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Worksheet

Worked examplesPractical 1: Polarisation of light

Practical 2: Speed of sound

End-of-chapter test

Marking scheme: Worksheet

Marking scheme: End-of-chapter test

Chapter 19

Waves

OCR (A) specifications: 5.3.2a,b,c,d,e,f,g,h,i,j

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172 Cambridge University Press 2005 19 Waves

Worksheetspeed of light in free space (vacuum) c= 3.0 108 ms1

speed of sound in air = 340ms1

Intermediate level1 For a progressive wave, define the following terms:

a amplitude; [1]

b wavelength; [1]

c frequency. [1]

2 Calculate the frequency of the following waves:

a red light of wavelength 6.5 107 m emitted from a light-emitting diode; [2]

b ultrasound of wavelength 7.0 mm emitted by a bat. [2]

3 In a water tank, a dipper oscillating at a frequency of 30Hz produces surface waterwaves of wavelength 2.5 cm.

a Calculate the speed of the water waves. [2]

b Determine the wavelength of the waves when the frequency of the dipper

is doubled. [2]

4 a Explain what is meant by plane polarised light. [1]

b Name the type of waves that can be polarised. [1]

Higher level5 Displacement against time graphs for two wavesAand B of the same frequency

are shown below.

a Determine the period and the frequency of the waves. [2]

b What is the phase difference between wavesAand B (in degrees)? [2]

A

B

0

Time (ms)1 96 10

+

2 3 4 5 7 8

Displacement

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19 Waves Cambridge University Press 2005 173

6 The diagram shows water waves travelling fromdeeper to shallower water.

a The wavefronts change direction when entering

the shallower water. What do we call the change

in direction of a wave at a boundary? [1]

b Describe what happens to the water waves

in shallower water in terms of wave speed,frequency and wavelength. [3]

7 An oscilloscope has its time base and Y sensitivity (Y gain) set on 0.5mscm1 and0.5Vcm1 respectively. A person whistles into a microphone connected to the

oscilloscope. The trace displayed on the oscilloscope screen is shown below.

a Determine the frequency of the soundwave. [2]

b Calculate the wavelength of the sound

produced by the whistle. [2]

c Describe how the oscilloscope trace would

change for a louder whistle of half the

frequency of a. [2]

ExtensionIn OCR Module 3, you do not have to learn how the intensity of light varies with

distance from a source. However, in the Synoptic Paperyou can be given information

that you have to analyse. Regard this question as preparation for such a paper.

8 The intensity of a wave is the power transmitted per unit area perpendicular to thewave. Intensity is measured in watts per square metre, W m2.

a For a point source of light, explain why the intensity Iat a distance raway from

the source obeys an inverse square law with distance. That is:

1I

r2 [2]

b The intensity of the solar radiation reaching the upper parts of our atmosphere

is about 1.4kWm2. The Sun is 1.5 1011m from the Earth.

i Calculate the intensity of solar radiation at the planet Neptune, which is

4.5 1012m from the Sun. [2]

ii Estimate the number of photons of visible light from the Sun arriving

per square metre at Neptune each second. [3]

(Average wavelength of visible light = 550nm; Planck constant h = 6.63 1034J s.)

deep

shallow

0.5

0.5

V

ms

Total: 34

Score: %

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Worked examplesExample 1Visible light has wavelength in the range 400 nm to 700 nm. Calculate the maximum

frequency of visible light.

The speed of light is constant. Since v =f and v is a constant, we have:

1f

The frequency is a maximum for the shortest wavelength of visible light.

v 3.0 108f=

=

400 109= 7.5 1014Hz

Example 2The diagram shows an oscilloscope trace when the time base is set at 5.0 ms cm1 and

the Y sensitivity is set at 2.0 V cm1. Determine the frequency and amplitude of the

signal displayed on the oscilloscope.

Frequency

The horizontal distance between

two neighbouring peaks = 4.0 cm.

period =distance

time base

between peaks setting

T= 4.0cm 5.0 ms cm1 = 20ms

The frequencyfof the signal is

related to the period Tby:

1f=

T

Hence:

1f=

20 103= 50Hz

Amplitude

amplitude = distance from zero to maximum height Y sensitivity

amplitude = 3.0 cm 2.0Vcm1 = 6.0V

TipYou can applyf=

vto each given wavelength and from the two answers quote the

greater frequency. However, there is a great deal of physics in remembering that

the frequency is inversely proportional to the wavelength.

2.0 V

5.0 ms

1nm = 109m

It is very important to convert the

time into seconds. If the period is left

in milliseconds, then the frequency

will be in kilohertz (kHz).

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Practical 1Polarisation of lightSafety

There are not likely to be any major hazards in carrying out this experiment. However,teachers and technicians should always refer to the departmental risk assessment before

carrying out any practical work.

Apparatus

two polarising filters (Polaroid) glass block

table lamp or ray box LCD display of a calculator

water with a few drops of milk small strip of plastic

in a plastic container protractor

Introduction

You will carry out a series of experiments on polarisation of light. At the end of the

experiments you will appreciate that light can be polarised either by reflection off shinysurfaces or when it is transmitted through a polarising filter.

Procedure

Details of polarisation of light using polarising filters is given on pages 163 and 164 of

Physics 1. The diagram below shows two further arrangements that may be used to show

the polarisation of light.

1 Look through one polarising filter at the light from the table lamp. Rotate the filter.Does this have any effect on the intensity of the light? Is the light from the lamp

plane polarised or unpolarised?

2 Now look at the light from the table lamp using two polarising filters (see figure19.10 on page 164 ofPhysics 1). Hold one filter still and rotate the other filter.

Measure the angle of rotation of this filter such that the intensity of light changes

from a maximum to a minimum. This angle should be 90. What can you say about

the transmitted light from the first filter?

3 Use one filter to look at the reflected light from the glass block. Try different anglesof reflection and rotate the filter. Is the reflected light from the glass block polarised?

4 Use one filter to look at the scattered light from the cloudy water at differentangles. Is there anything special for the light reflected through 90?

5 Finally, use a polarising filter to look at the transmitted light through a stressedlength of plastic strip and the light from an LCD display of a calculator. What do

you observe?

Guidance for teachers

Candidates do not have to know about polarisation of scattered or reflected light, but

the experiments are quick and highlight the practical applications of polarising filters.

light source

polarising filter glass block

light source

polarising filter

cloudy water in a container

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Practical 2Speed of soundSafety

There are not likely to be any major hazards in carrying out this experiment. However,teachers and technicians should always refer to the departmental risk assessment before

carrying out any practical work.

Apparatus

dual-beam oscilloscope two microphones

loudspeaker metre rule

signal generator

Introduction

An important item in this experiment is the dual-beam oscilloscope, which is used to

determine the time taken by sound to travel a certain distance. This information is used

to calculate the speed of sound in air.

Procedure

The experiment is described in detail on page 170 ofPhysics 1. The diagram below shows

the arrangement.

1 Set the signal generator to 1.0kHz.

2 Place the microphonesAand B at the same distance (about 30cm) from theloudspeaker.

3 Move microphone B a distance d further awayfrom the loudspeaker. One of the oscilloscope

traces is shifted by a time t. Use the time base

setting to determine the time t.

4 Obtain a range of values for d and t.

5 Record your results in a table.

6 Plot a graph of distance d against time t. Draw astraight line of best fit.

7 Determine the gradient of the line. This is equalto the speed of sound in air.

8 What is the uncertainty in your value for the speed of sound in air? How does yourvalue compare with the standard value of 340 ms1.

Guidance for teachers

As an extension exercise, students can investigate whether the frequency of the sound

has any effect on its speed in air.

signal generator

sound

loudspeaker

1

A

B

d

Y1

Y2

oscilloscopekHz

t

time

trace from A

trace from B

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End-of-chapter testAnswer all questions.

speed of sound in air = 340ms1

1 Explain what is meant by a transverse wave. [1]2 The flapping wings of a bee create a buzzing noise of frequency 250Hz.

a Determine the number of times the wings of the bee flap in an interval of

one minute. [2]

b Calculate the wavelength of sound due to this buzzing noise. [2]

3 The diagram shows the displacementagainst time graph for a wave on a rope.

State the phase difference between points:

a Aand B; [1]

b Aand D; [1]

c C and D. [1]

4 A person is listening to a radio station on a portable radio. The reception is excellentwhen the aerial of the radio is vertical but virtually no signal is registered when the

aerial is turned through an angle of 90. Explain this observation. [2]

5 A microphone is attached to the end of a metal rod. The microphone is connected toan oscilloscope. When the metal rod is hit with a hammer, a sound wave travels along

the length of the metal rod. The output from the microphone is shown below.

The sound within the rod has a wavelength of 2.5 m. The time base and Y sensitivity

(Y gain) settings for the oscilloscope are shown next to the trace.

Calculate:

a the frequency of the sound in the rod; [2]

b the speed of the sound in the metal. [2]

A

B

C

D

0

Time

+

Displacement

time base = 0.1

Y sensitivity = 0.2 V

ms cm1

cm1

Total: 14

Score: %

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Marking schemeWorksheet1 a Amplitude = maximum displacement from equilibrium position. [1]

b Wavelength = separation between two adjacent peaks (or troughs). [1]c Frequency = number of oscillations per unit time. [1]

2 a v =f; f=v

=3.0 108

[1] 6.5 107

f 4.6 1014Hz [1]

v 340[1]b v =f; f=

=

0.007

f 4.9 104 Hz [1]

3 a v =f; v = 30 0.025 [1]

v = 0.75ms1

[1]

b The speed of the wave is constant, therefore 1

. [1]f

The wavelength is halved because the frequency is doubled, so is 1.25 cm. [1]

4 a Plane polarised light has vibrations in just one plane. [1]

b Unlike longitudinal waves, all transverse waves (light, etc.) can be polarised. [1]

15 a f=

T; T= 4.0ms [1]

1f=

4.0

10

3 = 250 Hz [1]

b The adjacent peaks have a time difference of T/4; [1]

therefore the phase difference is 90. [1]

6 a Refraction [1]

b In shallow water:

speed of the wave decreases; [1]

the frequencyremains the same; [1]

thewavelength decreases (since v =fandf= constant, v). [1]

7 a T= 0.5 mscm1 3.0cm = 1.5ms [1]

1 1f=T

=1.5 103

670Hz [1]

b The speed of sound is given by v =f. Hence:

=v

=340

[1]f 670

= 0.51m (51cm) [1]

c The amplitude of the signal would increase. [1]

The peaks would be twice as far apart because the period increases to 3.0 ms.

[1]

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8 a The power spreads out uniformly over the surface of a sphere of radius r. [1]

PThe power per unit area = intensity =

4r2, wherePis the total power. [1]

SincePis a constant, the intensityIobeys an inverse square law with distance.

1b i SinceI

r

2 , we haveIr2 = constant.

Therefore 1.4 103 (1.5 1011)2 =INeptune (4.5 1012)2 [1]

INeptune 1.56 Wm2 [1]

ii Energy of photon,E =hc

=6.63 1034 3.0 108

550 109

E = 3.62 1019J [1]

1.56Number of photons per m2 =

3.62 1019[1]

= 4.3 1018 m2 1018 m2 [1]

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Marking schemeEnd-of-chapter test1 The vibrations are at 90 to the direction of travel of the wave. [1]

2 a The wings f lap 250 times every second. [1]Therefore, in one minute (60 s) there are 250 60 = 15000 flaps. [1]

b v =f; =v

=340

[1]; = 1.36m 1.4m [1]f 250

3 a 90 [1]; b 360 (or 0) [1]; c 180 [1]

4 The radio waves are plane polarised in the vertical plane. [1]

The vibrations have no horizontal component. This explains why a weak signal is

registered when the aerial is rotated through 90. [1]

5 a Period = 0.1mscm1 5.0cm = 0.5 ms [1]

1 1f=T

=5.0 104

= 2.0 103 Hz [1]

b v =f [1]; v = 2000 2.5 = 5000ms1 [1]

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