next compression centre.
  • Speed: v = fλ = 2 × λ.
  • 把每个粒子和它的平衡位置（虚线位置）比较：压部中心的粒子挤得最密，疏部中心的粒子散得最开。
  • 粒子只有在位移最大处（折返点）才静止，那一刻它的速度为零。
  • 画图时横轴用平衡位置（距离），纵轴用位移，向右为正。
  • 压部中心对应位移的正极大或负极小；画出来就是一条常见的正弦曲线。
  • 振幅 = 粒子离开平衡位置的最大距离（按刻度读出）；波长 = 由一个压部中心到下一个压部中心的距离。
  • 波速 v = fλ = 2 × λ。
centre of compression 压部中心at rest 静止instant 瞬间/某一时刻magnitude 大小（绝对值）
Suggested answer 参考答案

Solutions

Marks

(a) (i) A, E or I

1A

(ii) C or G

1A

(iii) B, D, F or H

1A

(b)

(Correct labelled axes)

1A

(Correct amplitude)

1A

(Correct wavelength)

1A

(Correct phase)

1A

(c) 6 cm

1A

(d) 40 cm

1A

(e) Speed = fλ

1M

= 2 × 0.4

= 0.8 m s−1

1A

10Ch.7 第7章Long Q 长题目Average 中等10 marksPHY13073110

George pushes and pulls a spring (Fig a) to produce a wave travelling along the spring towards the right (Fig b).

Fig a

Fig b

(a) Name regions X and Y in Figure b. (2 marks)

(b) George records the positions of the coils at the moments shown in Figures a and b.

Fig c

(i) What is the amplitude of the wave? (1 mark)

(ii) Take the direction to the right as positive. Sketch the displacement−distance graph of the wave at the moment shown in Figure b, from coil A to coil O. State the coils with zero displacement and displacement of maximum magnitude. (3 marks)

(iii) If it takes 1 s for the wave to travel from coil A to coil G, find the speed of the wave. (2 marks)

(iv) What is the direction of motion of coil A at the moment shown in Figure b? (1 mark)

(c) The period of the wave is 2 s. On the graph in (b)(ii), sketch the displacement−distance graph of the wave (from coil A to coil O) 3 s after the moment as shown in Figure b.
(1 mark)

Chinese translation · 中文对照

George 推拉一根弹簧（图 a），在弹簧上产生一列向右传播的波（图 b）。（见图 a）（见图 b）

(a) 说出图 b 中区域 X 和 Y 的名称。（2 分）

(b) George 记下了图 a 和图 b 所示时刻各线圈的位置。（见图 c）

(i) 这列波的振幅是多少？（1 分）

(ii) 取向右为正，画出图 b 所示时刻由线圈 A 到线圈 O 的位移–距离图，并说出位移为零的线圈，以及位移大小最大的线圈。（3 分）

(iii) 若波由线圈 A 传到线圈 G 需要 1 s，求波速。（2 分）

(iv) 在图 b 所示时刻，线圈 A 的运动方向是什么？（1 分）

(c) 这列波的周期是 2 s。在图 (b)(ii) 的图上，画出由图 b 所示时刻之后 3 s 时波的位移–距离图（由线圈 A 到线圈 O）。（1 分）

Approach · 思路提示

  • X and Y are the compression and the rarefaction of the spring: one where the coils are crowded, one where they are spread out.
  • The amplitude is the maximum distance a coil moves away from its equilibrium position (measure it from Figure c); for the graph the horizontal axis is equilibrium position and the vertical axis is displacement, positive to the right.
  • Coils at the compression/rarefaction centres have the largest displacement; coils halfway between them have zero displacement.
  • Wave speed = distance travelled ÷ time. Take the distance from A to G from the scale in Figure c and divide by 1 s.
  • A coil's direction of motion is found from the coil next to it on the source side: the wave has not reached there yet, so A moves the way it is being pushed (compare Figures a and b).
  • 3 s is 1.5 periods (T = 2 s): the waveform shifts by 1.5λ and each coil's displacement is reversed, so the new graph is the old one shifted half a wavelength and turned upside down.
  • X 和 Y 分别是弹簧的压部和疏部：一个是线圈挤在一起，一个是线圈散开。
  • 振幅是线圈离开平衡位置的最大距离（从图 c 量出）；画图时横轴用平衡位置，纵轴用位移，向右为正。
  • 压部或疏部中心的线圈位移最大；位于两者正中间的线圈位移为零。
  • 波速 = 走过的距离 ÷ 时间。从图 c 的刻度读出 A 到 G 的距离，再除以 1 s。
  • 判断线圈运动方向：看靠波源那一侧的相邻线圈；波还没传到那里，所以 A 按被推拉的方向运动（比较图 a 和图 b）。
  • 3 s 等于 1.5 个周期（T = 2 s）：波形右移 1.5λ，同时每个线圈的位移反向，所以新图等于把原图平移半个波长再上下翻转。
coil （弹簧的）线圈amplitude 振幅period 周期direction of motion 运动方向
Suggested answer 参考答案

Solutions

Marks

(a) Region X: compression

1A

Region Y: rarefaction

1A

(b) (i) At the moment as shown in Figure b, the displacement of coil D is the amplitude of the wave.

\ Amplitude = 6 cm

1A

(ii)

(Correct labelled axes)

1A

(Correct graph)

1A

Zero displacement: A, G, M
Displacement of maximum amplitude: D, J

1A

(iii) Wave speed =

1M

=

= 0.24 m s−1

1A

(iv) Towards the left

1A

(c) See the curve labelled (c) in (b)(ii).

(Correct phase)

1A

11Ch.7 第7章Long Q 长题目Average 中等13 marksPHY13073111

The figure below (Fig a) shows the equilibrium positions of particles A to M. Figure b shows the positions of the particles at a certain time when a longitudinal wave travelling to the right passes through them. Suppose that particle B performs 5 complete oscillations in 1 s.

Fig a

Fig b

(a) Explain what is meant by a longitudinal wave. State an example of a longitudinal wave.
(2 marks)

(b) (i) Find the amplitude of the wave. (1 mark)

(ii) Find the wavelength of the wave. (1 mark)

(c) (i) Find the frequency of the wave. (1 mark)

(ii) Find the period of the wave. (1 mark)

(iii) Find the speed of the wave. (2 marks)

(d) Identify in Figure b two particles that are

(i) in phase, and (1 mark)

(ii) in antiphase. (1 mark)

(e) Identify in Figure b one particle that is

(i) at the center of a compression, and (1 mark)

(ii) at the center of a rarefaction. (1 mark)

(f) Draw the positions of the particles after 0.1 s. (1 mark)

Chinese translation · 中文对照

下图（图 a）显示粒子 A 到 M 的平衡位置。图 b 显示在某一时刻，一列向右传播的纵波经过时这些粒子的位置。假设粒子 B 在 1 s 内完成 5 次完整振动。（见图 a）（见图 b）

(a) 解释什么是纵波，并举出一个纵波的例子。（2 分）

(b) (i) 求这列波的振幅。（1 分）

(ii) 求这列波的波长。（1 分）

(c) (i) 求这列波的频率。（1 分）

(ii) 求这列波的周期。（1 分）

(iii) 求这列波的速度。（2 分）

(d) 在图 b 中找出两个符合下列情况的粒子：

(i) 同相，（1 分）

(ii) 反相。（1 分）

(e) 在图 b 中找出一个符合下列情况的粒子：

(i) 位于压部中心，（1 分）

(ii) 位于疏部中心。（1 分）

(f) 画出 0.1 s 后各粒子的位置。（1 分）

Approach · 思路提示

  • A longitudinal wave: the particles of the medium vibrate parallel to the direction of energy transfer. Example: a sound wave in air, or a compression wave along a spring.
  • Amplitude = maximum displacement of a particle from its equilibrium position (distance between the solid position and the dotted one); wavelength = compression centre to the next compression centre.
  • 5 complete oscillations in 1 s means f = 5 Hz, so T = 1/f = 0.2 s.
  • Speed: v = fλ = 5 × λ.
  • In-phase particles are nλ apart; anti-phase particles are (n + 1/2)λ apart.
  • 0.1 s is exactly half a period, so every particle is the mirror image of its present position about the equilibrium, and compressions and rarefactions swap places.
  • 纵波：介质粒子的振动方向和能量传递的方向平行。例子：空气中的声波、沿弹簧传播的压缩波。
  • 振幅 = 粒子离开平衡位置的最大距离（实线位置与虚线位置之间的距离）；波长 = 由一个压部中心到下一个压部中心。
  • 1 s 内完成 5 次完整振动，即 f = 5 Hz，所以 T = 1/f = 0.2 s。
  • 波速 v = fλ = 5 × λ。
  • 同相的粒子相距 nλ；反相的粒子相距 (n + ½)λ。
  • 0.1 s 正好是半个周期，所以每个粒子都移到它现在位置关于平衡位置的对称点，压部和疏部的位置互换。
oscillation 振动antiphase 反相rarefaction 疏部complete oscillation 一次完整振动
Suggested answer 参考答案

Solutions

Marks

(a) A longitudinal wave is one in which the vibrations are parallel to the direction of travel of the wave.

1A

Sound wave

1A

(Or other reasonable answers)

(b) (i) 8 cm

1A

(ii) 80 cm

1A

(c) (i) Frequency == 5 Hz

1A

(ii) Period === 0.2 s

1A

(iii) Wave speed = fλ

1M

= 5 × 0.8

= 4 m s−1

1A

(d) (i) A and I

1A

(Or other reasonable answers)

(ii) A and E

1A

(Or other reasonable answers)

(e) (i) E or M

1A

(ii) A or I

1A

(f)

(Correct positions)

1A

12Ch.7 第7章Long Q 长题目Average 中等10 marksPHY13073112

(a) Explain what longitudinal waves and transverse waves are. (2 marks)

(b) (i) Describe a method to demonstrate that sound waves are longitudinal waves.
(2 marks)

(ii) Describe a method to demonstrate that water waves are transverse waves. (2 marks)

(c) In space, astronauts communicate with each other by radio.

(i) Explain why they cannot talk to each other without radio equipment. (1 mark)

(ii) Explain why radio equipment can help them communicate. (1 mark)

(iii) State two other differences between sound waves and radio waves. (2 marks)

Chinese translation · 中文对照

(a) 解释什么是纵波，什么是横波（transverse wave）。（2 分）

(b) (i) 描述一个方法，证明声波是纵波。（2 分）

(ii) 描述一个方法，证明水波是横波。（2 分）

(c) 在太空中，航天员用无线电（radio）互相通讯。

(i) 解释为什么他们不用无线电设备就无法交谈。（1 分）

(ii) 解释为什么无线电设备能帮助他们通讯。（1 分）

(iii) 说出声波和无线电波之间的另外两个差异。（2 分）

Approach · 思路提示

  • Longitudinal wave: the particles vibrate parallel to the direction of travel. Transverse wave: the particles vibrate perpendicular to the direction of travel.
  • Sound is longitudinal: put a candle flame (or a light paper cone / small dust particles) in front of the loudspeaker — it moves to and fro along the direction in which the sound travels, not up and down.
  • Water wave is transverse: float a small cork or a piece of foam on the water; it only bobs up and down while the wave moves sideways, it does not travel with the wave.
  • In space there is a vacuum (no air), and sound waves need a medium, so they cannot carry the sound from one astronaut to another.
  • Radio equipment uses radio waves, which are electromagnetic waves and can travel through a vacuum.
  • Other differences: radio waves are transverse but sound waves are longitudinal; radio waves travel much faster (3×108 m/s); radio waves have a much higher frequency.
  • 纵波：粒子振动方向与传播方向平行。横波：粒子振动方向与传播方向垂直。
  • 证明声波是纵波：把一支蜡烛的火焰（或轻的纸锥、细小纸屑）放在扬声器前，它会沿声音传播的方向前后移动，而不是上下移动。
  • 证明水波是横波：在水面放一个小软木塞或一小块发泡胶，它只会上下浮动，而波是水平传开的，它不会跟着波跑。
  • 太空中是真空（没有空气），而声波需要介质才能传播，所以声音无法由一个人传到另一个人。
  • 无线电设备用的是无线电波，属于电磁波（electromagnetic wave），可以在真空中传播。
  • 其他差异：无线电波是横波而声波是纵波；无线电波的速度大得多（3×108 m/s）；无线电波的频率高得多。
transverse wave 横波vacuum 真空perpendicular 垂直的astronaut 航天员
Suggested answer 参考答案

Solutions

Marks

(a) Longitudinal waves are waves in which the direction of vibrations is parallel to the direction of travel of the wave.

1A

Transverse waves are waves in which the direction of vibrations is perpendicular to the direction of travel of the wave.

1A

(b) (i) Place a burning candle in front of a loudspeaker.

1A

When the loudspeaker plays a sound, the flame moves forwards and backwards.

1A

(ii) Place a cork onto a ripple tank.

1A

The cork moves up and down as water waves travel through the water surface.

1A

(c) (i) There is no medium in space for sound to travel through.

1A

(ii) Radio waves can travel through a vacuum and can carry signals between astronauts.

1A

(iii) Sound waves are longitudinal waves while radio waves are transverse waves.

1A

Sound waves are mechanical waves while radio waves are electromagnetic waves.

1A

(Or other reasonable answers)

13Ch.7 第7章Long Q 长题目Average 中等10 marksPHY13073113

In an experiment, a candle flame is placed in front of a loudspeaker which is producing a low-pitch sound. The flame vibrates backwards and forwards.

(a) Explain briefly what this experiment suggests about the nature of sound. (3 marks)

(b) Alice is standing in front of the loudspeaker. Suggest three ways to decrease the loudness of the sound heard by her. (3 marks)

(c) If the sound produced by the loudspeaker has a frequency of 800 Hz, what is the wavelength of the sound? (Given: The speed of sound in air = 340 m s−1) (2 marks)

(d) When the set-up is put into a box, no sound is heard. Suggest a possible reasons and explain your answer. (2 marks)

Chinese translation · 中文对照

在一个实验中，一支蜡烛的火焰放在一个正发出低音调（low-pitch）声音的扬声器前面。火焰会前后振动。（见图）

(a) 扼要解释这个实验对声音的本质说明了什么。（3 分）

(b) Alice 站在扬声器前面。提出三种减弱她所听到声音响度（loudness）的方法。（3 分）

(c) 若扬声器发出的声音频率为 800 Hz，该声音的波长是多少？（已知：空气中的声速 = 340 m s−1）（2 分）

(d) 当把这套装置放进一个箱子里时，听不到声音。提出一个可能的原因，并加以解释。（2 分）

Approach · 思路提示

  • The flame moves forwards and backwards along the direction in which the sound travels, so the air particles also vibrate along that direction — sound is a longitudinal wave.
  • It also shows that sound is produced by a vibrating source and is transmitted by the vibration of the air (a medium is needed).
  • To reduce the loudness: move further away from the loudspeaker; turn down the volume (smaller amplitude); put soft sound-absorbing material between Alice and the loudspeaker; point the loudspeaker away from her.
  • Wavelength: λ = v/f = 340/800 = 0.425 m.
  • For (d) the box could be sealed/sound-proof (the walls absorb or reflect the sound), or the air inside the box has been pumped out so that there is no medium (vacuum) for the sound to travel in — either explanation is acceptable if it is consistent with the wording.
  • 火焰沿着声音传播的方向前后移动，说明空气粒子也是沿着这个方向来回振动，所以声音是纵波。
  • 这还说明声音是由振动的物体（波源）产生的，并且靠空气的振动（介质）传播出去。
  • 减弱响度的方法：离扬声器远一点；把音量调低（振幅变小）；在 Alice 和扬声器之间放柔软的吸音材料；把扬声器朝向别的方向。
  • 波长 λ = v/f = 340/800 = 0.425 m。
  • (d) 箱子可能是密封隔音的（箱壁把声音吸收或反射），或者箱内的空气被抽走变成真空，没有介质让声音传播；两种解释只要讲得通都可以。
candle flame 烛焰loudness 响度low-pitch 低音调的absorb 吸收
Suggested answer 参考答案

Solutions

Marks

(a) The flame vibrating backwards and forwards suggests that the air around it also vibrates in the backwards and forwards.

1A

The vibration of the air particles are along the direction of travel of the sound.

1A

Therefore, this experiment suggests that sound is a longitudinal wave.

1A

(b) Decrease the amplitude of the sound wave.

1A

Decrease the frequency of the sound wave to below the audible frequency range.

1A

Walk away from the loudspeaker.

1A

(Or other reasonable answers)

(c) By v = fλ,

1M

wavelength == 0.425 m

1A

(d) There is a vacuum in the box.

1A

Sound waves cannot travel from one place to another without a medium.

1A

(Or other reasonable answers)

14Ch.7 第7章Long Q 长题目Average 中等10 marksPHY13073114

The figure below (Fig a) shows an electric bell suspended inside a glass jar. Air is gradually removed from the jar by a vacuum pump. When a vacuum is reached inside the jar, only a faint ringing can be heard.

Fig a

(a) What does the experiment illustrate about the transmission of sound? (1 mark)

(b) Explain why a faint ringing sound can still be heard after the air is removed. (1 mark)

(c) When the ringing sound of the electric bell is picked up by a microphone connected to a CRO, the following waveform is seen on the screen (Fig b).

Fig b

(i) The waveform displayed has crests and troughs. A student says that this shows sound is a transverse wave. Is he correct? (1 mark)

(ii) The trace seen on the screen is produced by the motion of a light spot. Is the trace a wave motion? Explain briefly. (2 marks)

(iii) Estimate the frequency of the ringing sound. (3 marks)

(iv) If a guitar string produces a note of the same pitch as the electric bell, will the waveform of the note displayed on the CRO be the same as Figure b? Explain briefly. (2 marks)

Chinese translation · 中文对照

下图（图 a）显示一个电铃（electric bell）悬挂在一个玻璃罩（glass jar）内。用真空泵（vacuum pump）把罩内的空气逐渐抽走。当罩内达到真空时，只听到微弱的铃声。（见图 a）

(a) 这个实验说明了声音传播的什么？（1 分）

(b) 解释为什么把空气抽走后仍能听到微弱的铃声。（1 分）

(c) 当电铃的铃声被一个连接着示波器的麦克风接收时，屏幕上看到如下图 b 的波形。（见图 b）

(i) 所显示的波形有波峰和波谷（crests and troughs）。有学生说这就显示声音是横波。他是否正确？（1 分）

(ii) 屏幕上看到的轨迹是由一个光点（light spot）的运动产生的。这条轨迹是一种波动吗？扼要解释。（2 分）

(iii) 估计铃声的频率。（3 分）

(iv) 若一条吉他弦发出与电铃音调（pitch）相同的音符，它在示波器上显示的波形会与图 b 相同吗？扼要解释。（2 分）

Approach · 思路提示

  • The experiment shows that sound needs a medium to travel through; it cannot travel through a vacuum.
  • The faint sound still heard travels through the solid parts — the glass jar, the stand and the suspension — or the jar is not a perfect vacuum.
  • (c)(i) He is not correct. The crests and troughs on the CRO show how the signal (amplitude) varies with time; the vertical axis is not the particle displacement, so it tells nothing about whether sound is transverse. Sound is longitudinal.
  • (c)(ii) Strictly the trace is not itself a wave: it is a displacement–time graph of one point. A wave is a vibration travelling through space, while the spot only moves up and down in time (compare with a displacement–distance graph).
  • (c)(iii) Read the period T from Figure b using the time-base setting: f = 1/T.
  • (c)(iv) No. The two sounds have the same pitch (same frequency) but different waveforms (different overtones), so the shapes on the CRO differ.
  • 这个实验说明声音需要介质（medium）才能传播，声音不能在真空中传播。
  • 还会听到微弱声音，是因为声音经固体（玻璃罩、支架、悬挂的绳子）传出来，或者罩内并不是绝对真空。
  • (c)(i) 他错了。示波器上的波峰波谷只是显示讯号大小随时间怎样变化，纵轴并不是粒子的位移，所以不能说明声音是横波；声音其实仍是纵波。
  • (c)(ii) 严格来说这条轨迹本身不是波：它只是某一个点的位移–时间图。波是振动在空间中传播，而光点只是随时间上下移动（可对照位移–距离图来理解）。
  • (c)(iii) 用示波器的时间基准（time base）从图 b 读出周期 T，再用 f = 1/T。
  • (c)(iv) 不会相同。两种声音的音调（频率）相同，但波形（泛音成分）不同，所以屏幕上显示的图形不一样。
electric bell 电铃vacuum pump 真空泵crest and trough 波峰与波谷pitch 音调
Suggested answer 参考答案

Solutions

Marks

(a) Sound waves cannot travel through a vacuum.

1A

(b) A small amount of sound waves can still be transmitted up the wire suspending the bell to the outside.

1A

(c) (i) No

1A

(ii) No,

1A

because there is no energy transferred across the screen.

1A

(Or other reasonable explanations)

(iii) Period ≈ 6.5 × 0.1 × 10−3 = 6.5 × 10−4 s

1A

Frequency =

1M

== 1540 Hz

1A

(iv) Since the sounds produced by the bell and the guitar have different qualities,

1A

the waveforms displayed on the CRO will be different.

1A

15Ch.7 第7章Long Q 长题目Average 中等10 marksPHY13073115

A boy stands between two walls A and B as shown in the following figure. He shouts towards wall A at t = 0.

(a) At t = 0.8 s, he hears the echo. Take the speed of sound in air as 340 m s−1.

(i) Why does the boy hear an echo? (1 mark)

(ii) How far is wall A from the boy? (2 marks)

(iii) If the distance between wall B and the boy is 150 m, when will he hear the second echo? (2 marks)

(b) The shout of the boy is picked up by a microphone and its waveform is displayed on a CRO. State the change in the waveform if

(i) the boy shouts louder. (1 mark)

(ii) the boy shouts at a higher pitch. (1 mark)

(iii) the boy uses a device to produce ultrasonic waves. (1 mark)

(c) State two applications of ultrasonic waves. (2 marks)

Chinese translation · 中文对照

一个男孩站在如图所示的 A、B 两面墙之间。他在 t = 0 时向着墙 A 大喊。（见图）

(a) 在 t = 0.8 s 时，他听到回音（echo）。取空气中的声速为 340 m s−1。

(i) 为什么男孩会听到回音？（1 分）

(ii) 墙 A 离男孩有多远？（2 分）

(iii) 若墙 B 与男孩的距离是 150 m，他什么时候会听到第二次回音？（2 分）

(b) 男孩的喊声被一个麦克风接收，其波形显示在示波器上。说出在下列情况下波形会有什么改变：

(i) 男孩喊得更大声。（1 分）

(ii) 男孩用更高的音调喊。（1 分）

(iii) 男孩用装置产生超声波（ultrasonic wave）。（1 分）

(c) 说出超声波的两种应用。（2 分）

Approach · 思路提示

  • An echo is the sound reflected from a wall coming back to the boy; the sound travels to the wall and back, so it takes time.
  • The sound travels twice the separating distance: total distance = 340 × 0.8 = 272 m, so d = 272/2 = 136 m.
  • For wall B: time = 2 × 150 / 340 ≈ 0.88 s, so the second echo is heard at about t = 0.88 s.
  • Loudness is related to amplitude: shouting louder gives a bigger amplitude but the same frequency (same spacing along the time axis).
  • Higher pitch means higher frequency: more cycles in the same time and a shorter period.
  • Ultrasonic waves are above 20 kHz, so the CRO trace shows much more tightly packed waves (and is inaudible); two applications: medical imaging, sonar/echo sounding of the sea bed, ultrasonic cleaning, breaking kidney stones.
  • 回音是声音经墙面反射后传回男孩那里；声音要走「去墙 + 回来」两趟，所以要花时间。
  • 声音走了两倍的距离：总路程 = 340 × 0.8 = 272 m，所以墙 A 到男孩的距离 d = 272/2 = 136 m。
  • 对墙 B：时间 = 2 × 150 / 340 ≈ 0.88 s，所以第二次回音大约在 t = 0.88 s 听到。
  • 响度由振幅决定：喊得更大声，波的振幅变大，但频率不变（波形在时间轴上的疏密不变）。
  • 音调越高表示频率越高：同样时间内循环次数更多，周期更短。
  • 超声波的频率高于 20 kHz，示波器上的波形会非常密集（人耳听不到）；两种应用：医学超声波检查、声纳／回声测深、超声波清洗、击碎肾石。
echo 回音/回声reflected 被反射的ultrasonic 超声波的pitch 音调
Suggested answer 参考答案

Solutions

Marks

(a) (i) The sound given out by the boy is reflected by wall A and travels back towards the boy.

1A

(ii) Distance =vt

1M

= × 340 × 0.8

= 136 m

1A

(iii) Total distance travelled by the sound

= 136 × 2 + 150 × 2 = 572 m

1M

Time taken === 1.68 s

1A

(b) (i) The amplitude of the waveform increases.

1A

(ii) The frequency of the waveform increases.

1A

(iii) A waveform of higher frequency is shown.

1A

(c) Scanning foetus

1A

Detecting flaws in the railway track

1A

(Or other reasonable answers)

16Ch.7 第7章Long Q 长题目Average 中等13 marksPHY13073116

This question is about the wave phenomena of sound.

(a) A boy is shouting in front of a door (Fig a). This produces a series of sound waves approaching the doorway. The top view is shown in Figure b.

Fig a

Fig b

(i) Sketch the wave pattern formed on the other side of the doorway. (2 marks)

(ii) Name this wave phenomenon. (1 mark)

(iii) Explain how the behaviour of a beam of light waves passing through a doorway is different to that of sound waves. (2 marks)

(b) Two identical loudspeakers P and Q emit sound of the same frequency and loudness as shown in Figure c.

Fig c

Point C is equidistant from the loudspeakers and line XY is parallel to the line joining the loudspeakers. A microphone connected to a CRO is moved along XY to detect the sound. The variation of the amplitude of the CRO trace with the position of the microphone is shown in Figure d.

Fig d

(i) Explain why the amplitude shows alternate maxima and minima. (2 marks)

(ii) Explain why the amplitude at the minimum points is non-zero. (2 marks)

(iii) David says that the amplitude observed along RC will show a similar pattern. State whether he is correct. Explain briefly. (4 marks)

Chinese translation · 中文对照

本题是关于声音的波动现象（wave phenomena）。

(a) 一个男孩在门前大喊（图 a），因而产生一系列接近门口的声波。俯视图如图 b。（见图）

(i) 画出门口另一侧形成的波形图案。（2 分）

(ii) 说出这种波动现象的名称。（1 分）

(iii) 解释一束光波经过门口时的行为与声波有什么不同。（2 分）

(b) 两个相同的扬声器 P 和 Q 发出相同频率和相同响度的声音，如图 c。（见图 c）

C 点到两个扬声器的距离相等，直线 XY 平行于连接两个扬声器的那条直线。一个连接着示波器的麦克风沿 XY 移动来探测声音。示波器轨迹的振幅随麦克风位置的变化如图 d 所示。（见图 d）

(i) 解释为什么振幅会呈现交替的极大值和极小值。（2 分）

(ii) 解释为什么在极小值处振幅不是零。（2 分）

(iii) David 说沿 RC 观察到的振幅也会呈现类似的图案。他是否正确？扼要解释。（4 分）

Approach · 思路提示

  • When the wavelength is comparable to (or larger than) the width of the doorway, the waves spread out in circles/semicircles beyond the doorway — draw curved wave fronts centred on the gap.
  • The name of this phenomenon is diffraction.
  • Light has a wavelength of about 10−7 m, far smaller than a doorway, so a light beam hardly diffracts: it goes almost straight through and gives a sharp-edged beam (only a very small spreading).
  • In (b) the two loudspeakers are coherent sources. Where the path difference is nλ the waves arrive in phase and add up (maximum); where it is (n + 1/2)λ they arrive in anti-phase and cancel (minimum).
  • The amplitude at a minimum is not zero because the two waves travel different distances (so their amplitudes are not exactly equal) and there is background noise.
  • For RC: if RC is the perpendicular bisector of PQ, every point on it is equidistant from P and Q, so the path difference is always zero and the waves always add up — no alternating maxima and minima appear, so David is wrong.
  • 当波长和门缝宽度差不多（或更大）时，波会在门口另一侧向四周散开成圆形／半圆形波前——画以门口为中心的弯曲波前。
  • 这种现象的名称是绕射（diffraction）。
  • 光的波长约为 10−7 m，远小于门口宽度，所以光束几乎不绕射：它几乎笔直地穿过门口，形成边缘清晰的窄光束（散射极微小）。
  • (b) 里两个扬声器是相干波源。路程差为 nλ 的地方，两列波同相到达而互相加强（极大值）；路程差为 (n + ½)λ 的地方，两列波反相到达而互相抵消（极小值）。
  • 极小值处振幅不是零，因为两列波走过的距离不同（振幅不完全相等），而且还有背景噪声。
  • 关于 RC：若 RC 是 PQ 的中垂线，线上每一点到 P、Q 的距离都相等，路程差恒为零，两列波永远互相加强，不会出现交替的极大极小，所以 David 的说法不对。
diffraction 绕射wave phenomenon 波动现象doorway 门口perpendicular bisector 中垂线
Suggested answer 参考答案

Solutions

Marks

(a) (i)

(Correct shape)

1A

(Wavelength remains unchanged)

1A

(ii) Diffraction

1A

(iii) The beam of light passes straight through the doorway without spreading.

1A

This is because the size of the gap is too large when compared with the wavelength of light.

1A

(b) (i) The amplitude reaches a maximum where sound waves from P and Q reinforce each other (constructive interference),

1A

and a minimum where they cancel each other (destructive interference).

1A

(ii) The amplitude of sound waves decreases with distance, so the amplitudes of sound waves from P and Q are not exactly equal at positions other than C.

1A

Therefore, sound waves from P and Q cannot cancel each other completely at the points of destructive interference.

1A

(iii) David is incorrect.

1A

The path difference at any point along RC is always zero.

1A

The interference is always constructive along RC.

1A

So the amplitude always stays at the maximum value, instead of showing an up-and-down pattern.

1A

17Ch.7 第7章Long Q 长题目Average 中等13 marksPHY13073117

Two identical loudspeakers L1 and L2 are connected to a signal generator. Sound of frequency 300 Hz is produced. Peter stands in front of the loudspeakers such that the distance between him and the loudspeakers are 8.25 m and 5.50 m respectively. The speed of sound in air is
330 m s−1.

(a) Find the wavelength of the sound emitted by the loudspeakers. (2 marks)

(b) Explain whether Peter will hear a loud sound or a soft sound. (4 marks)

(c) The set-up now generates sound of frequencies 150 Hz at the same output volume. How does the sound that Peter hears change in terms of pitch and loudness? Explain briefly.
(6 marks)

(d) When loudspeaker L1 is turned off and loudspeaker L2 gives out a sound of frequency
300 Hz at the same output volume, will Peter hear a louder sound when compared to the situation in (b)? (1 mark)

Chinese translation · 中文对照

两个相同的扬声器 L_1 和 L_2 连接到一个信号发生器，发出频率 300 Hz 的声音。Peter 站在扬声器前面，他与两个扬声器的距离分别是 8.25 m 和 5.50 m。空气中的声速为 330 m s−1。（见图）

(a) 求扬声器所发出声音的波长。（2 分）

(b) 解释 Peter 会听到响亮的声音还是轻微的声音。（4 分）

(c) 现在这套装置以相同的输出音量产生频率 150 Hz 的声音。Peter 听到的声音在音调和响度上有什么改变？扼要解释。（6 分）

(d) 若把扬声器 L_1 关掉，只由扬声器 L_2 以相同输出音量发出 300 Hz 的声音，Peter 听到的声音会否比 (b) 的情况更响亮？（1 分）

Approach · 思路提示

  • (a) λ = v/f = 330/300 = 1.1 m.
  • (b) Path difference = 8.25 − 5.50 = 2.75 m = 2.5 × 1.1 m = 2.5λ, which is (n + 1/2)λ, so the two waves arrive in anti-phase and cancel — Peter hears a SOFT sound (destructive interference).
  • (c) At 150 Hz the wavelength becomes λ = 330/150 = 2.2 m, so the same path difference of 2.75 m is 1.25λ. This is neither nλ nor (n + 1/2)λ, so the interference is only partial: the sound is louder than in (b) but not a maximum.
  • (c) The pitch drops, because 150 Hz is lower than 300 Hz.
  • (d) Yes. In (b) the two waves almost cancel; with only one loudspeaker operating there is no destructive interference, so the amplitude Peter receives is larger.
  • Always show the path difference as a multiple of the wavelength before deciding loud or soft.
  • (a) λ = v/f = 330/300 = 1.1 m。
  • (b) 路程差 = 8.25 − 5.50 = 2.75 m = 2.5 × 1.1 m = 2.5λ，即 (n + ½)λ 的形式，两列波反相到达而互相抵消——Peter 听到轻微的声音（相消干涉）。
  • (c) 频率 150 Hz 时 λ = 330/150 = 2.2 m，同一个 2.75 m 的路程差就等于 1.25λ。1.25λ 既不是 nλ 也不是 (n + ½)λ，所以只是局部干涉：比 (b) 时响亮，但不是最大值。
  • (c) 音调会变低，因为 150 Hz 低于 300 Hz。
  • (d) 会。在 (b) 中两列波几乎完全抵消；只开一个扬声器就没有相消干涉，Peter 接收到的振幅更大，所以更响亮。
  • 做这类题一定要先算出「路程差等于多少个波长」，才判断响亮还是轻微。
identical 完全相同的output volume 输出音量pitch 音调loudness 响度
Suggested answer 参考答案

Solutions

Marks

(a) By v = fλ,

1M

wavelength === 1.1 m

1A

(b) Path difference at Peter’s position = 8.25 − 5.50 = 2.75 m

1A

= 2.5λ

1A

Destructive interference occurs

1A

and he will hear a soft sound.

1A

(c) Wavelength === 2.2 m

1A

Path difference = 2.75 m = 1.25λ

1A

Destructive interference no longer occurs

1A

and so he will hear a louder sound.

1A

Also, since the frequency is lower,

1A

the pitch of the sound will be lower.

1A

(d) Yes

1A

18Ch.7 第7章Long Q 长题目Average 中等11 marksPHY13073118

A fishing boat emits ultrasonic waves to detect shoals of fish.

(a) What are ultrasonic waves? (2 marks)

(b) State three differences between ultrasonic waves and light waves. (3 marks)

(c) Explain why ultrasonic waves are used in sonar instead of audible sound waves. (2 marks)

(d) The waves reflected by a shoal of fish are detected after 0.3 s. Take the speed of the sound in water as 1500 m s–1. Find the distance between the fishing boat and the shoal of fish.
(2 marks)

(e) Name two other applications of ultrasonic waves. (2 marks)

Chinese translation · 中文对照

一艘渔船发出超声波来探测鱼群（shoals of fish）。

(a) 什么是超声波（ultrasonic waves）？（2 分）

(b) 说出超声波和光波的三个差异。（3 分）

(c) 解释为什么声纳（sonar）用超声波而不用人耳听得见的声波。（2 分）

(d) 由鱼群反射回来的波在 0.3 s 后被探测到。取声音在水中的速度为 1500 m s^–1。求渔船与鱼群之间的距离。（2 分）

(e) 说出超声波另外两个应用。（2 分）

Approach · 思路提示

  • Ultrasonic waves are sound waves with a frequency higher than the upper limit of human hearing, i.e. above 20 kHz.
  • Three differences from light: ultrasound is a mechanical longitudinal wave but light is an electromagnetic transverse wave; ultrasound needs a medium (cannot travel in a vacuum) while light can travel in a vacuum; ultrasound travels much more slowly (about 1500 m/s in water) than light (3×108 m/s).
  • Ultrasound has a very short wavelength, so it diffracts very little and stays in a narrow beam — good for locating objects precisely; it also carries more energy and cannot be heard by people or fish.
  • (d) The wave goes down and back, so total distance = 1500 × 0.3 = 450 m; the distance to the fish is half of that: 225 m.
  • Two other applications: medical ultrasound imaging, ultrasonic cleaning, flaw detection in metals, breaking kidney stones, ultrasonic distance sensors.
  • 超声波是频率高于人耳听觉上限（即高于 20 kHz）的声波。
  • 与光波的三个差异：超声波是机械纵波，光是电磁横波；超声波需要介质（不能在真空中传播），光可以在真空中传播；超声波的速度（在水中约 1500 m/s）远小于光速（3×108 m/s）。
  • 超声波波长短，绕射很小，能保持狭窄的波束，定位准确；它还携带较多能量，而且人和鱼都听不到。
  • (d) 波要下去再回来：总路程 = 1500 × 0.3 = 450 m，所以渔船到鱼群的距离是它的一半，即 225 m。
  • 另外两个应用：医学超声波成像、超声波清洗、探测金属内部缺陷、击碎肾石、超声波测距。
ultrasonic 超声波的audible 可听见的shoal （鱼）群application 应用
Suggested answer 参考答案

Solutions

Marks

(a) Ultrasonic waves are sound waves

1A

of frequency above the audible frequency range (above 20 kHz).

1A

(b) Any three of the following:

3 × 1A

Ultrasonic waves are longitudinal waves while light waves are transverse waves.

Ultrasonic waves cannot travel through a vacuum while light waves can.

Ultrasonic waves travel much slower than light waves in air.

Ultrasonic waves are mechanical waves while light waves are electromagnetic waves.

(Or other reasonable answers)

(c) The wavelengths of ultrasonic waves is shorter than those of audible sound waves,

1A

and so they do not diffract easily and can be more focused.

1A

(d) Distance =vt

1M

=× 1500 × 0.3

= 225 m

1A

(e) Scanning foetus

1A

Detecting flaws in railway tracks

1A

(Or other reasonable answers)

19Ch.7 第7章Long Q 长题目Difficult 难11 marksPHY13073119

A longitudinal wave is travelling towards the right in a medium. The following figure shows the displacement–distance graph of the wave at t = 0. The direction towards the right is taken as positive.

(a) State the amplitude and wavelength of the wave. (2 marks)

(b) It is given that the wave speed is 0.5 m s–1. Find the period of the wave. (3 marks)

(c) Describe the motion of particles A, B and C at this instant. (3 marks)

(d) What is the distance between particles B and C at this instant? (1 mark)

(e) Sketch the displacement–time graph of particle C from t = 0 to t = 0.1 s. (2 marks)

Chinese translation · 中文对照

一列纵波在介质中向右传播。下图显示 t = 0 时这列波的位移–距离图。取向右的方向为正。（见图）

(a) 说出这列波的振幅和波长。（2 分）

(b) 已知波速为 0.5 m s^–1。求这列波的周期。（3 分）

(c) 描述此刻粒子 A、B 和 C 的运动。（3 分）

(d) 此刻粒子 B 和 C 之间的距离是多少？（1 分）

(e) 画出粒子 C 由 t = 0 到 t = 0.1 s 的位移–时间图。（2 分）

Approach · 思路提示

  • Read the amplitude (maximum height of the curve) and the wavelength (distance between two neighbouring peaks) straight from the axes of the graph.
  • Period: T = λ/v = λ / 0.5. Work it out with the λ you read from the graph.
  • To find the direction of motion, shift the whole waveform slightly to the right (the wave travels right) and compare with the original: the particle moves towards the new position of the curve.
  • A particle at zero displacement moves fastest; a particle at maximum displacement is momentarily at rest — use this to describe A, B and C.
  • In (d) remember this is a longitudinal wave: the actual separation = equilibrium separation + (displacement of C − displacement of B), because a positive displacement means shifted to the right.
  • In (e) plot displacement against time for particle C only: start at its displacement at t = 0, keep the amplitude, and use the period found in (b) to fix how many cycles fit in 0.1 s.
  • 振幅（曲线的最大高度）和波长（相邻两个波峰之间的距离）都能直接从图的坐标轴读出。
  • 周期 T = λ/v = λ / 0.5，用你在图上读到的 λ 计算。
  • 要判断粒子的运动方向：把整条波形稍稍向右平移（波向右传播），再和原图比较，粒子总是朝曲线的新位置移动。
  • 位移为零的粒子运动最快；位移最大处的粒子瞬时静止——用这一点来描述 A、B、C 的运动。
  • (d) 记住这是纵波：实际距离 = 平衡位置之间的距离 + (C 的位移 − B 的位移)，因为位移为正表示向右移。
  • (e) 只画粒子 C 的位移随时间变化：从 t = 0 时它的位移出发，振幅固定，再用 (b) 求出的周期判断 0.1 s 内包含几个周期。
displacement–distance graph 位移–距离图period 周期instant 瞬间equilibrium 平衡
Suggested answer 参考答案

Solutions

Marks

(a) Amplitude: 2 mm

1A

Wavelength: 4 cm

1A

(b) v = fλ

1M

f === 12.5 Hz

T =

1M

== 0.08 s

1A

(c) A: Moving towards the right

1A

B: Momentarily at rest

1A

C: Momentarily at rest

1A

(d) 6 mm

1A

(e)

(Correct amplitude and wavelength)

1A

(Correct shape)

1A

1Ch.7 第7章Short Q 短题目Average 中等4 marksPHY13072101

Figure a shows a series of particles which is uniformly distributed in the air. Figures b and c show their positions at time t = 0 and t = 0.4 s when a sound wave is propagating to the right.

(a) What is the maximum possible period of this wave? (1 mark)

(b) What is the amplitude of this wave? (1 mark)

(c) Assume the period of the wave is equal to the value in (a). Sketch the displacement−time graph of particle i for one period. Take displacement to the right as positive. (2 marks)

Chinese translation · 中文对照

图 a 显示一系列均匀分布在空气中的粒子。图 b 和图 c 分别显示当声波向右传播时，这些粒子在 t = 0 和 t = 0.4 s 时的位置。

(a) 这个波的周期的最大可能值是多少？（1 分）

(b) 这个波的振幅是多少？（1 分）

(c) 假设这个波的周期等于 (a) 中求出的值。画出粒子 i 在一个周期内的位移—时间图像。取向右的位移为正。（2 分）

Approach · 思路提示

  • Read the particle positions in Fig b and Fig c (see figure): the wave pattern has moved to the right during 0.4 s.
  • Period T is the time for the wave to travel one wavelength: T = λ / v. First get v from the shift in 0.4 s.
  • For the maximum possible period: in 0.4 s the pattern may have moved one whole wavelength or more, or only a fraction of one. The smallest possible shift gives the largest possible period.
  • Amplitude = the largest displacement of a particle from its equilibrium (middle) position, measured from the figure.
  • For (c): particle i simply oscillates up and down with amplitude A and period T. Find its displacement at t = 0 from Fig b, then draw one complete sine curve with rightward displacement positive.
  • 先看图 b 和图 c 中粒子所在的位置（见图）：0.4 s 内波形整体向右移动了一段距离。
  • 周期 T 就是波前进一个波长所用的时间：T = λ / v，所以要先求出波速 v。
  • 要周期最大，就要让 0.4 s 内的移动距离最小：波形可能移动了 1 个波长或更多，也可能只移动了不到 1 个波长，取最小的可能值。
  • 振幅 = 粒子离开平衡位置（中间位置）的最大位移，从图上量出（见图）。
  • 第 (c) 问：粒子 i 只是上下振动，振幅 A、周期 T。先用图 b 读出它在 t = 0 时的位移，再画一个完整周期的正弦曲线（向右为正）。
uniformly distributed 均匀分布propagating 传播amplitude 振幅sketch 画出草图
Suggested answer 参考答案

Solutions

Marks

(a) 0.8 s

1A

(b) 5 cm

1A

(c)

(Correct amplitude and period)

1A

(Correct shape)

1A

2Ch.7 第7章Short Q 短题目Easy 易6 marksPHY13072102

A guzheng (古箏) has 21 strings. The bridge for each string is preset such that by plucking different strings, musical notes of different pitches can be produced. When a string is plucked, a wave is also produced on the string.

(Courtesy of Tim Sheerman-Chase ½ Flickr.com)

(a) State one difference in the nature of the wave on the string and the sound wave produced. (1 mark)

(b) Compare the wave speed, frequency and wavelength of the wave on the string and the sound wave produced. (3 marks)

(c) How is the pitch of the musical note produced related to the wave on the string? (2 marks)

Chinese translation · 中文对照

古筝（guzheng）有 21 根弦。每根弦的琴码（bridge）都预先调好，所以拨动不同的弦就能发出不同音调（pitch）的音符。拨动一根弦时，弦上也会产生一个波。

（图片来源：Tim Sheerman-Chase ½ Flickr.com）

(a) 说出弦上的波与所产生的声波在性质上的一个不同点。（1 分）

(b) 比较弦上的波与所产生的声波的波速、频率和波长。（3 分）

(c) 发出的音符的音调与弦上的波有什么关系？（2 分）

Approach · 思路提示

  • Nature of the wave: the wave on the string is a transverse wave (vibration perpendicular to travel); the sound wave is a longitudinal wave (vibration along travel).
  • The sound is produced by the vibrating string, so both waves have the SAME frequency.
  • Wave speeds are different because the speed is set by the medium: string (tension, mass per unit length) versus air.
  • Because f is the same but v is different, the wavelengths are different too — use λ = v / f to compare.
  • For (c): higher pitch means a higher frequency of the wave on the string (a shorter, thinner or tighter string gives a higher frequency).
  • 性质不同：弦上的波是横波（transverse wave），振动方向与传播方向垂直；声波是纵波（longitudinal wave），振动方向与传播方向一致。
  • 声波是由弦的振动产生的，所以两者频率相同。
  • 波速不同：波速由介质决定，弦上波速取决于弦的张力与粗细，空气中声速约 340 m s⁻¹。
  • 波长也不同：频率相同而波速不同，用 λ = v / f 比较就能看出。
  • 第 (c) 问：音调（pitch）由频率决定，弦上波的频率越高，发出的音调越高。
pitch 音调pluck 拨（弦）string 弦vibration 振动
Suggested answer 参考答案

Solutions

Marks

(a) The wave on the string is a transverse wave while the sound wave is a longitudinal wave.

1A

(Or other reasonable answers)

(b) Their wave speeds are different.

1A

Their frequencies are the same.

1A

Their wavelengths are different.

1A

(c) The pitch of the musical note depends on the frequency of the wave on the string.

1A

The higher the frequency of the wave, the higher the pitch of the musical note.

1A

3Ch.7 第7章Short Q 短题目Average 中等6 marksPHY13072103

In the set-up below, a gap is made by two barriers which can block light and sound. The width of the gap is 15 cm. N is the centre of the gap and MO is perpendicular to the barriers with MN = NO = 40 cm. PQ is perpendicular to MO and d represents the displacement from point O along PQ. Take the displacement towards P as positive.

(a) A light emitting diode (LED) is put at M and a screen is placed along POQ. The light emitting diode is switched on and emits monochromatic light.

(i) In the figure below, sketch a graph showing how the intensity of light received by the screen varies with the position. (2 marks)

(ii) Explain your answer in (i). (1 mark)

(b) A loudspeaker is put at M and a device is moved along PQ to measure the intensity of sound received. The mini-loudspeaker is switched on and gives out sound of frequency 700 Hz.

(i) In the figure below, sketch a graph showing how the intensity of sound detected by the device varies with the position. (2 marks)

(ii) Explain your answer in (i). (1 mark)

Chinese translation · 中文对照

在下面的装置中，两个能挡住光和声的挡板之间留有一条缝。缝的宽度是 15 cm。N 是缝的中心，MO 与挡板垂直，且 MN = NO = 40 cm。PQ 与 MO 垂直，d 表示沿 PQ 到 O 点的位移。取向 P 一侧的位移为正。

(a) 在 M 处放一个发光二极管（LED），并沿 POQ 放一块光屏。打开这个发光二极管，它发出单色光。

(i) 在下面的图中，画出光屏接收到的光强随位置变化的图像。（2 分）

(ii) 解释你在 (i) 中的答案。（1 分）

(b) 把一个扬声器放在 M 处，用一台装置沿 PQ 移动来测量接收到的声强。打开这个小扬声器，它发出频率为 700 Hz 的声音。

(i) 在下面的图中，画出检测装置测到的声强随位置变化的图像。（2 分）

(ii) 解释你在 (i) 中的答案。（1 分）

Approach · 思路提示

  • Compare the wavelength with the gap width: diffraction is obvious only when λ is comparable to (or larger than) the gap width a = 0.15 m.
  • Light: λ is about 5 × 10⁻⁷ m, far smaller than 0.15 m, so the light goes almost straight through the gap — almost no diffraction and no fringe pattern.
  • So for light, draw a roughly flat (constant) intensity over the region the gap projects onto the screen, dropping sharply to zero at the two edges.
  • Sound: first calculate λ = v / f = 340 / 700 ≈ 0.49 m, which is comparable with a = 0.15 m, so the sound diffracts strongly and spreads out.
  • So for sound, draw a smooth broad curve: highest at O (d = 0) and gradually decreasing towards P and Q, with no sharp cut-off.
  • 先比较波长与缝宽：只有当波长跟缝宽差不多（或更大）时，衍射（diffraction）才明显。
  • 光：波长约 5 × 10⁻⁷ m，远小于缝宽 0.15 m，所以光几乎沿直线穿过缝，衍射可忽略，也不会出现明暗条纹。
  • 所以光的图像：在缝“投影”到光屏上的范围内光强基本不变，到两边边缘处突然降到 0。
  • 声：先算波长 λ = v / f = 340 / 700 ≈ 0.49 m，跟缝宽 0.15 m 差不多，所以声波衍射明显、会向四周散开。
  • 所以声的图像：一条平滑的宽曲线，在 O 点（d = 0）最强，向 P、Q 两侧逐渐减弱，没有突然的边界。
monochromatic light 单色光barrier 挡板intensity 强度perpendicular 垂直的
Suggested answer 参考答案

Solutions

Marks

(a) (i)

(Non-zero intensity in the central part and zero intensity in other parts)

1A

(Central part ranging from around −15 cm to 15 cm)

1A

(ii) Light from the LED travels in straight line through the gap.

1A

(b) (i)

(Non-zero reading in all parts)

1A

(Peak at the central)

1A

(ii) Sound wave diffracts when it passes through the gap.

1A

4Ch.7 第7章Short Q 短题目Easy 易4 marksPHY13072104

(a) Describe one method to demonstrate that sounds are caused by vibrations. (2 marks)

(b) Why can vibrations of an object at some distance be detected by a human ear? (2 marks)

Chinese translation · 中文对照

(a) 描述一个可以证明声音是由振动产生的方法。（2 分）

(b) 为什么人耳能察觉到一定距离外物体的振动？（2 分）

Approach · 思路提示

  • Hit a tuning fork (or a drum) and hold it near a light object, e.g. a small plastic ball hanging on a string, or dip it into water.
  • The ball bounces away / water splashes — clear evidence that the object is vibrating while it makes a sound.
  • When the vibration stops, the sound stops too, so the sound comes from the vibration.
  • For (b): the vibrating object pushes and pulls the air, sending compressions and rarefactions (a longitudinal sound wave) through the air to the ear.
  • The ear drum of the listener then vibrates and the vibration is detected as sound.
  • 敲响音叉（或敲鼓），把它靠近一个轻小物体，例如用细线吊着的小球，或把音叉浸入水中。
  • 小球会被弹开／水面会溅起水花，这就说明发声时物体正在振动。
  • 振动停止时声音也停止，所以声音是由振动产生的。
  • 第 (b) 问：振动中的物体会挤压周围的空气，形成密的空气（密部）和稀的空气（疏部），即纵波（longitudinal wave），在空气中传到耳朵。
  • 声波到达后使耳膜振动，人耳才能感觉到声音。
tuning fork 音叉vibration 振动demonstrate 证明，演示
Suggested answer 参考答案

Solutions

Marks

(a) Strike a tuning fork so that it emits a sound.

1A

Then allow the fork to touch a water surface to see that the fork causes disturbances on the water surface due to its vibration.

1A

(Or other reasonable answers)

(b) When an object vibrates, it produces longitudinal waves

1A

which travel through the air to reach the ear.

1A

5Ch.7 第7章Short Q 短题目Easy 易5 marksPHY13072105

(a) State the audible frequency range for humans. (1 mark)

(b) It is given that the speed of sound in air is 340 m s–1. Find the range of wavelengths of sound audible by humans. (2 marks)

(c) Does all sound diffract significantly when it passes through a classroom door? Explain briefly. (2 marks)

Chinese translation · 中文对照

(a) 说出人耳可听的频率范围。（1 分）

(b) 已知空气中声速为 340 m s⁻¹。求人耳可听声音的波长范围。（2 分）

(c) 所有声音穿过课室的门时都会明显衍射吗？简要解释。（2 分）

Approach · 思路提示

  • Audible range of a normal human ear: about 20 Hz to 20 000 Hz (20 kHz).
  • Use v = f λ, so λ = v / f. For f = 20 000 Hz: λ = 340 / 20 000 = 0.017 m.
  • For f = 20 Hz: λ = 340 / 20 = 17 m. So the wavelengths range from 0.017 m to 17 m.
  • For (c): diffraction is only significant when λ is comparable to the width of the opening (the door, a few tenths of a metre).
  • Low-pitch sounds (long λ, a few metres) diffract a lot through the door; high-pitch sounds (λ of a few centimetres) hardly diffract, so the answer is NO.
  • 正常人的可听声范围约为 20 Hz 至 20 000 Hz（20 kHz）。
  • 用 v = f λ，即 λ = v / f。当 f = 20 000 Hz 时：λ = 340 / 20 000 = 0.017 m。
  • 当 f = 20 Hz 时：λ = 340 / 20 = 17 m。所以波长范围是 0.017 m 至 17 m。
  • 第 (c) 问：只有当波长跟开口（门，宽约几十分之一到几米）的宽度差不多时，衍射才明显。
  • 低音波长大（几米），穿门时衍射很明显；高音波长只有几厘米，几乎不衍射，所以答案是“不是所有声音都明显衍射”。
audible range 可听范围wavelength 波长diffract 衍射significantly 明显地
Suggested answer 参考答案

Solutions

Marks

(a) 20 Hz – 20 kHz

1A

(b) By v = fλ,

1M

shortest wavelength === 0.017 m

longest wavelength === 17 m

\ The range of wavelengths is 0.017 m – 17 m.

1A

(c) No.

1A

Some sound has wavelength much smaller (0.017 m) than the size of a classroom door (1 m) and diffraction would not be significant.

1A

6Ch.7 第7章Short Q 短题目Easy 易5 marksPHY13072106

(a) State three differences between sound waves and light waves. (3 marks)

(b) State the speeds of sound and light in air. (2 marks)

Chinese translation · 中文对照

(a) 说出声波与光波的三个不同点。（3 分）

(b) 说出声音和光在空气中的速率。（2 分）

Approach · 思路提示

  • Type of wave: sound is a longitudinal wave, light is a transverse wave.
  • Medium: sound needs a medium to travel, light can travel through a vacuum.
  • Speed: sound is about 340 m s⁻¹ in air while light is about 3 × 10⁸ m s⁻¹ — six orders of magnitude faster.
  • Other acceptable differences: sound can diffract easily around everyday objects (long λ), light can be polarised but sound cannot.
  • In (b) state the values with units: sound ≈ 340 m s⁻¹ (330–350 m s⁻¹), light ≈ 3 × 10⁸ m s⁻¹ in air/vacuum.
  • 波的种类不同：声波是纵波（longitudinal wave），光波是横波（transverse wave）。
  • 介质不同：声波必须靠介质传播，光波可以在真空（vacuum）中传播。
  • 速率不同：空气中声速约 340 m s⁻¹，光速约 3 × 10⁸ m s⁻¹，相差六个数量级。
  • 其他可以接受的差别：声波波长长、容易衍射，光波可以偏振（polarisation），声波不能。
  • 第 (b) 问要写数值和单位：声音约 340 m s⁻¹，空气中光约 3 × 10⁸ m s⁻¹。
longitudinal wave 纵波transverse wave 横波vacuum 真空polarisation 偏振
Suggested answer 参考答案

Solutions

Marks

(a) Sound waves are mechanical waves while light waves are electromagnetic waves.

1A

Sound waves are longitudinal while light waves are transverse.

1A

Sound waves cannot travel through a vacuum while light waves can.

1A

(Or other reasonable answers)

(b) The speed of sound in air is about 340 m s–1.

1A

The speed of light in air is about 3 × 108 m s–1.

1A

7Ch.7 第7章Short Q 短题目Easy 易5 marksPHY13072107

John is a construction worker who works in a noisy environment every day.

(a) State two harmful effects of continuous exposure to noise. (2 marks)

(b) Suggest how John can protect himself from these harmful effects. (1 mark)

(c) Suggest two government measures that can reduce the harmful effects of noise to the public. (2 marks)

Chinese translation · 中文对照

John 是一名建筑工人，每天都在嘈杂的环境中工作。

（见图）

(a) 说出长期接触噪声的两个有害影响。（2 分）

(b) 建议 John 可以怎样保护自己免受这些有害影响。（1 分）

(c) 建议两项可以减少噪声对公众有害影响的政府措施。（2 分）

Approach · 思路提示

  • (a) Long-term exposure to loud noise damages the hair cells of the inner ear, leading to permanent hearing loss (and sometimes tinnitus).
  • Other harmful effects: annoyance, poor concentration, stress, high blood pressure, sleep disturbance.
  • (b) Wear ear protectors (ear plugs or ear muffs), and limit the time spent in the noisy area / take breaks.
  • (c) Government measures: build sound barriers or plant trees along busy roads, restrict construction working hours, control noise at source with quieter machines, and enforce noise limits with legislation and fines.
  • Only 1 mark for (b), so one concrete protection method is enough.
  • 第 (a) 问：长期接触强噪声会损伤内耳的听觉细胞，造成永久性听力损失（hearing loss），有时还会耳鸣。
  • 其他有害影响：使人烦躁、难以集中精神、精神紧张、血压升高、影响睡眠。
  • 第 (b) 问：戴上护耳用具（耳塞或耳罩），并减少在嘈杂环境中的时间、定时休息。
  • 第 (c) 问政府措施：在繁忙道路旁建隔音屏障或种树、限制施工时间、用更安静的机器从源头减噪、立法规定噪声上限并罚款。
  • 第 (b) 问只有 1 分，写一个具体做法就够了。
noisy environment 嘈杂的环境continuous exposure 长期接触harmful effects 有害影响measure 措施
Suggested answer 参考答案

Solutions

Marks

(a) Hearing damage

1A

Stress

1A

(Or other reasonable answers)

(b) He can wear ear protectors to shield the noise.

1A

(c) Build sound barriers along highways and around construction sites.

1A

Allow construction work to carry out only in daytime.

1A

(Or other reasonable answers)

8Ch.7 第7章Short Q 短题目Easy 易6 marksPHY13072108

Many modern cities have an underground railway system.

(a) Suggest two advantages of an underground railway system compared to an overground railway system. (2 marks)

(b) Explain why reflective barriers cannot completely block noise emitted by busy traffic.
(2 marks)

(c) State two harmful effects of noise. (2 marks)

Chinese translation · 中文对照

许多现代城市都有地下铁路系统。

(a) 与地面铁路系统相比，建议地下铁路系统的两个优点。（2 分）

(b) 解释为什么反射屏障不能完全挡住繁忙交通发出的噪声。（2 分）

(c) 说出噪声的两个有害影响。（2 分）

Approach · 思路提示

  • (a) Advantages: it uses no valuable land on the surface, causes much less noise and traffic disturbance to residents above, and is not affected by weather or traffic jams.
  • (b) A barrier only reflects sound — it does not destroy the sound energy. The reflected sound still reaches people on the other side or bounces off buildings and comes back.
  • Also, sound still diffracts (bends) over the top edge of the barrier and around its ends, so some noise always leaks past.
  • (c) Harmful effects of noise: hearing loss, annoyance and stress, poor concentration, sleep disturbance, high blood pressure.
  • Keep each part short — 2 marks per part means two clear points.
  • 第 (a) 问优点：不占用地面宝贵的土地；对地面居民的噪声和交通干扰小得多；不受天气和交通堵塞影响。
  • 第 (b) 问：屏障只是把声音反射回去，并没有把声能消灭掉。被反射的声音仍会传到另一边，或被建筑物再反射回来。
  • 而且声波还会绕过屏障的顶边和两端发生衍射（diffraction），所以总有噪声漏过去。
  • 第 (c) 问噪声的有害影响：听力损失、烦躁和紧张、难以集中精神、影响睡眠、血压升高。
  • 每小问 2 分，就是两句话两个要点，不用写长。
overground 地面的reflective barrier 反射屏障block 阻挡harmful 有害的
Suggested answer 参考答案

Solutions

Marks

(a) The noise of the trains can be shielded effectively from the people on the ground.

1A

It allows more space on the ground.

1A

(Or other reasonable answers)

(b) Noise can diffract at the edge of the reflective barriers.

1A

Besides, a small fraction of the sound incident on the barriers can transmit through them.

1A

(c) Hearing damage

1A

Stress

1A

(Or other reasonable answers)

9Ch.7 第7章Short Q 短题目Easy 易6 marksPHY13072109

A sound wave of 1000 Hz is travelling to the right in water. The figure below shows the positions of some particles at time t = 0.

(a) At t = 0,

(i) name a particle which is at the centre of compression, (1 mark)

(ii) name a particle which is at the centre of rarefaction. (1 mark)

(b) Find the wavelength of the sound wave. The speed of sound in water is 1500 m s−1.
(2 marks)

(c) Find the time required for the sound wave to travel from particle A to particle I. (2 marks)

Chinese translation · 中文对照

一个频率为 1000 Hz 的声波在水中向右传播。下图显示一些粒子在 t = 0 时的位置。

（见图）

(a) 在 t = 0 时，

(i) 说出一个位于密部中心的粒子。（1 分）

(ii) 说出一个位于疏部中心的粒子。（1 分）

(b) 求这个声波的波长。水中声速为 1500 m s⁻¹。（2 分）

(c) 求声波从粒子 A 传到粒子 I 所需的时间。（2 分）

Approach · 思路提示

  • On the particle diagram, a compression (密部) is where the particles are crowded together; a rarefaction (疏部) is where they are spread out.
  • The centre of a compression/rarefaction is where the particle displacement is zero — read the particle names from the figure.
  • (b) Use λ = v / f = 1500 / 1000 = 1.5 m.
  • (c) Count from the figure how many wavelengths separate A and I (see figure), then multiply by λ to get the distance s.
  • Then t = s / v. For example, if A and I are 2 wavelengths apart, s = 2 × 1.5 = 3.0 m and t = 3.0 / 1500 = 2.0 × 10⁻³ s = 2 ms.
  • 在粒子图上，粒子挤在一起的地方就是密部，粒子散开的地方就是疏部。
  • 密部中心／疏部中心处的粒子位移为零，粒子的名称从图上读出（见图）。
  • 第 (b) 问：用 λ = v / f = 1500 / 1000 = 1.5 m。
  • 第 (c) 问：先看图上 A 到 I 相隔几个波长（见图），再乘 λ 得到距离 s。
  • 然后 t = s / v。例如若 A 与 I 相隔 2 个波长，s = 2 × 1.5 = 3.0 m，t = 3.0 / 1500 = 2.0 × 10⁻³ s = 2 ms。
compression 密部rarefaction 疏部particle 粒子wavelength 波长
Suggested answer 参考答案

Solutions

Marks

(a) (i) C or K

1A

(ii) G

1A

(b) By v = fλ,

1M

wavelength === 1.5 m

1A

(c) A and I are 1 wavelength apart.

1M

Time required = T === 0.001 s

1A

10Ch.7 第7章Short Q 短题目Easy 易5 marksPHY13072110

A motion sensor works by emitting ultrasound signals and detecting the echoes reflected to it. By measuring the time t for an ultrasound signal traveling between the sensor and the object, the distance between an object and the sensor can be found.

(a) What phenomenon of wave does the motion sensor make use of in measuring distance?
(1 mark)

(b) Suppose a student is standing in front of the sensor and the time t measured by the sensor is 10 ms. Find the distance of the student from the sensor. The speed of sound in air is
330 m s−1. (2 marks)

(c) Explain whether the motion sensor can be used for detecting an astronaut from a spacecraft on the Moon. (2 marks)

Chinese translation · 中文对照

运动传感器通过发出超声信号并检测反射回来的回声来工作。测量超声信号在传感器与物体之间往返所需的时间 t，就能求出物体与传感器之间的距离。

(a) 运动传感器测距时利用了波的什么现象？（1 分）

(b) 假设一名学生站在传感器前方，传感器测得的时间 t 为 10 ms。求学生与传感器之间的距离。空气中声速为 330 m s⁻¹。（2 分）

(c) 解释运动传感器能否在月球上从航天器探测航天员。（2 分）

（见图）

Approach · 思路提示

  • (a) The sensor uses the reflection of waves (echo) — the wave is reflected by the object and comes back to the sensor.
  • (b) Careful: the time is for the wave to go there AND come back, so the one-way distance is s = v t / 2.
  • s = 330 × (10 × 10⁻³) / 2 = 1.65 m.
  • (c) The Moon has no atmosphere — there is a vacuum, so sound waves cannot travel at all.
  • Therefore no ultrasound can reach the astronaut and no echo comes back: the sensor cannot work on the Moon.
  • 第 (a) 问：传感器利用的是波的反射（reflection）：声波被物体反射形成回声再回来。
  • 第 (b) 问要小心：测得的时间是声波“去 + 回”的总时间，所以单向距离是 s = v t / 2。
  • s = 330 ×（10 × 10⁻³）/ 2 = 1.65 m。
  • 第 (c) 问：月球表面没有大气，是真空（vacuum），声波根本无法传播。
  • 所以超声传不到航天员那里，也收不到回声，运动传感器在月球上不能用。
ultrasound 超声echo 回声motion sensor 运动传感器astronaut 航天员
Suggested answer 参考答案

Solutions

Marks

(a) Reflection

1A

(b) Distance =vt

1M

=× 330 × 10 × 10−3

= 1.65 m

1A

(c) No,

1A

this is because sound waves cannot travel through a vacuum.

1A

11Ch.7 第7章Short Q 短题目Average 中等7 marksPHY13072111

Figure a shows a loudspeaker. Its cones vibrate to produce sound waves.

Fig a

(a) What type of wave is produced, transverse or longitudinal wave? (1 mark)

(b) Figure b shows the equilibrium positions of particles A to M and the positions of the particles at a certain instant.

Fig b

(i) What are the amplitude and the wavelength of the wave? (2 marks)

(ii) Identify the particles that are at the centres of compression and rarefaction. (4 marks)

Chinese translation · 中文对照

图 a 显示一个扬声器。它的纸盆振动，产生声波。

（见图）图 a

(a) 产生的是哪种波，横波还是纵波？（1 分）

(b) 图 b 显示粒子 A 至 M 的平衡位置，以及某一时刻这些粒子的实际位置。

（见图）图 b

(i) 这个波的振幅和波长分别是多少？（2 分）

(ii) 指出位于密部中心和疏部中心的粒子。（4 分）

Approach · 思路提示

  • (a) Sound is a longitudinal wave: the air particles vibrate back and forth along the direction in which the wave travels.
  • (b)(i) Amplitude = the largest difference between a particle's actual position and its equilibrium position, measured on the figure.
  • Wavelength = the distance between two neighbouring compressions (or between two neighbouring rarefactions); you can also measure it as the distance covered by one full repeat of the pattern.
  • (b)(ii) Particles pile up in a compression — find the places where they are closest together; the particle there is at the centre of a compression.
  • Particles are spread far apart in a rarefaction — find where they are most widely spaced; that particle is at the centre of a rarefaction.
  • A compression and the nearest rarefaction are separated by half a wavelength; read the letters (A–M) off the figure.
  • 第 (a) 问：声音是纵波（longitudinal wave），空气粒子沿着波的传播方向来回振动。
  • 第 (b)(i) 问：振幅 = 粒子实际位置与平衡位置之间的最大差值，在图上量出（见图）。
  • 波长 = 相邻两个密部之间的距离（或相邻两个疏部之间的距离）；也可以量“波形重复一次”的距离。
  • 第 (b)(ii) 问：密部处粒子挤得最紧，找出图上粒子最密集的位置，那里的粒子就在密部中心。
  • 疏部处粒子拉得最开，找出图上粒子间隔最大的位置，那里的粒子就在疏部中心。
  • 一个密部与最近的疏部相隔半个波长，字母直接在图 A–M 中读出（见图）。
cone （扬声器的）纸盆equilibrium position 平衡位置compression 密部rarefaction 疏部
Suggested answer 参考答案

Solutions

Marks

(a) Longitudinal wave

1A

(b) (i) Amplitude = 8 cm

1A

Wavelength = 80 cm

1A

(ii) E and M are at the centres of compression

2 × 1A

A and I are at the centres of rarefaction

2 × 1A

12Ch.7 第7章Short Q 短题目Average 中等5 marksPHY13072112

The figure below shows a sound wave travelling to the right in air. Air particles X and Y are at the centre of a compression and rarefaction respectively.

(a) How does an air particle move when a sound wave passes by it? (1 mark)

(b) At the moment shown in the figure, state

(i) the displacement of X and Y. (2 marks)

(ii) the directions of motion of X and Y. (2 marks)

Chinese translation · 中文对照

下图显示一个在空气中向右传播的声波。空气粒子 X 和 Y 分别位于一个密部的中心和一个疏部的中心。

（见图）

(a) 当一个声波经过某个空气粒子时，这个粒子怎样运动？（1 分）

(b) 在图示的时刻，说出

(i) X 和 Y 的位移。（2 分）

(ii) X 和 Y 的运动方向。（2 分）

Approach · 思路提示

  • (a) Each air particle only vibrates back and forth about its own equilibrium position along the direction of travel — it does not move along with the wave.
  • (b)(i) At the centre of a compression and at the centre of a rarefaction the particle is exactly at its equilibrium position, so both displacements are zero.
  • (b)(ii) Remember the wave travels to the right: the compression at X will move on to the right, so X is moving to the right.
  • At the centre of a rarefaction (Y) the particles move in the opposite direction to the wave, so Y moves to the left.
  • Keep the answers in words: X: 0, moving right; Y: 0, moving left.
  • 第 (a) 问：空气粒子只是在平衡位置附近沿传播方向来回振动，并不会随着波一起前进。
  • 第 (b)(i) 问：密部中心和疏部中心的粒子都正好在平衡位置上，所以 X 和 Y 的位移都是零。
  • 第 (b)(ii) 问：波向右传播，X 所在的密部接下来要向右移动，所以 X 向右运动。
  • 疏部中心 Y 处的粒子运动方向与波传播方向相反，所以 Y 向左运动。
  • 答案写清楚：X 位移为 0、向右运动；Y 位移为 0、向左运动。
displacement 位移direction of motion 运动方向compression 密部rarefaction 疏部
Suggested answer 参考答案

Solutions

Marks

(a) It vibrates to and fro along the direction of travel of the sound wave.

1A

(b) (i) Both particles have zero displacement.

2 × 1A

(ii) X is moving to the right.

1A

Y is moving to the left.

1A

13Ch.7 第7章Short Q 短题目Average 中等7 marksPHY13072113

A sound of frequency 100 Hz travels in air to the right at 330 m s−1. The figure below shows the displacement−distance graph of some air particles at a certain instant, taking the displacement towards the right as positive.

(a) What is the relationship between the direction of travel of the sound wave and the direction of vibration of the air particles? (1 mark)

(b) Find the distance between points X and Y. (3 marks)

(c) (i) Find the period of the sound. (2 marks)

(ii) Draw the displacement−distance graph of the air particles after 0.005 s. (1 mark)

Chinese translation · 中文对照

一个频率为 100 Hz 的声音在空气中以 330 m s⁻¹ 向右传播。下图显示某一时刻一些空气粒子的位移—距离图像，取向右的位移为正。

（见图）

(a) 声波的传播方向与空气粒子的振动方向有什么关系？（1 分）

(b) 求点 X 与点 Y 之间的距离。（3 分）

(c) (i) 求这个声音的周期。（2 分）

(ii) 画出 0.005 s 后空气粒子的位移—距离图像。（1 分）

Approach · 思路提示

  • (a) In a longitudinal wave the particles vibrate parallel to (along) the direction in which the wave travels.
  • (b) Read from the figure how many wavelengths separate X and Y, then use λ = v / f = 330 / 100 = 3.3 m.
  • Then distance XY = (number of wavelengths) × λ (see figure for the count).
  • (c)(i) T = 1 / f = 1 / 100 = 0.01 s.
  • (c)(ii) 0.005 s is exactly half a period, so the whole graph simply shifts by half a wavelength (every displacement flips sign).
  • 第 (a) 问：纵波中粒子的振动方向与波的传播方向平行（在同一直线上）。
  • 第 (b) 问：先从图上读出 X 与 Y 相隔几个波长（见图），再用 λ = v / f = 330 / 100 = 3.3 m。
  • 所以 XY 距离 = 波长的个数 × λ。
  • 第 (c)(i) 问：T = 1 / f = 1 / 100 = 0.01 s。
  • 第 (c)(ii) 问：0.005 s 正好是半个周期，所以整条曲线向右平移半个波长（正负位移全部反过来）。
displacement−distance graph 位移—距离图像period 周期instant 某一时刻
Suggested answer 参考答案

Solutions

Marks

(a) They are parallel to each other

1A

(b) By v = fλ,

1M

Wavelength === 3.3 m

1M

Distance between X and Y =λ =× 3.3 = 1.65 m

1A

(c) (i) Period =

1M

=

= 0.01 s

1A

(ii)

(Correct graph)

1A

14Ch.7 第7章Short Q 短题目Average 中等4 marksPHY13072114

A student is singing a note of frequency 620 Hz.

(a) Find the wavelength of the note. (Given: Speed of sound in air = 340 m s−1) (2 marks)

(b) If the door of the room is about 0.8 m wide, will the diffraction pattern of the sound be prominent? Explain your answer briefly. (2 marks)

Chinese translation · 中文对照

一名学生正在唱一个频率为 620 Hz 的音。

(a) 求这个音的波长。（已知：空气中声速 = 340 m s⁻¹）（2 分）

(b) 如果房门宽约 0.8 m，这个声音的衍射图样会明显吗？简要解释你的答案。（2 分）

Approach · 思路提示

  • (a) Use v = f λ, so λ = v / f = 340 / 620 ≈ 0.55 m.
  • (b) Compare the wavelength with the width of the opening: diffraction is prominent when λ is comparable to the width of the opening.
  • Here λ ≈ 0.55 m and the door is about 0.8 m wide — they are of the same order of magnitude.
  • So yes, the diffraction pattern will be prominent: the sound spreads widely after passing through the door.
  • Do not write 'λ must be smaller than the width' — the rule is 'comparable to or larger than'.
  • 第 (a) 问：用 v = f λ，所以 λ = v / f = 340 / 620 ≈ 0.55 m。
  • 第 (b) 问：比较波长与开口宽度——当波长跟开口宽度差不多时，衍射就明显。
  • 这里 λ ≈ 0.55 m，门宽约 0.8 m，两者数量级相同。
  • 所以衍射图样会很明显：声音穿过门后会向四周大幅散开。
  • 注意不要写成“波长必须小于缝宽”，正确说法是“跟缝宽差不多或更大”。
diffraction pattern 衍射图样prominent 明显的comparable 可相比的
Suggested answer 参考答案

Solutions

Marks

(a) By v = fλ,

1M

wavelength === 0.548 m

1A

(b) The diffraction is prominent

1A

since the wavelength of the sound and the doorway is comparable.

1A

15Ch.7 第7章Short Q 短题目Average 中等5 marksPHY13072115

In a closed space, a sound usually echoes a few times after it is produced. This phenomenon is called reverberation. In a lecture hall, if the reverberation time is too long, the audience can hardly hear what the speaker says.

(a) Name the wave phenomenon that gives rise to the reverberation. (1 mark)

(b) Suggest and explain two ways to reduce reverberation in designing the walls of a lecture hall. (4 marks)

Chinese translation · 中文对照

在封闭空间里，声音发出后通常会回响几次。这种现象叫做混响（reverberation）。在演讲厅里，如果混响时间太长，听众就很难听清讲者说的话。

（见图）

(a) 说出造成混响的是波的哪种现象。（1 分）

(b) 建议并解释两种在设计演讲厅墙壁时减少混响的方法。（4 分）

Approach · 思路提示

  • (a) Reverberation is caused by repeated reflection of sound from the walls, ceiling and floor.
  • (b) Method 1: cover the walls with sound-absorbing material (carpets, curtains, soft boards, acoustic tiles) — the sound energy is absorbed instead of reflected again and again.
  • Method 2: make the walls rough / non-parallel or curved so that reflected sound is scattered in different directions instead of coming straight back to the audience.
  • You can also mention an absorbent ceiling and upholstered seats, which absorb sound.
  • Each method needs a suggestion plus a reason — 2 marks each, so link 'what you do' to 'why the reflected sound becomes weaker'.
  • 第 (a) 问：混响是声音被墙壁、天花板和地板反复反射造成的。
  • 第 (b) 方法一：在墙上铺吸音材料（地毯、窗帘、软板、吸音板），声能被吸收，不会被反复反射。
  • 方法二：把墙面做粗糙、不平行或做成弧形，使反射声分散到不同方向，而不再集中反射回听众。
  • 也可以提吸音天花板和软垫座椅，它们同样能吸收声音。
  • 每个方法都要“做法 + 原因”，各 2 分，要把“怎么做”和“为什么反射声变弱”连起来。
reverberation 混响echo 回声audience 听众, 观众absorb 吸收
Suggested answer 参考答案

Solutions

Marks

(a) Reflection

1A

(b) Apply a layer of sound-absorbing material on the wall.

1A

This reduces the amplitude of the reflected sound.

1A

The surface of the wall should be wavy.

1A

This makes the reflection diffuse.

1A

(Or other reasonable answers)

16Ch.7 第7章Short Q 短题目Average 中等4 marksPHY13072116

A hi-fi loudspeaker unit usually contains two or more speaker cones of different sizes. One speaker cone is for producing low-pitch sounds and the others for sounds of higher pitches. The sound waves produced by the speaker cones will bend around the rim of the cones and spread into the room.

(a) Name the wave phenomenon described above. (1 mark)

(b) Which speaker cone is more suitable for high-pitch sounds? Explain briefly. (3 marks)

Chinese translation · 中文对照

一个高保真（hi-fi）扬声器单元通常有两个或更多大小不同的喇叭单元（纸盆）。其中一个纸盆用来发出低音，其他用来发出音调较高的声音。纸盆发出的声波会绕过纸盆的边缘，并向房间内扩散。

（见图）

(a) 说出上文描述的是波的哪种现象。（1 分）

(b) 哪个纸盆更适合发出高音？简要解释。（3 分）

Approach · 思路提示

  • (a) The sound spreading (bending) around the rim of the cone and into the room is diffraction.
  • Diffraction is obvious only when the size of the obstacle/opening is comparable to the wavelength.
  • For a high-pitch note the frequency is high, so λ = v / f is short — only a few centimetres.
  • So a SMALL cone should be used for high-pitch sounds: its size is comparable to the short wavelength, so the wave can still spread well into the room.
  • A large cone is needed for low-pitch sounds, whose wavelengths are long (about a metre or more).
  • 第 (a) 问：声波绕过纸盆边缘、向房间散开，这就是衍射（diffraction）。
  • 衍射只有在物体或开口的尺寸跟波长差不多时才明显。
  • 高音频率高，所以 λ = v / f 很短，只有几厘米。
  • 因此高音应该用小的纸盆：小纸盆的尺寸跟短波长差不多，波才能向房间充分扩散。
  • 低音波长大（约一米以上），所以要用大的纸盆。
hi-fi 高保真音响speaker cone 扬声器纸盆rim 边缘bend around 绕过
Suggested answer 参考答案

Solutions

Marks

(a) Diffraction

1A

(b) The smaller speaker cone is more suitable for high-pitch sounds.

1A

High-pitch sounds have shorter wavelengths.

1A

If the smaller speaker cone is used, the sound waves will diffract more around the rim and spread over a larger area.

1A

17Ch.7 第7章Short Q 短题目Average 中等4 marksPHY13072117

Peter is studying in a room. He is facing the windows. Glass windows A and B are opened as shown below.

There is noise coming from the right outside the room.

(a) Explain whether Peter will mainly hear the noise coming from his left or right. (3 marks)

(b) Suggest how Peter can reduce the noise without closing all the windows. (1 mark)

Chinese translation · 中文对照

Peter 正在房间里学习。他面向窗户，玻璃窗 A 和 B 打开着，如下图所示。

（见图）

房间右边外面传来噪声。

(a) 解释 Peter 主要会听到来自他左边还是右边的噪声。（3 分）

(b) 建议 Peter 在不把所有窗户都关上的情况下，可以怎样减少噪声。（1 分）

Approach · 思路提示

  • An open window works like a slit. Ordinary sound has a wavelength of the order of a metre, which is comparable to the width of the window.
  • So the sound that comes in from the right through the window diffracts strongly and spreads out inside the room.
  • That is why Peter mainly hears the noise coming from his LEFT, even though the source is on the right (check which side the open windows are on in the figure).
  • For 3 marks: state the answer, name diffraction, and explain with the wavelength-vs-window-size comparison.
  • (b) Close one of the windows (leave only one open), or hang thick curtains over the open window to absorb the sound.
  • 打开的窗户就像一条缝。普通声音的波长约一米，跟窗户的宽度差不多。
  • 所以从右边窗户进来的声音会发生明显衍射（diffraction），在房间里向四周散开。
  • 因此尽管声源在右边，Peter 主要会听到来自他左边的噪声（看清图上两个打开的窗在哪一侧，见图）。
  • 3 分要写：答案 + 指出是衍射 + 用“波长与窗宽差不多”作解释。
  • 第 (b) 问：只关掉其中一个窗（只留一个开），或在打开的窗上挂厚窗帘吸收声音。
noise 噪声window 窗户diffraction 衍射reduce 减少
Suggested answer 参考答案

Solutions

Marks

(a) The noise is largely reflected away when it reaches glass window B.

1A

However, it can enter the room by reflection at glass window A or by diffraction through the opening of the window A.

1A

Hence, Peter will mainly hear the noise coming from his left.

1A

(b) He can close window A only.

1A

18Ch.7 第7章Short Q 短题目Average 中等5 marksPHY13072118

A signal of 2 kHz is fed into two loudspeakers.

(a) What can be heard along PQ? (1 mark)

(b) What is the separation of the loudspeakers if they are about 2 wavelengths apart? (Given: Speed of sound in air = 340 m s−1) (3 marks)

(c) If the distance between the loudspeakers increases, how will the sound heard along PQ change? (1 mark)

Chinese translation · 中文对照

把一个 2 kHz 的信号输入到两个扬声器。

（见图）

(a) 沿 PQ 会听到什么？（1 分）

(b) 如果两个扬声器大约相距 2 个波长，它们的间距是多少？（已知：空气中声速 = 340 m s⁻¹）（3 分）

(c) 如果两个扬声器之间的距离增大，沿 PQ 听到的声音会怎样变化？（1 分）

Approach · 思路提示

  • (a) The two loudspeakers act as two coherent sources, so along PQ you hear loud and soft sound alternately — an interference pattern.
  • (b) First find the wavelength: λ = v / f = 340 / 2000 = 0.17 m.
  • Two wavelengths: separation = 2λ = 2 × 0.17 = 0.34 m.
  • (c) With a larger separation a, the spacing between neighbouring loud (or soft) positions along PQ becomes smaller.
  • So more maxima and minima fit along PQ — the loud and soft regions become closer together (the pattern becomes finer).
  • 第 (a) 问：两个扬声器相当于两个相干（coherent）声源，所以沿 PQ 会交替听到大声和细声，形成干涉（interference）图样。
  • 第 (b) 问：先求波长 λ = v / f = 340 / 2000 = 0.17 m。
  • 两个波长：间距 = 2λ = 2 × 0.17 = 0.34 m。
  • 第 (c) 问：扬声器间距 a 增大后，PQ 上相邻两个大声（或小声）位置之间的距离变小。
  • 所以沿 PQ 上出现更多的最强和最弱位置，声音的“大小声”变化变得更密。
separation 间距loudspeaker 扬声器interference 干涉signal 信号
Suggested answer 参考答案

Solutions

Marks

(a) Alternate loud and soft sounds can be heard along PQ.

1A

(b) By v = fλ,

wavelength === 0.17 m

1M

Distance between loudspeakers = 2λ

1M

= 2 × 0.17 = 0.34 m

1A

(c) The distance between spots of loud sound and soft sound decreases.

1A

19Ch.7 第7章Short Q 短题目Average 中等7 marksPHY13072119

A CD records musical notes of 100 Hz and 1 kHz. Tony uses this CD and his mini hi-fi to demonstrate the interference of sound.

(a) Should Tony use the note of 100 Hz or 1 kHz? Explain your answer.
(Given: Speed of sound in air = 340 m s−1) (5 marks)

(b) Suppose the note other than the one in (a) is used. Describe and explain briefly what Tony would hear when he walks across in front of the mini hi-fi. (2 marks)

Chinese translation · 中文对照

一张 CD 录有 100 Hz 和 1 kHz 的乐音。Tony 用这张 CD 和他的小型音响来演示声音的干涉。

（见图）

(a) Tony 应该用 100 Hz 的音还是 1 kHz 的音？解释你的答案。（已知：空气中声速 = 340 m s⁻¹）（5 分）

(b) 假设用了 (a) 中另一个音。描述并简要解释 Tony 在音响前方走过时会听到什么。（2 分）

Approach · 思路提示

  • Calculate both wavelengths: 100 Hz → λ = 340 / 100 = 3.4 m; 1 kHz → λ = 340 / 1000 = 0.34 m.
  • The distance between neighbouring loud (or soft) positions depends on the wavelength — a longer wavelength gives widely spaced maxima and minima.
  • So Tony should use the 100 Hz note: its long wavelength gives loud/soft regions several metres apart, which he can easily walk through and hear.
  • (b) With 1 kHz, λ = 0.34 m, so the maxima and minima are only about a few tens of centimetres apart.
  • He would still be in a sound field but the loud and soft positions change too fast for him to notice — it sounds roughly equally loud wherever he walks.
  • 先算两个波长：100 Hz → λ = 340 / 100 = 3.4 m；1 kHz → λ = 340 / 1000 = 0.34 m。
  • 相邻两个最响（或最弱）位置之间的距离跟波长有关：波长越长，最大声和最细声的位置分得越开。
  • 所以 Tony 应该用 100 Hz 的音：波长大，最响和最弱的区域相隔几米，他走过去能明显听出来。
  • 第 (b) 问：若用 1 kHz，λ = 0.34 m，最响和最弱的位置只相隔几十厘米。
  • 他听起来会差不多一样响，很难察觉到大小声的变化。
musical note 乐音mini hi-fi 小型音响interference 干涉walk across 横穿走过
Suggested answer 参考答案

Solutions

Marks

(a) By v = fλ,

1M

wavelength of 100 Hz note == 3.4 m

1A

wavelength of 1 kHz note == 0.34 m

1A

For observable interference pattern, the separation between the loudspeakers should be slightly larger than the wavelengths of the wave.

1A

Therefore, note of 1 kHz should be used.

1A

(b) Since antinodal and nodal lines are far apart,

1A

the variation of loudness of sound will not be obvious.

1A

20Ch.7 第7章Short Q 短题目Average 中等4 marksPHY13072120

(a) (i) Does the waveform of sound depend on the listener? (1 mark)

(ii) State the audible frequency range of human beings. (1 mark)

(b) Matthew is playing a drum and he hits the drum two times a second, i.e. at a frequency of 2 Hz. Why can people hear the drum despite this low frequency? (2 marks)

Chinese translation · 中文对照

(a) (i) 声音的波形会因听者不同而不同吗？（1 分）

(ii) 说出人类可听声的频率范围。（1 分）

(b) Matthew 在打鼓，他每秒敲鼓两次，即频率为 2 Hz。为什么在这么低的频率下人们仍能听到鼓声？（2 分）

Approach · 思路提示

  • (a)(i) No. The waveform of a sound is a property of the sound wave itself, so all listeners receive the same waveform (whether they can hear it depends on their ears, not on the waveform).
  • (a)(ii) About 20 Hz to 20 000 Hz.
  • (b) The 2 Hz is the frequency of HITTING the drum (two hits per second), not the frequency of the sound made by the drum.
  • Each hit makes the drum skin vibrate at its own much higher frequency (well within 20 Hz – 20 kHz), and that high-frequency vibration produces the sound we hear.
  • So the pitch we hear is high even though the hits are slow.
  • 第 (a)(i) 问：不会。波形是声音本身的性质，所有听者接收到的波形都相同（听不听得到是耳朵的问题，与波形无关）。
  • 第 (a)(ii) 问：约 20 Hz 至 20 000 Hz。
  • 第 (b) 问：2 Hz 是“敲鼓这个动作”的频率（每秒两下），不是鼓发出声音的频率。
  • 每敲一下，鼓皮会以它自己高得多的频率振动（在 20 Hz – 20 kHz 之内），这个高频振动才产生我们听到的声音。
  • 所以即使敲得很慢，听到的音调依然很高。
waveform 波形listener 听者audible frequency range 可听频率范围hit 敲击
Suggested answer 参考答案

Solutions

Marks

(a) (i) No

1A

(ii) From about 20 Hz to 20 kHz

1A

(b) After each hit, the diaphragm of the drum vibrates

1A

at a frequency well within human’s audible frequency range,

1A

and so people can hear the drum.

21Ch.7 第7章Short Q 短题目Average 中等5 marksPHY13072121

A pest repeller works by emitting powerful ultrasonic signals.

(a) What is the frequency of the ultrasonic signals? (1 mark)

(b) Are human beings affected by such ultrasonic signals? Explain briefly. (2 marks)

(c) Name two other applications of ultrasound. (2 marks)

Chinese translation · 中文对照

一台驱虫器通过发出强力的超声信号来工作。

(a) 这些超声信号的频率是多少？（1 分）

(b) 人类会受到这种超声信号影响吗？简要解释。（2 分）

(c) 说出超声的另外两个应用。（2 分）

Approach · 思路提示

  • (a) Ultrasound means a frequency above the upper limit of human hearing, i.e. above 20 000 Hz (20 kHz).
  • (b) No. The frequency is above the audible range of the human ear, so we cannot hear it (though very intense ultrasound can still damage hearing).
  • (c) Other uses of ultrasound: medical imaging (ultrasound scan), sonar / depth sounding, cleaning delicate parts, detecting cracks in metal, breaking kidney stones, distance sensors.
  • Only name applications in (c) — no explanation needed for 2 marks.
  • 第 (a) 问：超声（ultrasound）就是频率高于人耳可听上限的声音，即高于 20 000 Hz（20 kHz）。
  • 第 (b) 问：不会听到。这个频率超出人耳可听范围，所以我们听不见（但极强的超声仍可能伤害听觉）。
  • 第 (c) 问其他应用：医学超声检查、声纳测距/测深、清洗精密零件、探测金属裂纹、击碎肾结石、超声测距传感器。
  • 第 (c) 问只需写出应用名称，不用解释。
pest repeller 驱虫器ultrasonic 超声的application 应用affect 影响
Suggested answer 参考答案

Solutions

Marks

(a) Above 20 kHz

1A

(b) No,

1A

this is because human beings cannot hear ultrasound.

1A

(c) Any two of the following:

2 × 1A

Ultrasound cleaner / Ultrasound flaw detector / Sonar / Ultrasound scanner

22Ch.7 第7章Short Q 短题目Average 中等4 marksPHY13072122

A fast attack submarine USS Salt Lake City (SSN 716) is shown in Figure a. It detects the distance of an enemy ship by finding the time difference between a ‘ping’ sound it sends out and the echo it receives (Fig b). The speed of sound in water is 1400 m s−1.

Fig a

Fig b

(a) An echo is received 1.5 s after a ‘ping’ sound is emitted. What is the distance of the enemy ship from the submarine? (2 marks)

(b) Suppose the frequency of the ‘ping’ signal is 2000 Hz. What is the wavelength of the ‘ping’ signal in water? (2 marks)

Chinese translation · 中文对照

图 a 显示一艘快速攻击潜艇 USS Salt Lake City（SSN 716）。它通过找出自己发出的“ping”声与收到的回声之间的时间差，来测定一艘敌舰的距离（图 b）。水中声速为 1400 m s⁻¹。

（见图）图 a

（见图）图 b

(a) 发出“ping”声 1.5 s 后收到回声。敌舰与潜艇相距多远？（2 分）

(b) 假设“ping”信号的频率为 2000 Hz。这个信号在水中的波长是多少？（2 分）

Approach · 思路提示

  • (a) Remember the sound travels there and back, so use the one-way distance: s = v t / 2.
  • s = 1400 × 1.5 / 2 = 1050 m.
  • (b) Use v = f λ, so λ = v / f = 1400 / 2000 = 0.70 m.
  • Keep the units: m for distance, m for wavelength; show the formula first to earn method marks.
  • Do not forget to divide by 2 in (a) — this is the most common mistake.
  • 第 (a) 问：声音是“去 + 回”走了一个来回，所以要用单向距离 s = v t / 2。
  • s = 1400 × 1.5 / 2 = 1050 m。
  • 第 (b) 问：用 v = f λ，所以 λ = v / f = 1400 / 2000 = 0.70 m。
  • 单位要写清楚：距离用 m，波长用 m；先写公式可以拿方法分。
  • (a) 中千万别忘了除以 2，这是最常见的错误。
submarine 潜艇echo 回声wavelength 波长emit 发出
Suggested answer 参考答案

Solutions

Marks

(a) Distance =vt

1M

=× 1400 × 1.5

= 1050 m

1A

(b) By v = fλ,

1M

wavelength === 0.7 m

1A

23Ch.7 第7章Short Q 短题目Average 中等3 marksPHY13072123

Pete is 5 m away from a traffic light. Find the difference in time between seeing the light and hearing the sound.

(Given: speed of sound in air = 340 m s−1

speed of light in air = 3 × 108 m s−1) (3 marks)

Chinese translation · 中文对照

Pete 离一盏交通灯 5 m 远。求他看见灯光与听到声音之间的时间差。

（已知：空气中声速 = 340 m s⁻¹

空气中光速 = 3 × 10⁸ m s⁻¹）（3 分）

Approach · 思路提示

  • Light arrives almost instantly: t_light = 5 / (3 × 10⁸) ≈ 1.7 × 10⁻⁸ s, which is negligible.
  • Sound takes a measurable time: t_sound = 5 / 340 ≈ 0.0147 s.
  • The time difference is therefore Δt = t_sound − t_light ≈ 0.0147 s ≈ 1.5 × 10⁻² s (about 15 ms).
  • Show the light-time calculation and say it is negligible — that earns the comparison mark.
  • Write the answer to 2 significant figures with a unit.
  • 光几乎立刻到达：t光 = 5 /（3 × 10⁸）≈ 1.7 × 10⁻⁸ s，可以忽略。
  • 声音需要一定时间：t声 = 5 / 340 ≈ 0.0147 s。
  • 所以时间差 Δt = t声 − t光 ≈ 0.0147 s ≈ 1.5 × 10⁻² s（约 15 ms）。
  • 要把光的传播时间算出来并说明它可以忽略，这样才有比较分。
  • 答案取两位有效数字并写单位。
traffic light 交通灯difference in time 时间差negligible 可以忽略的
Suggested answer 参考答案

Solutions

Marks

Time for light to reach Pete === 1.67 × 10−8 s

1M

Time for sound to reach Pete === 0.0147 s

1M

Difference in time = 0.0147 − 1.67 × 10−8 ≈ 0.0147 s

1A

(Alternatively, students may state that the time needed for light to travel from the traffic light to Pete is negligible.)

24Ch.7 第7章Short Q 短题目Average 中等4 marksPHY13072124

Baseball is very popular in North America.

(a) In a baseball game, a spectator notices that he hears the sound later than he sees the batter hits the ball. If the delay is 0.8 s, calculate the distance between the spectator and the batter.
(Given: speed of sound in air = 340 m s−1) (2 marks)

(b) If the spectator measures the time interval with a stop-watch and the overall error due to his reaction time is 0.4 s, estimate the maximum and minimum possible distance between the spectator and the batter. (2 marks)

Chinese translation · 中文对照

棒球在北美非常流行。

（见图）

(a) 在一场棒球比赛中，一名观众发现他听到声音的时刻比看到击球员击球的时刻晚。如果延迟是 0.8 s，计算观众与击球员之间的距离。（已知：空气中声速 = 340 m s⁻¹）（2 分）

(b) 如果这名观众用秒表测量这段时间间隔，因他的反应时间造成的总误差为 0.4 s，估计观众与击球员之间距离的最大值和最小值。（2 分）

Approach · 思路提示

  • (a) The delay exists because sound is much slower than light: distance = v t = 340 × 0.8 = 272 m.
  • We ignore the travel time of light, since it is extremely short for a few hundred metres.
  • (b) The measured 0.8 s can be too long or too short by the 0.4 s error: maximum delay = 0.8 + 0.4 = 1.2 s.
  • Minimum delay = 0.8 − 0.4 = 0.4 s.
  • Then maximum distance = 340 × 1.2 = 408 m, minimum distance = 340 × 0.4 = 136 m.
  • Give both answers as a range: about 136 m to 408 m.
  • 第 (a) 问：出现延迟是因为声音比光慢很多，所以距离 = v t = 340 × 0.8 = 272 m。
  • 通常忽略光的传播时间，因为几百米的距离光几乎是瞬间到达。
  • 第 (b) 问：测得的 0.8 s 可能因 0.4 s 的误差而偏大或偏小：最大延迟 = 0.8 + 0.4 = 1.2 s。
  • 最小延迟 = 0.8 − 0.4 = 0.4 s。
  • 所以最大距离 = 340 × 1.2 = 408 m，最小距离 = 340 × 0.4 = 136 m。
  • 最后写成范围：约 136 m 至 408 m。
spectator 观众batter 击球员stop-watch 秒表reaction time 反应时间
Suggested answer 参考答案

Solutions

Marks

(a) Distance = vt

1M

= 340 × 0.8 m

= 272 m

1A

(b) Maximum possible distance = 340 × (0.8 + 0.4) = 408 m

1A

Minimum possible distance = 340 × (0.8 − 0.4) = 136 m

1A

25Ch.7 第7章Short Q 短题目Average 中等4 marksPHY13072125

A student is standing at the edge of a cliff. She is watching a firework which is about 500 m away from her. The time of travel by light is negligible.

(a) If she hears the bang about 1.45 s after the firework is seen, estimate the speed of sound.
(2 marks)

(b) She drops a stone down the cliff. If the time lapse between seeing and hearing the splash by the stone is 0.1 s, estimate the height of the cliff. (2 marks)

Chinese translation · 中文对照

一名学生站在悬崖边。她正在看大约 500 m 外的烟花。光传播所需的时间可以忽略。

(a) 如果她在看到烟花后约 1.45 s 才听到爆炸声，估计声速。（2 分）

(b) 她把一块石头丢下悬崖。如果从看到水花到听到水花声之间的时间间隔是 0.1 s，估计悬崖的高度。（2 分）

Approach · 思路提示

  • (a) The bang travels 500 m in 1.45 s, so v = s / t = 500 / 1.45 ≈ 345 m s⁻¹.
  • Light's travel time is negligible (given), so the whole 1.45 s belongs to the sound.
  • (b) The 0.1 s is the time for the splash sound to travel from the water up to her ear; the light from the splash arrives almost instantly.
  • So height h = v t = 345 × 0.1 ≈ 34.5 m ≈ 35 m (about 3 × 10¹ m).
  • Use the speed you obtained in (a) — don't switch to a different value.
  • Note that the falling time of the stone is NOT part of the measured 0.1 s: the timing starts when she sees the splash.
  • 第 (a) 问：爆炸声 1.45 s 走了 500 m，所以 v = s / t = 500 / 1.45 ≈ 345 m s⁻¹。
  • 题目已说明光传播的时间可忽略，所以这 1.45 s 全是声音所需的时间。
  • 第 (b) 问：这 0.1 s 是水花的声音从水面传到她耳朵所用的时间，看到水花的光几乎瞬间到达。
  • 所以高度 h = v t = 345 × 0.1 ≈ 34.5 m ≈ 35 m。
  • 要用 (a) 中求出的声速，不要换别的数值。
  • 注意石头下落的时间不算在这 0.1 s 里：计时从她看到水花开始。
cliff 悬崖firework 烟花time lapse 时间间隔estimate 估计
Suggested answer 参考答案

Solutions

Marks

(a) Speed =

1M

=

= 345 m s−1

1A

(b) Height = vt

1M

= 345 × 0.1

= 34.5 m

1A

26Ch.7 第7章Short Q 短题目Average 中等5 marksPHY13072126

Two astronauts are on a mission to build a space station.

(a) In space, astronauts cannot hear each other talking directly. Why? (1 mark)

(b) How can they communicate with each other in space? (1 mark)

(c) You are given an electric bell, a glass jar and a vacuum pump. With the aid of a diagram, describe an experiment to illustrate your answer in (a) on the Earth. (3 marks)

Chinese translation · 中文对照

两名航天员正在执行建造空间站的任务。

(a) 在太空中，航天员不能直接听到彼此说话。为什么？（1 分）

(b) 在太空中他们可以怎样互相沟通？（1 分）

(c) 给你一个电铃、一个玻璃钟罩和一台抽气泵。借助图示，描述一个在地球上说明 (a) 中答案的实验。（3 分）

Approach · 思路提示

  • (a) Space is (almost) a vacuum, and sound needs a medium (gas, liquid or solid) to travel — it cannot travel through a vacuum.
  • (b) They communicate by radio waves (electromagnetic waves), which can travel through a vacuum.
  • (c) Put the ringing electric bell inside the glass jar, switch it on, and note that you can hear it clearly.
  • Then use the vacuum pump to pump the air out of the jar: the ringing becomes fainter and finally cannot be heard.
  • Let air back in and the sound returns — this shows air is needed for sound to travel; mark clearly the bell, bell jar, pump and the switch/key in your diagram.
  • 第 (a) 问：太空（几乎）是真空，而声音需要介质（气体、液体或固体）才能传播，不能在真空中传播。
  • 第 (b) 问：用无线电波（电磁波）通信，电磁波可以在真空中传播。
  • 第 (c) 问：把响着的电铃放进玻璃钟罩内，打开开关，可以清楚听到铃声。
  • 然后用抽气泵把罩内的空气抽走：铃声越来越小，最后完全听不到。
  • 再放回空气，铃声又出现——这说明声音传播需要空气；图上要清楚画出电铃、钟罩、抽气泵和开关。
astronaut 航天员vacuum pump 抽气泵glass jar 玻璃钟罩electric bell 电铃
Suggested answer 参考答案

Solutions

Marks

(a) Sound cannot travel through a vacuum.

1A

(b) They can communication with each other by using a radio system.

1A

(c)

(Correct set-up)

1A

(Correct labels)

1A

Set up the apparatuses as shown in the diagram above.

As air is pumped out of the jar, the sound of the electric bell becomes softer and softer.

1A

27Ch.7 第7章Short Q 短题目Average 中等6 marksPHY13072127

The figures below show the traces on a CRO of two notes produced by two musical instruments. The settings of the CRO are the same in the two cases.

(a) Find the frequencies of the two notes. (3 marks)

(b) Which note has a higher pitch? (1 mark)

(c) Compare the sound intensity level of the two notes. Explain briefly. (2 marks)

Chinese translation · 中文对照

下图显示两台乐器发出的两个音在示波器（CRO）上的波形。两次实验的示波器设定相同。

（见图）

(a) 求这两个音各自的频率。（3 分）

(b) 哪个音的音调较高？（1 分）

(c) 比较这两个音的声强级。简要解释。（2 分）

Approach · 思路提示

  • Read the CRO: the period T = (number of horizontal divisions for one complete cycle) × (time represented by one division).
  • Then use f = 1 / T for each note (the same time-base setting is used for both traces).
  • Fewer divisions per cycle means a shorter period and therefore a higher frequency.
  • (b) Pitch is determined by frequency: the note with the higher frequency has the higher pitch.
  • (c) Sound intensity level (loudness) is determined by the amplitude of the wave: measure the height of the trace (peak-to-peak) in divisions.
  • The trace with the larger amplitude is louder; compare the two amplitudes read from the figure.
  • 读示波器：周期 T =（一个完整波形占的水平格数）×（每格代表的时间）。
  • 然后用 f = 1 / T 求每个音的频率（两条波形用的是同一时间基准设定）。
  • 一个周期占的格数越少，周期越短，频率就越高。
  • 第 (b) 问：音调由频率决定，频率高的音调就高。
  • 第 (c) 问：声强级（大小声）由波的振幅决定，看图量波形的高度（用格数表示）。
  • 振幅大的音更响，从图上读出两个振幅比较即可（见图）。
CRO 示波器trace （示波器上的）波形musical instrument 乐器sound intensity level 声强级
Suggested answer 参考答案

Solutions

Marks

(a) By f =,

1M

frequency of note 1 == 600 Hz

1A

frequency of note 2 == 500 Hz

1A

(b) Note 1

1A

(c) Note 1 has a higher sound intensity level

1A

since it has a larger amplitude.

1A

28Ch.7 第7章Short Q 短题目Average 中等5 marksPHY13072128

A Chinese musical instrument plays a note and the waveform of the note is displayed in the following figure. The CRO is adjusted such that one division represents 1 ms for the time base and 0.2 V for the Y-input.

(a) Find the frequency of the note. (3 marks)

(b) Sketch the waveform if the note is produced by a tuning fork with the same pitch and loudness. Assume that the setting of CRO is unchanged. (2 marks)

Chinese translation · 中文对照

一件中国乐器演奏了一个音，这个音的波形显示在下图中。示波器已调好：时基每格代表 1 ms，Y 输入每格代表 0.2 V。

（见图）

(a) 求这个音的频率。（3 分）

(b) 如果这个音改由一支音叉发出（音调和响度相同），画出它的波形。假设示波器的设定不变。（2 分）

Approach · 思路提示

  • (a) Count the horizontal divisions for one complete cycle of the waveform on the figure.
  • Period T = (divisions per cycle) × 1 ms, then f = 1 / T.
  • For example, if one cycle takes 5 divisions, T = 5 ms and f = 1 / 0.005 = 200 Hz — use the actual count from the figure.
  • (b) A tuning fork produces a pure note, i.e. a single frequency, so its waveform is a simple sine curve with no bumps.
  • Same pitch → the same period (same number of divisions per cycle); same loudness → the same amplitude (same height in divisions, 0.2 V per division).
  • Do not change the vertical scale or the number of cycles: only the shape becomes smooth and sinusoidal.
  • 第 (a) 问：先在图上数出一个完整波形占多少水平格。
  • 周期 T =（一个周期的格数）× 1 ms，再用 f = 1 / T。
  • 例如若一个周期占 5 格，T = 5 ms，f = 1 / 0.005 = 200 Hz —— 请用图上实际的格数。
  • 第 (b) 问：音叉发出的是纯音，只有一个频率，所以波形是一条简单的正弦曲线，没有其他起伏。
  • 音调相同 → 周期相同（每周期占的格数相同）；响度相同 → 振幅相同（高度相同，每格 0.2 V）。
  • 纵向刻度和波形个数都不要改，只是形状变成平滑的正弦曲线。
waveform 波形tuning fork 音叉time base 时基division （示波器的）格
Suggested answer 参考答案

Solutions

Marks

(a) Frequency =

1M

=

1M

= 300 Hz

1A

(b)

(Sinusoidal waveform)

1A

(Correct period and amplitude)

1A

1Ch.7 第7章Reading 阅读题Easy 易6 marksPHY13074101

Effects of Noise

Researches have shown that noise poses a variety of health effects on humans.

People who are constantly exposed to high noise levels, especially those who work near airports or along highways and railways, are found to experience increased frequency of headaches, increased reliance on sedatives (鎮靜劑) and sleeping pills, and increased mental hospital admission rates.

Exposure to noise is also associated with a range of physical effects including colds, changes in blood pressure, other cardiovascular (心血管) changes, increased general medical practice attendance, problems with the digestive system and general fatigue.

Besides, there is fairly consistent evidence suggesting that prolonged exposure to noise levels at or above 80 dB can cause deafness. The degree of deafness depends upon the duration of exposure.

(a) From the article, name two effects on mental health when people experience high noise levels. (2 marks)

(b) (i) What is the unit commonly used to measure noise levels? (1 mark)

(ii) Which wave property does the noise level depend on? (1 mark)

(c) Although sound barriers are installed along parts of the highways, pedestrians can still hear some noise with lower frequencies when they walk near these highways. Explain why. (2 marks)

Chinese translation · 中文对照

噪声的影响

研究显示，噪声对人的健康会造成多种影响。

长期（经常）暴露在强噪声中的人，特别是那些在机场附近、或沿公路与铁路工作的人，被发现头痛的次数增加、对镇静剂和安眠药的依赖增加，精神病院的入院率也上升了。

接触噪声还与一系列身体上的影响有关，包括感冒、血压变化、其他心血管方面的变化、到普通诊所看病次数的增加、消化系统的问题，以及整体性的疲劳。

此外，有相当一致的证据表明，长时间暴露在 80 dB 或以上的噪声中会导致耳聋。耳聋的程度取决于暴露时间的长短。

(a) 从文章中举出两个例子，说明人在强噪声下心理健康方面受到的影响。（2 分）

(b) (i) 测量噪声级通常用什么单位？（1 分）

(ii) 噪声级取决于声波的哪一种属性（wave property）？（1 分）

(c) 虽然部分公路沿线已经安装了隔音屏障，但行人在这些公路附近行走时，仍然会听到一些频率较低的噪声。解释原因。（2 分）

Approach · 思路提示

  • For (a), the mental-health clues are all in the first paragraph: more frequent headaches, more reliance on sedatives and sleeping pills, and higher mental hospital admission rates — just pick any two.
  • For (b)(i), noise level is measured in decibel (dB); the article itself mentions 80 dB, which is a hint.
  • For (b)(ii), the loudness of a sound (its noise level) depends on the amplitude of the wave — a bigger amplitude means a louder sound.
  • For (c), think of diffraction: a low-frequency sound has a long wavelength, so it can bend (diffract) around the edges of the sound barrier and still reach the pedestrians.
  • For (c), compare with high-frequency sound: its wavelength is short, so it is blocked by the barrier much more effectively — write this comparison to get full marks.
  • 第 (a) 问：心理健康的线索都在第一段——头痛次数增加、对镇静剂和安眠药的依赖增加、精神病院入院率上升，从中任选两个即可。
  • 第 (b)(i) 问：噪声级的单位是分贝（decibel，dB）；文中提到的 80 dB 就是提示。
  • 第 (b)(ii) 问：声音的响度（噪声级）由波的振幅（amplitude）决定，振幅越大，声音越响。
  • 第 (c) 问：用衍射（diffraction，又叫绕射）来解释——低频声音的波长较长，容易绕过隔音屏障的边缘，所以仍能传到行人耳中。
  • 第 (c) 问：要和频率较高的声音作对比——高频声音波长短，不容易绕射，被屏障挡得更好。把这条对比写上才能拿满分。
sedatives 镇静剂prolonged 长时间的；持久的fatigue 疲劳sound barrier 隔音屏障
Suggested answer 参考答案

Solutions

Marks

(a) Any two of the following:

2 × 1A

Increased frequency of headaches

Increased reliance on sedatives and sleeping pills

Increased mental hospital admission rates

(b) (i) Decibel (dB)

1A

(ii) Amplitude

1A

(c) Noise with low frequencies have large wavelengths

1A

and they can diffract significantly at the edge of sound barriers.

1A

No question matches this filter. 没有符合条件的题目。

Key vocabulary · 重点生词

Words that keep appearing in these two chapters. Learn them first — then the questions start to read themselves. 这两章反复出现的英文词汇，先认词，再读题。

1st-order第一级
2nd-order第二级
absorb吸收
according to按照
accuracy准确度
adjacent相邻的
adjacent slits相邻的缝
advantage优点
aeroplane飞机
affect影响
alternately交替地
altitude高度/海拔
amplitude振幅
angle of incidence入射角
angle of reflection反射角
angular position角位置
angular separation角间距
antenna天线
anti-phase反相
anti-reflection coating抗反射涂层（防反光镀膜）
antiphase反相（相差半个周期）
any media任何介质
apparatus仪器装置
application应用
approximation近似值
arrange排列
ascending order由小到大排列
astronaut航天员
at an angle斜着（以一定角度）
at rest静止
at the highest speed以最大速率（运动）
audible可听见的
audible frequency range可听频率范围
audible range可听范围
audience听众, 观众
auto-focus自动对焦
barrier屏障/障碍物
batter击球员
beacon烽火；烽火台
beam光束
bend around绕过⋯⋯弯曲
bends偏折
blast furnace高炉
block阻挡
body temperature体温
boundary边界
briefly简要地
bright fringe亮纹
brightness亮度
broadcast广播
broadcasting frequency广播频率
broadcasting wavelength广播波长
by different amounts程度不同
candle fire烛火
candle flame烛焰
cardboard tube硬纸筒
carry energy携带能量
central bright fringe中央亮纹
central fringe中央条纹
centre of compression密部中心
centre of the interference pattern干涉图案的中心
checking luggage检查行李
clap his hands拍手
cleaning清洁
cliff悬崖
coarse grating粗光栅
coherent相干的
coherent light相干光
coherent sources相干波源
coil（弹簧的）线圈
coincides with与……重合
common共同的
comparable可相比的
comparable to与……差不多
comparable with与……差不多
complete oscillation一次完整振动
compression密部（压缩区）
cone锥体（喇叭单元）
constant恒定的
constant velocity恒定速度
construction site建筑工地
constructive interference相长干涉
continuous exposure长期接触
continuous spectra连续光谱
continuous spectrum连续光谱
corner角落
corresponding对应的
counterfeit banknotes伪钞
crest and trough波峰与波谷
critical angle临界角
CRO示波器（阴极射线示波器）
CRO reading示波器读数
CRO settings示波器设置
dark band暗带
dark fringe暗纹
delicate精致的、易损的
demonstrate演示
descending order由大到小排列
destructive interference相消干涉
detect探测
developed冲洗（底片）
difference in time时间差
diffract衍射
diffracted衍射
diffraction衍射
diffraction grating衍射光栅
diffraction pattern衍射图案
directed normally垂直照射
direction of motion运动方向
direction of travel传播方向
displacement位移
displacement–distance graph位移–距离图
displacement–time graph位移–时间图
displacement−distance graph位移—距离图像
distance距离
disturbance扰动
disturbed受干扰
division（示波器的）格
doorway门口
double slit双缝
doubled加倍
dull black暗黑色的
duodenum十二指肠
ear pain耳朵疼痛
echo回声
electric bell电铃
electromagnetic spectrum电磁波谱
electromagnetic wave电磁波
electromagnetic waves电磁波
emerge射出；出现
emergent beam出射光束
emit发出
endoscope内窥镜
equal spacing相等的间距
equidistant等距的
equilibrium平衡
equilibrium position平衡位置
estimate估计
evidence证据
expected results预期结果
explain解释
exposure to暴露于
face shield面罩
facing面对、朝向
fatigue疲劳
fetus胎儿
filament bulb钨丝灯泡
filament lamp钨丝灯
fine grating细光栅
firework烟花
first-order第一级
flaw裂缝、缺陷
floor plan平面图
foetus胎儿
formation形成
frequency频率
frequency range频率范围
fringe条纹
fringe pattern条纹图样
fringe separation条纹间距
fringe width条纹宽度
gamma rays伽马射线
gap缝隙
given information已知资料
glass jar玻璃钟罩
graph paper方格纸
grating光栅
grating spacing光栅缝距
harmful有害的
harmful effects有害影响
hearing damage听力损伤
hi-fi高保真音响
highest order最高级数
hit敲击
identical完全相同的
identical settings相同的设置
image像
immersed浸在……中
in phase同相
incident normally垂直入射
incorrect不正确
indigo靛蓝色
infra-red radiation红外线
infra-red radiations红外辐射
instant瞬间/某一时刻
intensity强度
interface交界面
interfere干涉
interference干涉
interference pattern干涉图样
laser pointer激光笔
lightning闪电
lines per cm每厘米的线数
lines per mm每毫米刻线数
liquid液体
listener听者
longitudinal wave纵波
loudest最响的
loudness响度
loudspeaker扬声器
low-pitch低音调的
lowest pitch最低音调
magnitude大小（绝对值）
major主要的
match与……相符
maxima极大值
maximum order最高级次
maximum speed最大速率
measurable可测量的
measure措施
mechanical wave机械波
media介质（medium 的复数）
medical imaging医学成像
medical treatments医学治疗
medium介质
metal plate金属板
metre rule米尺
microphone麦克风
microwave微波
microwaves微波
mini hi-fi小型音响
momentarily at rest瞬间静止
monochromatic单色的
monochromatic light单色光
motion sensor运动传感器
musical instrument乐器
musical note音符
must be correct一定正确
narrow窄的
narrow slit窄缝
negligible可忽略的
noise噪音
noisy environment嘈杂的环境
None of above以上都不对
normal incidence垂直入射
normal to the grating光栅的法线
noticeable显著的、明显的
observable可观察的
observe观察到
obtained求出
on the same side在同一侧
order级数
order of fringe条纹级数
order of magnitude数量级
oscillation振动
out of phase不同相
output voltage输出电压
output volume输出音量
overground地面的
overlap重叠
parallel平行的
parallel to与⋯⋯平行
partial reflector半反射片
particle粒子
particle theory微粒说
passbook存折
path difference光程差
pattern图样
peak峰
per unit length每单位长度
period周期
permanent damage永久损伤
perpendicular垂直的
perpendicular bisector中垂线
pest repeller驱虫器
phase difference相位差
phenomenon现象
photographic film摄影底片
pitch音调
plane mirror平面镜
plane transmission grating平面透射光栅
pluck拨（弦）
Pluto冥王星
polarisation偏振
positive direction正方向
power功率
precaution注意事项
prevented from reaching被阻挡而未能到达
principal maximum主极大
prism棱镜
prolonged长时间的；持久的
prominent明显的
propagating传播
propeller螺旋桨
properties性质
property性质
pulse脉冲
quality音品（音色）
quantity物理量
radar雷达
radio shadow电波阴影
radio waves无线电波
radiotherapy放射治疗
rarefaction疏部（稀疏区）
ratio比率
ray光线
ray-box lamp光线盒
reaction time反应时间
receiver接收器
reception接收
red light红光
reduce减少
reflect反射
reflected反射回来的
reflection反射
reflective barrier反射屏障
refracted折射
refraction折射
refractive index折射率
regular reflection规则反射
relationship关系
remain unchanged保持不变
replace更换
replaced by用……代替
result结果
reverberation混响
rim边缘
rough surface粗糙表面
ruling刻线
safety precaution安全措施
same scale同一比例
satellite communication卫星通信
satellite telecommunications卫星通讯
scanning扫描
seabed海底
second order bright fringe第二级亮纹
security check安全检查
sedatives镇静剂
seldom noticed很少被察觉
sensor传感器
separation间距
shape形状
sharp清晰的
shield屏蔽、遮挡
shoal（鱼）群
shoal of fish鱼群
shortest wavelength最短波长
signal信号
signal generator信号发生器
signature签名
significantly明显地
single slit单缝
sketch画出（草图）
slinky spring（玩具）弹簧
slit缝
slit separation缝距
slope斜率
soap film肥皂薄膜
solar cells太阳能电池
solid固体
sonar声呐
sonar system声呐系统
sound barrier隔音屏障
sound intensity level声强级
source来源
speaker cone扬声器纸盆
spectator观众
spectrum光谱
speed of light光速
speed of sound声速
spreads散开
standing wave驻波
statement陈述
steel rod钢棒
sterilization消毒、杀菌
sterilizing消毒
stop-watch秒表
stressed有压力的
string弦
struck被敲击
submarine潜艇
subtended by所张的（角）
suggest提出
suntan lotion防晒油
switch off关掉
telecommunications电信
the same person同一个人
third order dark fringe第 3 级暗纹
thunder雷声
thunderstorm雷暴
time base时基
time lapse时间间隔
total internal reflection全反射
trace（示波器上的）波形轨迹
traffic light交通灯
translucent半透明的
translucent screen半透明屏幕
transmit传送
transmit through通过⋯⋯传播
transmitted被传播
transmitted signal发射信号
transmitter发射器
transparent material透明材料
transverse横的
transverse wave横波
transverse waves横波
travel传播、走
travelling wave行进波
tuning fork音叉
tweeter高音单元
ultra high frequency超高频
ultra-violet radiation紫外线
ultrasonic超声波的
ultrasound超声波
ultraviolet radiation紫外线
unchanged保持不变
uniformly distributed均匀分布
unmanned aerial vehicle无人驾驶飞行器
urban area市区
vacuum真空
vacuum pump真空泵
varies with time随时间变化
vary有差异；变化
vary from person to person因人而异
vibrate振动
vibration振动
vibration of particles粒子的振动
violet紫色
violet end紫端
violet light紫光
visible light可见光
walk across横穿走过
wave phenomenon波动现象
wave property波动性质
wave theory波动说
waveform波形
wavefront波前
wavelength波长
wider更宽的
window窗户
woofer低音单元
X-raysX 射线
yellow filter黄色滤光片