WAVES

AQA GCSE Physics · Lessons 1–11
⬇ Student Booklet ⬇ Teacher Booklet

Key Terms · Waves

All Terms

01
Wave
DefinitionTransfers energy from one place to another without transferring matter.
02
Oscillation (vibration)
DefinitionA repeated back-and-forth movement about a fixed point. All waves are caused by something oscillating.
03
Transverse wave
DefinitionA wave whose vibrations are at right angles (perpendicular) to the direction of energy transfer, e.g. water waves, all electromagnetic waves and S-waves.
04
Longitudinal wave
DefinitionA wave whose vibrations are parallel to the direction of energy transfer, e.g. sound, ultrasound and P-waves.
05
Compression
DefinitionA region of a longitudinal wave where the particles are close together.
06
Rarefaction
DefinitionA region of a longitudinal wave where the particles are spread out.
07
Amplitude
DefinitionThe maximum displacement of a point on a wave away from its undisturbed position. For sound, a bigger amplitude means a louder sound.
08
Wavelength (λ)
DefinitionThe distance from a point on one wave to the equivalent point on the adjacent wave, e.g. crest to crest. Measured in metres (m).
09
Frequency (f)
DefinitionThe number of waves passing a point each second. Measured in hertz (Hz). For sound, a higher frequency means a higher pitch.
10
Period (T)
DefinitionThe time taken for one complete wave to pass a point. Measured in seconds (s). T = 1 / f.
11
Wave speed (v)
DefinitionHow far a wave travels each second, in m/s. Found from the wave equation v = fλ.
12
Hertz (Hz)
DefinitionThe unit of frequency. 1 Hz means one complete wave passes a point every second.
13
Standing wave
DefinitionA pattern of fixed nodes and antinodes formed on a stretched string by a vibration generator. The distance between alternate nodes is one wavelength.
14
Normal
DefinitionAn imaginary line at 90° to a surface at the point where a ray hits it. Angles of incidence, reflection and refraction are all measured from the normal.
15
Reflection
DefinitionA wave bouncing off a surface. The angle of incidence equals the angle of reflection.
16
Specular reflection
DefinitionReflection from a smooth surface in a single direction, giving a clear image (e.g. a mirror).
17
Diffuse reflection
DefinitionReflection from a rough surface: the light is scattered in all directions, so no clear image is formed.
18
Transmission
DefinitionWhen a wave passes through a material at a boundary. Whether a wave is absorbed, transmitted, reflected or refracted depends on its wavelength and the material.
19
Absorption
DefinitionWhen a wave is taken in by a material at a boundary and its energy is transferred to the material.
20
Transparent
DefinitionA material that transmits light with no scattering, so you can see through it clearly.
21
Translucent
DefinitionA material that transmits light but scatters it, so you cannot see clearly through it.
22
Refraction
DefinitionThe change in direction (and speed) of a wave as it passes from one material into another. Light bends towards the normal when it slows down.
23
Electromagnetic spectrum
DefinitionA family of seven transverse waves that all travel at the speed of light in a vacuum: radio, microwave, infrared, visible, ultraviolet, X-ray and gamma.
24
Ionising radiation
DefinitionRadiation with enough energy to knock electrons off atoms, forming ions. X-rays and gamma rays are ionising and can damage cells and cause cancer.
25
Infrared
DefinitionThe part of the EM spectrum that transfers heat. All objects emit and absorb infrared; matt black surfaces are the best absorbers and emitters.
26
Perfect black body
DefinitionAn object that absorbs all the radiation that hits it (reflecting and transmitting none). It is also the best possible emitter.
27
Ultrasound
DefinitionSound with a frequency above the upper limit of human hearing (above 20 000 Hz). Used for pre-natal scans and finding flaws in materials.
28
Echo
DefinitionA reflected sound or ultrasound pulse. Timing the echo (and halving the time) gives the distance to a boundary.
29
Echo sounding (sonar)
DefinitionUsing the time between a sound pulse and its echo to find a distance, e.g. the depth of the sea below a ship or a bat finding an insect.
30
Radiation dose
DefinitionA measure of the risk of harm from exposure to ionising radiation, measured in sieverts (Sv). 1000 millisieverts (mSv) = 1 Sv.
31
Leslie cube
DefinitionA hollow metal cube with four different faces (matt black, shiny black, matt white, shiny silver), filled with hot water, used to compare how much infrared each surface emits.
32
P-wave
DefinitionA longitudinal seismic wave that travels through solids and liquids.
33
S-wave
DefinitionA transverse seismic wave that travels through solids only. The S-wave shadow zone is evidence for a liquid outer core.
34
Convex (converging) lens
DefinitionA lens that bulges outwards and brings parallel rays together to the principal focus.
35
Concave (diverging) lens
DefinitionA lens that caves inwards and spreads parallel rays out. It always produces a virtual, upright, diminished image.
36
Principal focus (F)
DefinitionThe point where rays parallel to the axis are brought together by a convex lens (or appear to come from for a concave lens).
37
Focal length
DefinitionThe distance from the centre of the lens to the principal focus.
38
Real image
DefinitionAn image that can be projected onto a screen.
39
Virtual image
DefinitionAn image that cannot be projected onto a screen, e.g. the image in a magnifying glass.
40
Magnification
Definitionimage height ÷ object height. It has no units because it is a ratio.
41
Colour filter
DefinitionTransmits (lets through) light of its own colour and absorbs all the other colours.
42
Opaque object
DefinitionAn object you cannot see through. Its colour is the colour of light it reflects; it absorbs the other colours.

Lesson 1 · Transverse and Longitudinal Waves

Do Now

Q1
What do you think a wave carries from place to place?
Model AnswerEnergy (not matter).
Q2
Give one example of a wave.
Model AnswerAny sensible example, e.g. sound, light, water waves, a wave on a rope.
Q3
What is a vibration?
Model AnswerA repeated back-and-forth movement (an oscillation) about a fixed point.

Part 1 · What waves do

Read the passage, then answer the questions.

A wave transfers energy from one place to another without transferring matter.

Waves are caused by something vibrating (oscillating).

When a wave travels through a material, the particles vibrate about a fixed point but do not travel with the wave.

Key ideaWaves transfer energy and information, not the material itself.

Questions · What waves do

Q1
What do all waves transfer? (1 mark)
Model AnswerEnergy (from one place to another).
Q2
Do waves transfer matter? Explain. (2 marks)
Model AnswerNo: the particles vibrate about a fixed point; only energy is transferred, not the material.
Q3
What causes a wave? (1 mark)
Model AnswerA vibration (oscillation).
Q4
A cork floats on water as a wave passes. Describe and explain its motion. (2 marks)
Model AnswerIt bobs up and down about a fixed point but does not move along with the wave, because the wave transfers energy, not water.
Q5
Give two examples of things waves can transfer. (2 marks)
Model AnswerEnergy and information.
Q6
Explain why a wave can travel a long way but the water itself stays roughly in place. (2 marks)
Model AnswerEach particle only vibrates about its position and passes energy to the next; the energy moves along while the particles stay put.

Part 2 · Transverse and longitudinal

Read the passage, then answer the questions.

In a transverse wave, the vibrations are at right angles (perpendicular) to the direction of energy transfer. Example: water waves and all EM waves.

In a longitudinal wave, the vibrations are parallel to the direction of energy transfer. Example: sound waves.

In a longitudinal wave, regions where particles are close together are compressions; regions where they are spread out are rarefactions.

Fig 1.1: A transverse wave and a longitudinal wave
Fig 1.1: A transverse wave and a longitudinal wave

Questions · Transverse and longitudinal

Q7
In a transverse wave, how do the vibrations compare with the direction of energy transfer? (1 mark)
Model AnswerThey are at right angles (perpendicular) to it.
Q8
In a longitudinal wave, how do the vibrations compare with the direction of energy transfer? (1 mark)
Model AnswerThey are parallel to it.
Q9
Name the two regions in a longitudinal wave. (2 marks)
Model AnswerCompressions (particles close together) and rarefactions (particles spread out).
Q10
Give one example of (a) a transverse wave (b) a longitudinal wave. (2 marks)
Model Answera) water wave / any EM wave  b) sound wave.
Q11
State whether each is transverse or longitudinal: light, sound, water waves. (3 marks)
Model AnswerLight: transverse. Sound: longitudinal. Water waves: transverse.
Q12
Describe what happens to the air particles when a sound wave passes through air. (1 mark)
Model AnswerThey vibrate backwards and forwards (parallel to the wave), creating compressions and rarefactions.

Part 3 · Examples of longitudinal and transverse waves

Read the passage, then answer the questions.

Both types transfer energy and have the same wave properties (wavelength, frequency, amplitude).

The key difference is the direction of vibration relative to the wave’s travel.

You need to be able to recall examples of each type of wave for AQA GCSE Physics:

Transverse wavesLongitudinal waves
▸ All electromagnetic waves: radio, microwave, infrared, visible light, ultraviolet, X-ray, gamma ray
▸ Ripples/waves on the surface of water
▸ S-waves (secondary seismic waves)
▸ A rope or slinky spring shaken side to side
▸ Sound waves (in air, liquids and solids), including ultrasound
▸ P-waves (primary seismic waves)
▸ A slinky spring pushed and pulled along its length, producing compressions and rarefactions

Questions · Examples of longitudinal and transverse waves

Q13
State one thing transverse and longitudinal waves have in common. (1 mark)
Model AnswerBoth transfer energy (without transferring matter).
Q14
State the key difference between transverse and longitudinal waves. (2 marks)
Model AnswerThe direction of vibration: perpendicular to travel (transverse) vs parallel to travel (longitudinal).
Q15
A slinky spring is pushed and pulled along its length. State which type of wave is produced and explain. (2 marks)
Model AnswerLongitudinal: the coils vibrate back and forth parallel to the wave’s direction, making compressions and rarefactions.
Q16
A slinky spring is shaken side to side. State which type of wave is produced and explain. (2 marks)
Model AnswerTransverse: the coils vibrate at right angles to the direction the wave travels.
Q17
Explain why sound cannot travel through a vacuum but light can. (3 marks)
Model AnswerSound is longitudinal and needs particles to vibrate, so it cannot travel through a vacuum; light is an EM (transverse) wave that does not need a medium, so it can.

Exam-style questions

Q1
(a) Complete the sentence. Tick one box. (1 mark)The oscillations producing a light wave are ____________ to the direction of energy transfer by the light wave.☐ parallel☐ perpendicular(b) Describe the difference between a longitudinal wave and a transverse wave. (2 marks)(c) Describe one piece of evidence that shows that when a sound wave travels through the air it is the wave, and not the air itself, that travels. (1 mark)(d) Sound waves are longitudinal. Describe a longitudinal sound wave. (2 marks)(e) Radio waves are electromagnetic waves. Describe how radio waves are different from sound waves. (4 marks)Total: 10 marks
Mark Scheme(a) perpendicular.(b) In a longitudinal wave the oscillations (vibrations) are parallel to the direction of energy transfer; in a transverse wave the oscillations are perpendicular (at right angles) to the direction of energy transfer.(c) Any one: a candle flame in front of a loudspeaker flickers but is not blown away; you can hear a sound without feeling any wind / movement of air from the source; smoke or dust near a loudspeaker vibrates back and forth but does not travel away.(d) The air particles oscillate (vibrate) backwards and forwards parallel to the direction in which the wave transfers energy, producing regions of compression (particles close together) and rarefaction (particles spread out).(e) Any four: radio waves are transverse, sound waves are longitudinal; radio waves can travel through a vacuum, sound needs a medium (particles); radio waves travel much faster (3.0 × 10⁸ m/s in a vacuum) than sound (about 330 m/s in air); radio waves are electromagnetic (oscillating electric and magnetic fields), sound is a mechanical vibration of particles; radio waves have much higher frequencies / shorter wavelengths than sound.

Lesson 2 · Wave Properties and Period

Do Now

Q1
What is the highest point of a wave called?
Model AnswerThe crest (peak).
Q2
What is the unit of time?
Model AnswerThe second (s).
Q3
What does frequency measure?
Model AnswerHow many waves (vibrations) pass a point each second.

Part 1 · Describing a wave

Read the passage, then answer the questions.

Amplitude is the maximum displacement of a point on a wave away from its undisturbed position.

Wavelength (λ) is the distance from a point on one wave to the equivalent point on the adjacent wave (e.g. crest to crest). It is measured in metres (m).

For a sound wave, a bigger amplitude means a louder sound.

Fig 2.1: Amplitude and wavelength of a wave (displacement–distance graph)
Fig 2.1: Amplitude and wavelength of a wave (displacement–distance graph)

Questions · Describing a wave

Q1
Define amplitude. (1 mark)
Model AnswerThe maximum displacement of a point on a wave away from its undisturbed position.
Q2
Define wavelength and give its unit. (2 marks)
Model AnswerThe distance from a point on one wave to the equivalent point on the adjacent wave (e.g. crest to crest); measured in metres (m).
Q3
On a wave diagram, between which two points would you measure the wavelength? (1 mark)
Model AnswerFrom one crest to the next crest, or from one trough to the next trough.
Q4
Two waves have the same wavelength but different amplitudes. State which wave quantity is the same and which is different. (2 marks)
Model AnswerSame: wavelength. Different: amplitude.
Q5
Explain how the amplitude of a sound wave relates to how loud the sound is. (1 mark)
Model AnswerA larger amplitude means a louder sound.
Q6
State the labels used on the two axes of a graph used to show the wavelength and amplitude of a wave. (1 mark)
Model AnswerDisplacement (y-axis) against distance (x-axis).
Q7
Wave X has an amplitude of 2 cm and wave Y (of the same type) has an amplitude of 6 cm. Which wave is transferring more energy, and how would this appear on a displacement-distance graph? (2 marks)
Model AnswerWave Y: a bigger amplitude means more energy is being transferred; on the graph, wave Y's peaks and troughs would be further from the rest position (taller wave).

Part 2 · Frequency and time period

Read the passage, then answer the questions.

Frequency (f) is the number of waves passing a point each second, measured in hertz (Hz).

The period (T) is the time taken for one complete wave to pass a point, measured in seconds (s).

For a sound wave, a higher frequency means a higher pitch.

Fig 2.2: Period and amplitude of a wave (displacement–time graph)
Fig 2.2: Period and amplitude of a wave (displacement–time graph)

Period and frequency are linked by the AQA equation:

period = 1frequency    T = 1f

A wave has a frequency of 50 Hz. Find its period.

Worked Example · Finding period
V
Variables f = 50 Hz   T = ?
E
Equation T = 1f
S
Substitute T = 150
S
Solve T = 0.02
U
Units s (seconds)
Answer: T = 0.02 s

A wave has a period of 0.2 s. Find its frequency.

Worked Example · Finding frequency
V
Variables T = 0.2 s   f = ?
E
Equation T = 1f
S
Substitute 0.2 = 1f
R
Rearrange f = 10.2
S
Solve f = 5
U
Units Hz (hertz)
Answer: f = 5 Hz

Questions · Frequency and time period

Q8
Define frequency and give its unit. (2 marks)
Model AnswerThe number of waves passing a point each second; measured in hertz (Hz).
Q9
A wave has a frequency of 50 Hz. State what this means. (1 mark)
Model Answer50 complete waves pass a point each second.
Q10
Define the period of a wave and give its unit. (2 marks)
Model AnswerThe time for one complete wave to pass a point; measured in seconds (s).
Q11
Calculate the period for each frequency:(a) 10 Hz(b) 5 Hz(c) 0.2 Hz(d) 1200 Hz (4 marks)
Model Answer(a) 0.1 s(b) 0.2 s(c) 5 s(d) 0.000833 s
Q12
A wave has a period of 0.25 s. Calculate its frequency. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables T = 0.25 s   f = ?
E
Equation T = 1f
S
Substitute 0.25 = 1f
R
Rearrange f = 10.25
S
Solve f = 4
U
Units Hz (hertz)
Answer: f = 4 Hz
Q13
A typical adult male speaking voice has a frequency of about 125 Hz. Calculate the period of this sound wave. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables f = 125 Hz   T = ?
E
Equation T = 1f
S
Substitute T = 1125
S
Solve T = 0.008
U
Units s (seconds)
Answer: T = 0.008 s
Q14
Middle C on a piano has a frequency of about 262 Hz. Calculate the period of the note. Give your answer to two significant figures. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables f = 262 Hz   T = ?
E
Equation T = 1f
S
Substitute T = 1262
S
Solve T = 0.0038
U
Units s (seconds)
Answer: T = 0.0038 s
Q15
UK mains electricity is an alternating current with a frequency of 50 Hz. Calculate its period. (2 marks)
Model AnswerT = 1/f = 1/50 = 0.02 s.
Q16
A hummingbird beats its wings about 50 times every second. State the frequency of the wing beats and calculate the period of one beat. (2 marks)
Model AnswerFrequency = 50 Hz; period = 1/50 = 0.02 s.
Q17
A resting heart beats 72 times in one minute. Calculate (a) the frequency of the heartbeat in hertz, (b) the period of one beat. (3 marks)
Model Answera) 72 ÷ 60 = 1.2 Hz  b) T = 1/1.2 = 0.83 s.
Q18
The whine of a mosquito comes from its wings, which beat with a period of 0.0017 s. Calculate the frequency of the sound. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables T = 0.0017 s   f = ?
E
Equation T = 1f
S
Substitute 0.0017 = 1f
R
Rearrange f = 10.0017
S
Solve f ≈ 590
U
Units Hz (hertz)
Answer: f ≈ 590 Hz
Q19
A child on a swing completes one full swing every 2.5 s. Calculate the frequency of the swing. (2 marks)
Model Answerf = 1/T = 1/2.5 = 0.4 Hz.
Q20
A bat emits ultrasound at a frequency of 50 kHz (50 000 Hz). Calculate the period of the wave, giving your answer in standard form. (3 marks)
Model AnswerT = 1/f = 1/50 000 = 0.00002 s = 2 × 10⁻⁵ s.

Part 3 · Determining Wave Properties from a Graph

Read the passage, then answer the questions.

You can find a wave’s properties directly from a graph: wavelength and amplitude from a displacement–distance graph; period from a displacement–time graph (then use T = 1/f to find frequency).

Worked Example: Finding Wavelength from a Graph

  1. Look at the displacement–distance graph (Fig 2.3).
  2. Identify two identical points on the wave, e.g. two neighbouring peaks (crests).
  3. Read off the distance between these two points on the x-axis.
  4. This distance is the wavelength: λ = 4 cm.

Worked Example: Finding Amplitude from a Graph

  1. Look at the same graph.
  2. Find the maximum displacement of the wave from the rest position (where the line crosses 0 on the y-axis).
  3. Read this value off the y-axis.
  4. This is the amplitude: amplitude = 2 cm.
Fig 2.3: A displacement–distance graph
Fig 2.3: A displacement–distance graph

Use Fig 2.4 to find the frequency of the wave.

Worked Example · Finding Frequency from a Graph
V
Variables read from the graph: T = 4 s   f = ?
E
Equation T = 1f
S
Substitute 4 = 1f
R
Rearrange f = 14
S
Solve f = 0.25
U
Units Hz (hertz)
Answer: f = 0.25 Hz
Fig 2.4: A displacement–time graph
Fig 2.4: A displacement–time graph

Questions · Determining Wave Properties from a Graph

Q21
The two graphs below each show a wave. Use Graph 1 to state the wavelength of the wave.
Graph 1
Graph 1
Graph 2
Graph 2
(1 mark)
Model Answerλ = 6 cm.
Q22
Using Graph 1 above, state the amplitude of the wave. (1 mark)
Model AnswerAmplitude = 3 cm.
Q23
Using Graph 2 above, state the period of the wave. (1 mark)
Model AnswerT = 5 s.
Q24
Using Graph 2 above, calculate the frequency of the wave. (2 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables T = 5 s   f = ?
E
Equation T = 1f
S
Substitute 5 = 1f
R
Rearrange f = 15
S
Solve f = 0.2
U
Units Hz (hertz)
Answer: f = 0.2 Hz
Q25
The two graphs below each show a wave. Use Graph 3 to state the wavelength and the amplitude of the wave.
Graph 3
Graph 3
Graph 4
Graph 4
(2 marks)
Model Answerλ = 2 cm; amplitude = 1.5 cm.
Q26
Using Graph 4 above, calculate the frequency of the wave. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables read from the graph: T = 0.02 s   f = ?
E
Equation T = 1f
S
Substitute 0.02 = 1f
R
Rearrange f = 10.02
S
Solve f = 50
U
Units Hz (hertz)
Answer: f = 50 Hz
Q27
The two graphs below each show a wave. Use Graph 5 to state the wavelength of the wave.
Graph 5
Graph 5
Graph 6
Graph 6
(1 mark)
Model Answerλ = 5 cm.
Q28
Using Graph 5 above, state the amplitude of the wave. (1 mark)
Model AnswerAmplitude = 2 cm.
Q29
Using Graph 6 above, state the period of the wave. (1 mark)
Model AnswerT = 3 s.
Q30
Using Graph 6 above, calculate the frequency of the wave. Give your answer to two significant figures. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables T = 3 s   f = ?
E
Equation T = 1f
S
Substitute 3 = 1f
R
Rearrange f = 13
S
Solve f = 0.33
U
Units Hz (hertz, 2 s.f.)
Answer: f = 0.33 Hz
Q31
The two graphs below each show a wave. Use Graph 7 to state the wavelength of the wave.
Graph 7
Graph 7
Graph 8
Graph 8
(1 mark)
Model Answerλ = 1.5 cm.
Q32
Using Graph 7 above, state the amplitude of the wave. (1 mark)
Model AnswerAmplitude = 1 cm.
Q33
Using Graph 8 above, state the period of the wave. (1 mark)
Model AnswerT = 0.004 s.
Q34
Using Graph 8 above, calculate the frequency of the wave. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables T = 0.004 s   f = ?
E
Equation T = 1f
S
Substitute 0.004 = 1f
R
Rearrange f = 10.004
S
Solve f = 250
U
Units Hz (hertz)
Answer: f = 250 Hz

Exam-style questions

Q1
Figure 1 shows two transverse waves, wave 1 and wave 2, drawn on the same axes.
exam l1 transverse
(a) Name one wave quantity that is the same for wave 1 and wave 2. (1 mark)(b) Name one wave quantity that is different for wave 1 and wave 2. (1 mark)(c) The waves in Figure 1 are transverse. Which one of the following is not a transverse wave? Tick one box. (1 mark)☐ gamma rays☐ sound☐ visible lightFigure 2 shows a longitudinal wave being produced in a stretched spring.
exam l1 spring
(d) Which letter on Figure 2, J, K, L or M, marks the centre of the rarefaction closest to the hand? (1 mark)(e) Which two letters on Figure 2 have a distance of one wavelength between them? Tick one box. (1 mark)☐ J and K☐ K and L☐ L and M☐ J and L(f) Describe how the end of the spring should be moved instead, in order to produce a transverse wave. (1 mark)Total: 6 marks
Mark Scheme(a) Wavelength (accept frequency).(b) Amplitude.(c) Sound (it is longitudinal; gamma rays and visible light are both transverse EM waves).(d) K.(e) J and L.(f) Move the end from side to side (or up and down) at right angles to the length of the spring, instead of pushing and pulling along its length.

Lesson 3 · The Wave Equation

Do Now

Q1
What is the unit of wavelength?
Model AnswerThe metre (m).
Q2
What is the unit of frequency?
Model AnswerThe hertz (Hz).
Q3
Calculate 6 × 0.5.
Model Answer6 × 0.5 = 3.

Part 1 · The wave equation

Read the passage, then answer the questions.

Wave speed is the speed at which the energy is transferred (or the wave moves) through the medium.

The speed of a wave is linked to its frequency and wavelength by the AQA equation:

wave speed = frequency × wavelength    v = f λ

where v is wave speed in m/s, f is frequency in Hz and λ is wavelength in m.

A wave has frequency 12 Hz and wavelength 3 m. Find its speed.

Worked Example · Finding speed
V
Variables f = 12 Hz   λ = 3 m   v = ?
E
Equation v = f λ
S
Substitute v = 12 × 3
S
Solve v = 36
U
Units m/s
Answer: v = 36 m/s

Questions · The wave equation

Q1
Define wave speed. (1 mark)
Model AnswerThe speed at which the energy is transferred (or the wave moves) through the medium.
Q2
Write down the wave equation. (1 mark)
Model Answerv = f λ.
Q3
State the symbols and units for wave speed, frequency and wavelength. (3 marks)
Model Answerv (m/s), f (Hz), λ (m).
Q4
Calculate the wave speed for each:(a) f = 5 Hz, λ = 1 m(b) f = 6 Hz, λ = 0.25 m(c) f = 10 Hz, λ = 0.2 m(d) f = 2000 Hz, λ = 4 m (4 marks)
Model Answer(a) 5 m/s(b) 1.5 m/s(c) 2 m/s(d) 8000 m/s
Q5
A wave on a string has a frequency of 50 Hz and a wavelength of 0.4 m. Calculate its speed. (2 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables f = 50 Hz   λ = 0.4 m   v = ?
E
Equation v = f λ
S
Substitute v = 50 × 0.4
S
Solve v = 20
U
Units m/s
Answer: v = 20 m/s

Part 2 · Rearranging the wave equation

Read the passage, then answer the questions.

If you know v and need to find f or λ, substitute the numbers into the given equation v = fλ first, then rearrange to solve; see the worked examples in Part 1.

Always check the units match (speed in m/s needs wavelength in m and frequency in Hz).

A 170 Hz sound wave travels at 340 m/s. Find λ.

Worked Example · Finding wavelength
V
Variables v = 340 m/s   f = 170 Hz   λ = ?
E
Equation v = f λ
S
Substitute 340 = 170 × λ
R
Rearrange λ = 340170
S
Solve λ = 2
U
Units m
Answer: λ = 2 m

A wave travels at 20 m/s with a wavelength of 4 m. Find f.

Worked Example · Finding frequency
V
Variables v = 20 m/s   λ = 4 m   f = ?
E
Equation v = f λ
S
Substitute 20 = f × 4
R
Rearrange f = 204
S
Solve f = 5
U
Units Hz
Answer: f = 5 Hz

Questions · Rearranging the wave equation

Q6
A water wave has a speed of 1.5 m/s and a wavelength of 0.3 m. Calculate its frequency. (2 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables v = 1.5 m/s   λ = 0.3 m   f = ?
E
Equation v = f λ
S
Substitute 1.5 = f × 0.3
R
Rearrange f = 1.50.3
S
Solve f = 5
U
Units Hz
Answer: f = 5 Hz
Q7
A wave has a speed of 6 m/s and a wavelength of 2 m. Calculate its frequency. (2 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables v = 6 m/s   λ = 2 m   f = ?
E
Equation v = f λ
S
Substitute 6 = f × 2
R
Rearrange f = 62
S
Solve f = 3
U
Units Hz
Answer: f = 3 Hz
Q8
A keyboard note has frequency 440 Hz and the sound travels at 340 m/s. Calculate the wavelength. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables v = 340 m/s   f = 440 Hz   λ = ?
E
Equation v = f λ
S
Substitute 340 = 440 × λ
R
Rearrange λ = 340440
S
Solve λ ≈ 0.77
U
Units m
Answer: λ ≈ 0.77 m
Q9
A radio wave travels at 3 × 10⁸ m/s with a wavelength of 300 m. Calculate its frequency. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables v = 3 × 10⁸ m/s   λ = 300 m   f = ?
E
Equation v = f λ
S
Substitute 3 × 108 = f × 300
R
Rearrange f = 3 × 108300
S
Solve f = 1 × 106
U
Units Hz (1 MHz)
Answer: f = 1 × 106 Hz
Q10
Calculate the frequency for each:(a) v = 10 m/s, λ = 2 m(b) v = 340 m/s, λ = 0.5 m(c) v = 1500 m/s, λ = 3 m(d) v = 3 m/s, λ = 0.1 m (4 marks)
Model Answer(a) 5 Hz(b) 680 Hz(c) 500 Hz(d) 30 Hz
Q11
Calculate the wavelength for each:(a) v = 20 m/s, f = 4 Hz(b) v = 340 m/s, f = 170 Hz(c) v = 3 × 10⁸ m/s, f = 1 × 10⁶ Hz(d) v = 0.6 m/s, f = 3 Hz (4 marks)
Model Answer(a) 5 m(b) 2 m(c) 300 m(d) 0.2 m
Q12
An ultrasound wave of frequency 40 kHz travels through air at 340 m/s. Calculate its wavelength. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables v = 340 m/s   f = 40 kHz = 40 000 Hz   λ = ?
E
Equation v = f λ
S
Substitute 340 = 40 000 × λ
R
Rearrange λ = 34040 000
S
Solve λ = 0.0085
U
Units m
Answer: λ = 0.0085 m
Q13
Microwaves of wavelength 0.12 m travel at 3 × 10⁸ m/s. Calculate their frequency. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables v = 3 × 10⁸ m/s   λ = 0.12 m   f = ?
E
Equation v = f λ
S
Substitute 3 × 10⁸ = f × 0.12
R
Rearrange f = 3 × 10⁸0.12
S
Solve f = 2.5 × 10⁹
U
Units Hz
Answer: f = 2.5 × 10⁹ Hz
Q14
State what a frequency of 440 Hz means. (2 marks)
Model Answer440 complete waves (vibrations) pass a point each second.

Part 3 · Measuring wave speed

Read the passage, then answer the questions.

Required practical: measure the speed of waves in a ripple tank and waves in a solid (e.g. a stretched string).

Ripple tank: measure the wavelength by photographing the waves against a ruler. Measure ten wavelengths and divide by ten for accuracy.

Measure the frequency by counting the waves passing a point in 10 s and dividing by 10. Then use v = fλ.

Method: Ripple Tank

  1. Photograph waves over a ruler; measure 10 λ and divide by 10.
  2. Count waves passing a point in 10 s; divide by 10 for f.
  3. Calculate speed using v = fλ.

Stretched string: attach a string to a vibration generator, with the other end passing over a pulley with a weight to keep it taut.

Turn on the generator and adjust its frequency until a clear standing wave pattern (with fixed nodes) forms along the string.

Method: Stretched String

  1. Set up a string between a vibration generator and a pulley, with a weight keeping it taut.
  2. Adjust the signal generator’s frequency until a clear standing wave pattern appears.
  3. Measure the wavelength directly from the pattern (the distance between alternate nodes is one wavelength).
  4. Read the frequency from the signal generator.
  5. Calculate speed using v = fλ.

Speed of sound in air: two students stand a measured distance apart, at least 100 m, on an open field. One bangs two blocks of wood together; the other starts a stopwatch when they see the blocks hit and stops it when they hear the sound.

Speed = distance ÷ time. Repeat several times and take a mean, because the reaction time of the person with the stopwatch is a large source of random error.

Method: Speed of Sound in Air

  1. Measure a large distance (e.g. 100 m) between two students with a trundle wheel or tape.
  2. Student A bangs two wooden blocks together; student B starts the stopwatch on seeing them hit and stops it on hearing the bang.
  3. Repeat and calculate a mean time.
  4. Calculate speed = distance ÷ time.

Questions · Measuring wave speed

Q15
How is the wavelength of waves in a ripple tank measured? (2 marks)
Model AnswerPhotograph the waves against a ruler and measure the distance across several (e.g. ten) wavelengths, then divide.
Q16
How is the frequency of the waves measured? (2 marks)
Model AnswerCount the number of waves passing a point in a fixed time (e.g. 10 s) and divide by that time.
Q17
Why is it better to measure ten wavelengths rather than one? (2 marks)
Model AnswerMeasuring ten and dividing by ten reduces the percentage error in the measurement, giving a more accurate value.
Q18
Once frequency and wavelength are known, how is the wave speed found? (1 mark)
Model AnswerUsing v = fλ.
Q19
In a ripple tank, 8 waves pass in 4 s and the wavelength is 0.05 m. Calculate the wave speed. (3 marks)
Model Answerf = 8/4 = 2 Hz; v = fλ = 2 × 0.05 = 0.1 m/s.
Q20
Describe how a standing wave is produced on a stretched string for this required practical. (2 marks)
Model AnswerA vibration generator makes one end of the string oscillate; its frequency is adjusted until a clear pattern of nodes and antinodes (a standing wave) forms along the string.
Q21
How is the wavelength found from the standing wave pattern on the string? (2 marks)
Model AnswerIt is read directly from the pattern: the distance between alternate nodes (points that do not move) is one wavelength.
Q22
Describe a method to measure the speed of sound in air. (3 marks)
Model AnswerTwo people stand a large measured distance apart (e.g. 100 m); one makes a loud sound (bangs blocks together) and the other times the gap between seeing the blocks hit and hearing the sound; speed = distance ÷ time; repeat and take a mean to reduce the random error from reaction time.
Q23
A student 150 m away hears the bang 0.45 s after seeing the blocks hit. Calculate the speed of sound. (2 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables distance = 150 m   time = 0.45 s   speed = ?
E
Equation speed = distancetime
S
Substitute speed = 1500.45
S
Solve speed ≈ 330
U
Units m/s
Answer: speed ≈ 330 m/s

Exam-style questions

Q1
A note was played on an electric keyboard. The frequency of the note was 440 Hz.(a)(i) What does a frequency of 440 Hz mean? (1 mark)(ii) The sound waves produced by the keyboard travel at a speed of 340 m/s. Calculate the wavelength of the note. Give your answer to three significant figures. (3 marks)(b) Figure 2 shows a graph of the displacement of the note against time.
exam l3 fig2
A second note, of different wavelength, was then played on the keyboard. Figure 3 shows a graph of the second note, drawn on the same axes and to the same scale.
exam l3 fig3
(c) What two conclusions should be made about the second note compared with the first? Give a reason for each conclusion. (4 marks)Total: 8 marks
Mark Scheme(a)(i) 440 complete waves (vibrations) pass a fixed point every second.(ii)
Worked Answer · VESSU
V
Variables v = 340 m/s   f = 440 Hz   λ = ?
E
Equation v = f λ
S
Substitute 340 = 440 × λ
R
Rearrange λ = 340440
S
Solve λ = 0.773 (3 s.f.)
U
Units m
(c) Conclusion 1: the second note has a higher frequency (shorter wavelength). Reason: Figure 3 shows more complete waves in the same space as Figure 2, drawn to the same scale.
Conclusion 2: the second note is quieter. Reason: the waves in Figure 3 have a smaller amplitude (height) than in Figure 2.

Lesson 4 · Reflection and Refraction

Do Now

Q1
What is the normal in a ray diagram?
Model AnswerAn imaginary line at 90° to the surface at the point where the ray hits.
Q2
What happens to light at a mirror?
Model AnswerIt is reflected: the angle of incidence equals the angle of reflection.
Q3
Does light speed up or slow down in glass?
Model AnswerIt slows down.

Part 1 · Reflection Triple Only

Read the passage, then answer the questions.

When a wave hits a surface it can be reflected. We measure angles from the normal (a line at 90° to the surface).

The law of reflection: the angle of incidence equals the angle of reflection.

angle of incidence (i) = angle of reflection (r)
Fig 4.1: Reflection at a plane mirror
Fig 4.1: Reflection at a plane mirror

Questions · Reflection

Q1
What is the normal? (1 mark)
Model AnswerAn imaginary line at right angles (90°) to the surface at the point the ray hits.
Q2
State the law of reflection. (1 mark)
Model AnswerThe angle of incidence equals the angle of reflection.
Q3
From where are the angles of incidence and reflection measured? (1 mark)
Model AnswerFrom the normal.
Q4
A ray hits a mirror at an angle of incidence of 30°. State the angle of reflection. (1 mark)
Model Answer30°.
Q5
A ray of light has an angle of incidence of 50°. State the angle between the incident ray and the mirror surface. (2 marks)
Model Answer40° (90° − 50°, since the angle is measured from the normal).
Q6
Describe an experiment to show that the angle of incidence equals the angle of reflection. (4 marks)
Model AnswerShine a ray box at a plane mirror on paper; draw the normal; mark the incident and reflected rays; measure i and r with a protractor for several angles; the results show i = r each time.

Part 2 · Refraction

Read the passage, then answer the questions.

When a wave passes into a different material it can change speed and direction: this is refraction.

Light slows down when it enters a denser material (e.g. air → glass), so it bends towards the normal.

When light speeds up (glass → air) it bends away from the normal.

Fig 4.2: Refraction at an air–glass boundary
Fig 4.2: Refraction at an air–glass boundary

Questions · Refraction

Q7
What is refraction? (2 marks)
Model AnswerThe change in direction (and speed) of a wave as it passes from one material into another.
Q8
Which way does light bend when it enters a denser material? (1 mark)
Model AnswerTowards the normal.
Q9
Why does light bend when it enters glass from air? (2 marks)
Model AnswerIt slows down in the denser glass, which changes its direction (towards the normal).
Q10
Which way does light bend going from glass into air? Explain. (2 marks)
Model AnswerAway from the normal, because it speeds up in the less dense air.
Q11
A ray enters a glass block along the normal (0°). State what happens to its direction and explain. (2 marks)
Model AnswerIt does not bend (carries straight on), because there is no angle to the normal even though it slows down.
Q12
Describe how you could investigate how the angle of incidence affects the angle of refraction. (4 marks)
Model AnswerShine a single ray at a glass block on paper; draw the normal; vary the angle of incidence; mark and measure the refracted angle each time with a protractor; record results and look for the pattern.

Part 3 · Reflection, transmission and absorption Triple Only

Read the passage, then answer the questions.

When a wave reaches a boundary between two materials, three things can happen: it can be reflected (bounces back), transmitted (passes through, possibly refracted) or absorbed (its energy is transferred to the material).

Usually a mixture happens: for example, light hitting a window is mostly transmitted, partly reflected (you can see a faint reflection) and slightly absorbed.

The law of reflection applies to all types of wave, not just light.

Refraction happens at a boundary only if the wave changes speed; the bigger the speed change, the more it bends.

Questions · Reflection, transmission and absorption

Q13
State the three things that can happen to a wave when it meets a boundary between two materials. (3 marks)
Model AnswerIt can be reflected, transmitted or absorbed.
Q14
Explain what happens to the energy of a wave that is absorbed by a material. (1 mark)
Model AnswerIt is transferred to the material (e.g. heating it).
Q15
Does the law of reflection apply only to light? (1 mark)
Model AnswerNo: it applies to all types of wave.
Q16
A wave hits a boundary but carries straight on without bending. What can you conclude? (2 marks)
Model AnswerEither it hit along the normal, or its speed did not change at the boundary.
Q17
For light passing from air into glass, an angle of incidence of 20° gives an angle of refraction of 13°, and 40° gives 25°. Describe the pattern between the angle of incidence and the angle of refraction. (2 marks)
Model AnswerAs the angle of incidence increases, the angle of refraction also increases (but stays smaller than the angle of incidence).
Q18
Explain why a straw in a glass of water looks bent. (3 marks)
Model AnswerLight from the straw refracts (changes direction) as it passes from water to air, so the straw appears to be in a different position, making it look bent.
Q19
State two ways the behaviour of a wave at a mirror differs from its behaviour entering a glass block. (2 marks)
Model AnswerAt a mirror the wave bounces back (reflects) with i = r; entering glass it passes through and changes direction (refracts) by slowing down.

Exam-style questions

Q1
Required Practical: Reflection Triple OnlyA student investigated the reflection of light by a plane mirror using a ray box, a plane mirror and a protractor. For each angle of incidence, the student repeated the measurement four times. The table shows the student’s results.
Angle of
incidence
Angle of reflection
Test 1Test 2Test 3Test 4
20°19°22°20°19°
30°31°28°32°30°
40°42°40°43°41°
50°56°49°53°46°
(a) For each angle of incidence, the angle of reflection has a range of values. This is caused by an error. What type of error will have caused this range of values? (1 mark)(b) Suggest what the student may have done during the investigation to cause each angle of reflection to have a range of values. (1 mark)(c) Estimate the uncertainty in the angle of reflection when the angle of incidence is 50°. Show how you determine your estimate. (2 marks)(d) The student concluded that for a plane mirror, the angle of incidence is equal to the angle of reflection. Explain whether you agree, using examples from the table. (2 marks)(e) What extra evidence could be collected to support the student’s conclusion? (1 mark)(f) State one change the student should make to the apparatus if he wants to use the same method to investigate diffuse reflection. (1 mark)Total: 8 marks
Mark Scheme(a) A random error.(b) E.g. slightly misread the protractor each time, or did not line up the ray box/paper in exactly the same way for each repeat (parallax error).(c) Range at 50° = 56 − 46 = 10°; uncertainty = half the range = ±5°.(d) Largely agree: at 20°, 30° and 40° the readings are close to the angle of incidence (e.g. mean at 20° ≈ 20°); at 50° the readings are more spread out (46–56°) so are less reliable there, but the results still broadly support i = r.(e) Repeat each angle more times and calculate a mean, and/or test more (in-between) angles of incidence.(f) Replace the plane mirror with a rough/matte surface (e.g. white card) that scatters the light diffusely.
Q2
Required Practical: Refraction Triple OnlyThe data below was obtained from an investigation into the refraction of light at an air-to-glass boundary.
Angle of incidenceAngle of refraction
20°13°
30°19°
40°25°
50°30°
(a) Describe an investigation a student could complete in order to obtain similar data to that given in the table above. Your answer should consider any cause of inaccuracy in the data. (6 marks)(b) State the reason why light is refracted as it crosses from air into glass. (1 mark)Total: 7 marks
Mark Scheme(a) Shine a ray from a ray box at a rectangular glass block placed on paper; draw round the block, and mark the incident ray and the point where it emerges; remove the block and use a ruler to join the points to show the refracted ray inside the outline; draw the normal at the point of entry; measure the angle of incidence and angle of refraction with a protractor; repeat for a range of angles of incidence (e.g. 20–50° in steps of 10°), repeating each angle and taking a mean to reduce random error. Possible causes of inaccuracy: difficulty judging the exact centre of a wide ray box beam, and parallax error when reading the protractor.(b) Light changes speed (slows down) as it crosses the boundary between two materials of different density, which changes its direction (unless it travels along the normal).

Lesson 5 · The Electromagnetic Spectrum

Do Now

Q1
What type of wave is light?
Model AnswerA transverse (electromagnetic) wave.
Q2
What is the fastest thing in the universe?
Model AnswerLight (all EM waves travel at 3 × 10⁸ m/s in a vacuum).
Q3
Name a wave used by a TV remote.
Model AnswerInfrared.

Part 1 · The EM family

Read the passage, then answer the questions.

The electromagnetic (EM) spectrum is a family of 7 types of wave.

All EM waves are transverse and travel at the same speed (the speed of light, 3.0 × 10⁸ m/s) in a vacuum.

In order of increasing frequency (and energy): radio, microwave, infrared, visible, ultraviolet, X-ray, gamma.

EM waves are generated or absorbed when changes happen inside atoms or their nuclei. Gamma rays come from changes in the nucleus.

Fig 5.1: The electromagnetic spectrum
Fig 5.1: The electromagnetic spectrum

Questions · The EM family

Q1
How many types of wave are in the EM spectrum? (1 mark)
Model AnswerSeven.
Q2
Are EM waves transverse or longitudinal? (1 mark)
Model AnswerTransverse.
Q3
At what speed do all EM waves travel in a vacuum? (1 mark)
Model AnswerThe speed of light, 3.0 × 10⁸ m/s.
Q4
List the EM spectrum from longest to shortest wavelength. (3 marks)
Model AnswerRadio, microwave, infrared, visible, ultraviolet, X-ray, gamma.
Q5
Which EM wave has (a) the longest wavelength (b) the highest frequency? (2 marks)
Model Answera) radio waves  b) gamma rays.
Q6
Where do gamma rays originate? (1 mark)
Model AnswerFrom changes in the nucleus of an atom.
Q7
State the relationship between frequency and energy across the EM spectrum. (2 marks)
Model AnswerThe higher the frequency, the higher the energy of the EM wave.

Part 2 · Wavelength, frequency and energy

Read the passage, then answer the questions.

Across the spectrum, as wavelength decreases, frequency increases and so does energy.

X-rays and gamma rays are ionising radiation and can be harmful. UV is also harmful (it can damage skin and eyes) even though it carries less energy than X-rays or gamma rays.

Key ideaRadio has the longest wavelength and lowest energy; gamma has the shortest wavelength and highest energy.

Questions · Wavelength, frequency and energy

Q8
As wavelength decreases across the spectrum, what happens to frequency? (1 mark)
Model AnswerIt increases.
Q9
Name the two ionising EM waves. (2 marks)
Model AnswerX-rays and gamma rays.
Q10
Why are X-rays and gamma rays dangerous? (3 marks)
Model AnswerThey are ionising and carry a lot of energy, so they can damage cells and cause cancer.
Q11
Convert 1 200 000 000 Hz into gigahertz (GHz). (1 mark)
Model Answer1.2 GHz.
Q12
A radio wave and a gamma ray travel through space. Compare their speed, wavelength and energy. (3 marks)
Model AnswerSame speed (speed of light); the radio wave has a much longer wavelength and lower energy; the gamma ray has a much shorter wavelength and higher energy.

Part 3 · Refraction of EM waves

Read the passage, then answer the questions.

Like all waves, EM waves can be refracted when they pass into a different material, changing speed and direction.

Different materials change the speed by different amounts, which is why a prism can separate white light into colours.

(HT only) Explaining refraction with wave fronts: when a wave front meets the boundary at an angle, the part that enters the glass first slows down while the rest is still travelling faster in the air. This swings the wave front round, so the wave changes direction towards the normal. The frequency does not change, so the slower wave has a shorter wavelength: the wave fronts are closer together in the glass (Fig 5.2).

Fig 5.2: (HT) Wave fronts crossing an air–glass boundary: the wave slows, its wavelength shortens and its direction turns towards the normal
Fig 5.2: (HT) Wave fronts crossing an air–glass boundary: the wave slows, its wavelength shortens and its direction turns towards the normal

Questions · Refraction of EM waves

Q13
What happens to an EM wave when it passes into a different material? (2 marks)
Model AnswerIt can be refracted: it changes speed and may change direction.
Q14
Explain why different colours of light are refracted by different amounts in a prism. (2 marks)
Model AnswerEach colour (wavelength) travels at a slightly different speed in glass, so each is bent by a different amount.
Q15
State what stays the same and what changes when light slows down entering glass. (2 marks)
Model AnswerFrequency stays the same; speed and wavelength decrease (and direction can change).
Q16
Glass is transparent to visible light but absorbs most ultraviolet. What does this tell you about how materials interact with EM waves? (2 marks)
Model AnswerWhether a substance absorbs, transmits, reflects or refracts an EM wave depends on the wavelength (frequency) of the wave.
Q17
Explain why all EM waves travel at the same speed in a vacuum but light slows in glass. (2 marks)
Model AnswerIn a vacuum there is nothing to interact with, so all EM waves travel at the speed of light; in glass they interact with the material, which slows them down.
Q18
(HT only) A beam of light crosses from air into glass at an angle. Use the idea of wave fronts to explain why the light changes direction, and state what happens to its wavelength and frequency. (4 marks)
Model AnswerThe part of each wave front that enters the glass first slows down while the rest is still moving faster in air, so the wave front swings round and the direction of travel turns towards the normal. The frequency stays the same; the wavelength gets shorter (the wave fronts are closer together in the glass).

Exam-style questions

Q1
Figure 4 shows an incomplete diagram of the electromagnetic spectrum.
exam l5 spectrum
(a) What name is given to the group of waves at the position labelled A? Tick one box. (1 mark)☐ infrared☐ radio☐ visible light☐ X-ray(b) Electromagnetic waves have many practical uses. State the use that best matches each type of wave below. (3 marks)
Electromagnetic waveUse
Gamma rays
Microwaves
Ultraviolet
Choose from: fibre optic communications · communicating with a satellite · seeing security markings · sterilising surgical instruments (one use is not needed).(c) Complete the sentence, using an answer from the box. (1 mark)black body · ionising · nuclearX-rays can be dangerous to people because X-rays are ______________________ radiation.Total: 5 marks
Mark Scheme(a) Radio.(b) Gamma rays → sterilising surgical instruments. Microwaves → communicating with a satellite. Ultraviolet → seeing security markings. (‘Fibre optic communications’ is a distractor; that is a use of visible light/infrared.)(c) Ionising.

Lesson 6 · Uses and Dangers of EM Waves

Do Now

Q1
Which EM wave do we see with?
Model AnswerVisible light.
Q2
Name an EM wave used in a microwave oven.
Model AnswerMicrowaves.
Q3
Which EM waves are ionising?
Model AnswerUltraviolet, X-rays and gamma rays.

Part 1 · Uses of EM waves

Read the passage, then answer the questions.

Each EM wave is used for jobs that suit its properties:

Radio: TV and radio broadcasts (long wavelength, so they diffract around hills and buildings and travel long distances).

Microwaves: mobile phones, satellite communications and microwave ovens.

Infrared: electrical heaters, cooking food, infrared (thermal imaging) cameras and remote controls.

Visible light: seeing and fibre optic communications. UV: energy-efficient lamps and sun tanning. X-rays: medical images of bones. Gamma: sterilising equipment and treating cancer.

Questions · Uses of EM waves

Q1
State a use of radio waves. (1 mark)
Model AnswerTelevision and radio broadcasting.
Q2
State two uses of microwaves. (2 marks)
Model AnswerMobile phone/satellite communications and cooking in microwave ovens.
Q3
State a use of infrared and a use of X-rays. (2 marks)
Model AnswerInfrared: electrical heaters / cooking / infrared cameras / remote controls. X-rays: medical imaging of bones.
Q4
Why are radio waves good for broadcasting over long distances? (2 marks)
Model AnswerThey have long wavelengths, so they diffract around hills and buildings and travel long distances.
Q5
Explain why X-rays are suitable for imaging bones. (2 marks)
Model AnswerX-rays pass through soft tissue but are absorbed by dense bone, so bones show up as shadows on the image.
Q6
Match each use to a wave: seeing, broken bones, remote controls. (3 marks)
Model AnswerSeeing → visible light; broken bones → X-rays; remote controls → infrared.

Part 2 · Dangers of EM waves

Read the passage, then answer the questions.

X-rays and gamma rays are ionising radiation: they can cause cell mutation and cancer.

Ultraviolet is also harmful: it can cause sunburn and skin cancer and damage eyes.

In general, the higher the frequency (and energy) of an EM wave, the more harmful it can be.

The harm done by UV, X-rays and gamma rays depends on the type of radiation and the size of the dose.

Radiation dose, measured in sieverts (Sv), is a measure of the risk of harm from exposure to radiation. 1000 millisieverts (mSv) = 1 sievert (Sv). You do not need to recall the unit, but you may be asked to compare doses from given data.

Questions · Dangers of EM waves

Q7
What does 'ionising' mean? (1 mark)
Model AnswerThe radiation can knock electrons off atoms, forming ions.
Q8
State one danger of ultraviolet radiation. (1 mark)
Model AnswerSunburn / skin cancer / eye damage.
Q9
Why are X-rays and gamma rays dangerous to people? (2 marks)
Model AnswerThey are ionising, so they can damage or mutate cells, which can cause cancer.
Q10
Explain why exposure to UV from sunbeds should be limited. (2 marks)
Model AnswerUV can damage skin cells, increasing the risk of skin cancer and premature ageing, so exposure should be kept low.
Q11
Suggest two ways radiographers reduce their exposure when taking X-rays. (2 marks)
Model AnswerStand behind a lead screen / wear a lead apron, and limit the time and number of exposures.
Q12
Explain why the danger of an EM wave is generally linked to its frequency. (2 marks)
Model AnswerHigher-frequency waves carry more energy; the most energetic types (X-rays and gamma rays) are ionising, so they cause more damage to cells.
Q13
A dental X-ray gives a dose of 0.005 mSv and a CT scan of the chest gives 7 mSv. How many times greater is the risk from the CT scan? (2 marks)
Model Answer70.005 = 1400 times greater.

Part 3 · Producing radio waves HT Only

Read the passage, then answer the questions.

Radio waves can be produced by oscillations (alternating current) in an electrical circuit / aerial.

When radio waves are absorbed by an aerial, they create an alternating current with the same frequency, which carries the signal.

Key ideaThe frequency of the radio waves equals the frequency of the alternating current that makes them.
Fig 6.1: An alternating current makes charges in the transmitting aerial oscillate and emit radio waves; the waves make charges in the receiving aerial oscillate, producing an alternating current of the same frequency
Fig 6.1: An alternating current makes charges in the transmitting aerial oscillate and emit radio waves; the waves make charges in the receiving aerial oscillate, producing an alternating current of the same frequency

Questions · Producing radio waves

Q14
What produces radio waves in a transmitter? (1 mark)
Model AnswerOscillations / an alternating current in a wire (aerial).
Q15
What happens when radio waves are absorbed by a receiving aerial? (2 marks)
Model AnswerThey produce an alternating current in the aerial with the same frequency as the radio waves.
Q16
State the link between the frequency of the radio wave and the alternating current that produces it. (1 mark)
Model AnswerThey have the same frequency.
Q17
Explain how a signal can be sent and received using radio waves. (3 marks)
Model AnswerAn a.c. in the transmitter aerial produces radio waves; these travel to a receiving aerial where they are absorbed and create an a.c. of the same frequency, recreating the signal.
Q18
Suggest why radio waves rather than gamma rays are used to carry broadcast signals. (3 marks)
Model AnswerRadio waves are not ionising (safe), have long wavelengths that diffract around obstacles and travel long distances; gamma rays are dangerous and impractical for broadcasting.

Exam-style questions

Q1
Scientists have detected radio waves emitted from a distant galaxy.(a) Which one of the following is not an electromagnetic wave? Tick one box. (1 mark)☐ Gamma rays☐ Sound☐ Ultraviolet☐ X-rays(b) What type of electromagnetic wave do our eyes detect? (1 mark)(c) Some of the radio waves from the distant galaxy have a frequency of 1 200 000 000 hertz. Which is the same as this? Tick one box. (1 mark)☐ 1.2 gigahertz☐ 1.2 kilohertz☐ 1.2 megahertz☐ 1.2 millihertz(d) Radio waves travel through space at 300 000 kilometres per second (km/s). How is 300 000 km/s converted to metres per second (m/s)? Tick one box. (1 mark)☐ 300 000 ÷ 1000 = 300 m/s☐ 300 000 × 1000 = 300 000 000 m/s☐ 300 000 + 1000 = 301 000 m/s☐ 300 000 − 1000 = 299 000 m/s(e) Write the equation which links frequency, wavelength and wave speed. (1 mark)(f) Calculate the wavelength of the radio waves emitted from the distant galaxy. Give your answer in metres. (3 marks)Total: 8 marks
Mark Scheme(a) Sound.(b) Visible light.(c) 1.2 gigahertz.(d) 300 000 × 1000 = 300 000 000 m/s.(e) v = f × λ.(f)
Worked Answer · VESSU
V
Variables v = 300 000 000 m/s   f = 1 200 000 000 Hz   λ = ?
E
Equation v = f λ
S
Substitute 300 000 000 = 1 200 000 000 × λ
R
Rearrange λ = 300 000 0001 200 000 000
S
Solve λ = 0.25
U
Units m

Lesson 7 · Infrared and Black Body Radiation Triple Only

Do Now

Q1
Which EM wave do we feel as heat?
Model AnswerInfrared.
Q2
What colour absorbs heat best?
Model AnswerMatt black.
Q3
What does a hotter object emit more of?
Model AnswerMore radiation (with the peak at a shorter wavelength).

Part 1 · Infrared and surfaces

Read the passage, then answer the questions.

Infrared (IR) radiation transfers energy; all objects, whatever their temperature, emit and absorb infrared.

Matt black surfaces are the best absorbers and emitters of infrared.

Shiny, light-coloured surfaces are the worst absorbers and emitters (they reflect IR).

Questions · Infrared and surfaces

Q1
What does infrared radiation transfer? (1 mark)
Model AnswerEnergy (transferred by radiation).
Q2
Which type of surface is the best absorber of infrared? (1 mark)
Model AnswerA matt black surface.
Q3
Which type of surface is the best emitter of infrared? (1 mark)
Model AnswerA matt black surface.
Q4
Which type of surface is the worst absorber, and what does it do instead? (2 marks)
Model AnswerA shiny/light surface; it reflects most of the infrared away.
Q5
Explain why the inside of a solar cooker is lined with shiny metal foil. (2 marks)
Model AnswerThe shiny surface reflects the Sun’s infrared radiation, focusing it onto the food/water to heat it.
Q6
Suggest why a hot drink stays warm longer in a shiny flask than a black one. (3 marks)
Model AnswerA shiny surface is a poor emitter of infrared, so it radiates energy away more slowly than a matt black surface, keeping the drink warm longer.

Part 2 · Black body radiation

Read the passage, then answer the questions.

A perfect black body absorbs all radiation that hits it (reflecting and transmitting none) and is the best possible emitter.

All bodies emit a range of wavelengths. As an object gets hotter, it emits more radiation and the peak shifts to shorter wavelengths.

Fig 7.1: Intensity of emitted radiation against wavelength at different temperatures
Fig 7.1: Intensity of emitted radiation against wavelength at different temperatures

Questions · Black body radiation

Q7
What is a perfect black body? (2 marks)
Model AnswerAn object that absorbs all the radiation that hits it (and is the best possible emitter).
Q8
What happens to the amount of radiation emitted as an object gets hotter? (1 mark)
Model AnswerIt emits more radiation.
Q9
What happens to the peak wavelength as an object gets hotter? (1 mark)
Model AnswerThe peak moves to shorter wavelengths.
Q10
Explain why a small hole in a tin can looks black. (2 marks)
Model AnswerRadiation entering the hole is absorbed inside and very little is reflected back out, so it behaves like a black body and looks black.
Q11
Using Fig 7.1, describe how the radiation curve changes as temperature increases. (3 marks)
Model AnswerThe whole curve rises (more intensity at every wavelength), the total emitted radiation increases, and the peak moves to a shorter wavelength.
Q12
Explain why a body at constant temperature emits and absorbs radiation at equal rates. (2 marks)
Model AnswerIf it absorbed more than it emitted it would warm up, and vice versa; constant temperature means the rates are balanced.

Part 3 · Temperature of objects HT Only

Read the passage, then answer the questions.

The temperature of an object depends on the balance between the radiation it absorbs and the radiation it emits.

If an object absorbs more than it emits, its temperature rises; if it emits more than it absorbs, it cools.

Key ideaConstant temperature → radiation absorbed each second = radiation emitted each second.

Questions · Temperature of objects

Q13
What does the temperature of an object depend on? (2 marks)
Model AnswerThe balance between the radiation it absorbs and the radiation it emits.
Q14
What happens to an object’s temperature if it absorbs more radiation than it emits? (1 mark)
Model AnswerIts temperature rises.
Q15
Explain how the Earth can stay at a roughly steady temperature. (3 marks)
Model AnswerOn average it emits as much radiation (mainly infrared) back into space as it absorbs from the Sun, so the rates balance and the temperature stays steady.
Q16
Explain how increasing gases that trap infrared could raise the Earth’s temperature. (3 marks)
Model AnswerThey absorb more of the outgoing infrared, so the Earth emits less to space than it absorbs from the Sun; the imbalance makes the temperature rise.
Q17
A metal block is left in bright sunlight. Explain why it eventually reaches a steady temperature. (1 mark)
Model AnswerWhen it emits radiation as fast as it absorbs it, the temperature stops rising.

Exam-style questions

Q1
Figure 1 shows what happens when a ray of light enters a tin can through a small hole.
exam l7 tincan
(a) Explain why the small hole looks black. (2 marks)(b) All objects absorb and emit radiation. What is meant when an object is described as a perfect black body? (1 mark)Figure 2 shows how the intensity of different wavelengths of radiation from a hot object varies with temperature.
exam l7 graph
(c) What can be concluded from Figure 2 about how the distribution of the intensity of radiation from an object changes as the temperature of the object increases? (3 marks)(d) Figure 3 shows how the balance between incident radiation from space and the radiation emitted by the Earth has changed over the last 200 years.
exam l7 earth
Explain how the temperature of the Earth and its atmosphere has changed over the last 200 years. Use the information in Figure 3. (3 marks)Total: 9 marks
Mark Scheme(a) Almost all the light entering the hole is absorbed inside the can (any that is reflected bounces around and is absorbed before it can escape), so very little light comes back out and the hole looks black.(b) A perfect black body absorbs all of the radiation incident on it (reflects and transmits none), and is also the best possible emitter of radiation.(c) As temperature increases: the intensity increases at every wavelength (the whole curve gets higher); the total amount of radiation emitted increases; and the peak (the wavelength of maximum intensity) shifts to shorter wavelengths.(d) 200 years ago, the incident radiation (170 000 TW) was balanced by the reflected (50 000 TW) plus emitted (120 000 TW) radiation, so the Earth’s temperature was roughly steady. Now, the emitted radiation has fallen to 116 620 TW while the incident and reflected radiation are unchanged, so the Earth is absorbing more energy than it emits; this net gain of energy means the Earth and its atmosphere have warmed up.

Lesson 8 · Required Practical 10: Infrared Radiation

Do Now

Q1
Which electromagnetic wave do warm objects give off?
Model AnswerInfrared.
Q2
Which type of surface is the best absorber of infrared?
Model AnswerA matt black surface.
Q3
What is a control variable?
Model AnswerA variable that is kept the same throughout the investigation so that it does not affect the results.
Q4
Why is a mean calculated from repeat readings?
Model AnswerTo reduce the effect of random errors (and make anomalies easy to spot), giving a more accurate mean value.

Part 1 · Aim, apparatus and variables

Read the passage, then answer the questions.

This required practical investigates how the nature of a surface affects the amount of infrared radiation it emits.

A Leslie cube is a hollow metal cube with four different outer faces: matt black, shiny black, matt white and shiny silver. It is filled with hot water so that all four faces are at the same temperature.

An infrared detector (or a thermometer with a blackened bulb) is held at a fixed distance from each face in turn and the reading is recorded.

Independent variable: the variable you change in an investigation.

Dependent variable: the variable you measure to see the effect of the change.

Control variables: the variables you keep the same so that the test is fair (valid).

Hypothesis: a prediction of what you expect to happen, with a scientific reason for it.

Safety: use water at or below 60 °C (hot water can scald), wear eye protection, do not overfill the cube, carry it on a heatproof mat and do not touch the metal faces.

Fig 8.1: The Leslie cube and infrared detector, seen from above
Fig 8.1: The Leslie cube and infrared detector, seen from above

Questions · Aim, apparatus and variables

Q1
State the aim of this required practical. (1 mark)
Model AnswerTo investigate how the nature (type) of a surface affects the amount of infrared radiation it emits.
Q2
Name the independent variable and the dependent variable. (2 marks)
Model AnswerIndependent: the type of surface (matt black, shiny black, matt white, shiny silver). Dependent: the amount of infrared radiation detected (the detector reading).
Q3
Explain why all four faces of the Leslie cube are at the same temperature. (2 marks)
Model AnswerThe cube is a single metal container filled with the same hot water; the metal conducts, so every face is heated by the same water to the same temperature.
Q4
Explain why the detector must be held the same distance from each face. (2 marks)
Model AnswerInfrared spreads out as it travels, so a detector that is closer gives a higher reading; keeping the distance the same (a control variable) makes the comparison between the faces fair (valid).
Q5
Write a hypothesis for this practical and give a reason for it. (2 marks)
Model AnswerThe matt black face will give the highest reading and the shiny silver face the lowest, because matt black surfaces are the best emitters of infrared and shiny, light surfaces are the worst.
Q6
Give one hazard in this practical and a precaution that reduces the risk. (2 marks)
Model AnswerHot water can scald: use water at or below 60 °C, wear eye protection, do not overfill the cube, pour carefully and do not touch the metal faces (or: the hot cube can burn: carry it on a heatproof mat, allow it to cool before emptying).

Part 2 · Method and results

Read the passage, then answer the questions.

Method: the distance between the detector and the face, and the time the detector is held there, must be the same for every face.

1. Place the Leslie cube on a heatproof mat and fill it with hot water from a kettle. Put the lid on.

2. Wait about one minute so that all four faces reach the same temperature.

3. Use a ruler to position the infrared detector 10 cm from the matt black face. Record the reading.

4. Repeat for the other three faces, keeping the distance and the time the same.

5. Repeat the whole set of readings twice more and calculate a mean for each face.

6. Record the results in a table and compare the readings for the four faces.

Taking repeat readings and calculating a mean makes the results more precise and makes it easier to spot an anomalous result, which should be left out when the mean is calculated.

If a thermometer with a blackened bulb is used instead of a detector, record the temperature rise of the thermometer after a fixed time (e.g. 2 minutes) at each face.

Results table: record your results during the practical below:

SurfaceReading 1Reading 2Reading 3Mean
matt black
shiny black
matt white
shiny silver

Questions · Method and results

Q7
Explain why the readings should be repeated and a mean calculated. (2 marks)
Model AnswerRepeats allow anomalous results to be identified (and ignored); the mean of the repeats is more precise / reduces the effect of random error.
Q8
If the readings for the matt white face were 18.2, 18.6 and 18.4 units, calculate the mean reading. (1 mark)
Model Answer(18.2 + 18.6 + 18.4) ÷ 3 = 18.4 units.
Q9
A thermometer with a blackened bulb can be used instead of an infrared detector. Explain why the bulb is painted matt black. (2 marks)
Model AnswerMatt black is the best absorber of infrared, so the bulb absorbs more of the radiation from the face and its temperature rises more (and faster), giving a larger, easier-to-measure change.

Part 3 · Analysis and conclusions

Read the passage, then answer the questions.

The independent variable (the type of surface) is categoric, so the results are shown on a bar chart, not a line graph.

Typical results are shown in Fig 8.2: matt black gives the highest reading and shiny silver the lowest, with matt white and shiny black in between.

Conclusion: matt black surfaces are the best emitters of infrared and shiny, light-coloured surfaces are the worst emitters. Matt surfaces emit more than shiny ones, and dark surfaces emit more than light ones.

Evaluation: the cube cools during the experiment, so later readings are slightly lower than they should be. Take the readings quickly, repeat them in a different order, or top up the hot water.

Fig 8.2: Typical detector readings for the four faces of a Leslie cube
Fig 8.2: Typical detector readings for the four faces of a Leslie cube

Questions · Analysis and conclusions

Q10
State which type of graph should be used to display the results and explain why. (2 marks)
Model AnswerA bar chart, because the independent variable (type of surface) is categoric, not continuous.
Q11
Using Fig 8.2, rank the four surfaces from the best emitter to the worst emitter of infrared. (2 marks)
Model AnswerMatt black (best), matt white, shiny black, shiny silver (worst).
Q12
Write a conclusion for this practical. (3 marks)
Model AnswerMatt black is the best emitter of infrared and shiny silver the worst; matt surfaces emit more infrared than shiny surfaces, and black surfaces emit more than white ones. This supports the hypothesis.
Q13
The Leslie cube cools down while the readings are being taken. Explain how this could affect the results and describe one way to reduce the effect. (3 marks)
Model AnswerFaces measured later are at a lower temperature, so they emit less infrared and give a lower reading than they should; this makes the comparison unfair. Reduce the effect by taking the readings quickly, repeating them in a different order, topping up with hot water, or checking the water temperature stays the same.
Q14
Survival blankets are made from thin, shiny silver material. Use the results of this practical to explain why. (2 marks)
Model AnswerA shiny silver surface is a very poor emitter of infrared, so little energy is radiated away from the person; it also reflects infrared back towards the body.

Exam-style questions

Q1
Required Practical: Infrared RadiationA student used the apparatus in Figure 1 to investigate how the nature of a surface affects the amount of infrared radiation it emits. The Leslie cube was filled with hot water and the detector was placed 10 cm from each face in turn.
leslie cube
(a) State the independent variable in this investigation. (1 mark)(b) Give two control variables. (2 marks)The student’s results are shown in the table.
SurfaceDetector reading (arbitrary units)
matt black20.5
shiny black15.0
matt white16.5
shiny silver8.0
(c) Which surface is the best emitter of infrared radiation? Give a reason for your answer. (2 marks)(d) The student took the reading for the shiny silver face last. Explain why this could make the comparison unfair, and suggest how the student could overcome the problem. (2 marks)(e) The inside of a vacuum flask is coated with shiny silver. Use the results to explain why. (1 mark)Total: 8 marks
Mark Scheme(a) The type (nature) of the surface.(b) Any two of: the distance of the detector from the face; the temperature of the water; the time the detector is held in front of each face; the same detector / cube.(c) Matt black, because it gives the highest detector reading (20.5), so the most infrared is emitted from that face.(d) The cube cools as time passes, so by the time the shiny silver face is measured it is at a lower temperature and emits less infrared than it would have at the start; the reading is lower than it should be. Overcome by taking the readings quickly, repeating them in a different order, or refilling the cube with hot water between readings.(e) Shiny silver is a very poor emitter (and a good reflector) of infrared, so little energy is radiated away from the hot liquid and it stays hot for longer.

Lesson 9 · Sound, Ultrasound and Seismic Waves Triple Only, HT Only

Do Now

Q1
Is sound transverse or longitudinal?
Model AnswerLongitudinal.
Q2
What is an echo?
Model AnswerA reflected sound wave.
Q3
What causes an earthquake wave?
Model AnswerSudden movement of rock in the Earth’s crust (an earthquake) sends out vibrations as seismic waves.

Part 1 · Sound and hearing

Read the passage, then answer the questions.

Sound is a longitudinal wave caused by vibrations; it needs a medium (it cannot travel through a vacuum).

Sound entering the ear makes the eardrum vibrate; these vibrations are passed on and converted into nerve signals.

The human hearing range is about 20 Hz to 20 000 Hz.

The eardrum, and other solids, can only be made to vibrate by sound over a limited range of frequencies. This is why human hearing is limited to roughly 20 Hz to 20 kHz.

Questions · Sound and hearing

Q1
What type of wave is sound? (1 mark)
Model AnswerA longitudinal wave.
Q2
Why can sound not travel through a vacuum? (2 marks)
Model AnswerSound needs particles to vibrate and pass on the energy; a vacuum has no particles.
Q3
State the approximate range of human hearing. (2 marks)
Model AnswerAbout 20 Hz to 20 000 Hz.
Q4
Describe what happens in the ear when a sound wave arrives. (2 marks)
Model AnswerThe sound makes the eardrum vibrate; these vibrations are passed through the ear and converted into electrical signals sent to the brain.
Q5
What is ultrasound? (1 mark)
Model AnswerSound with a frequency above the upper limit of human hearing (above 20 000 Hz).
Q6
Explain why an astronaut on the airless Moon could not hear a sound directly. (2 marks)
Model AnswerThere is no air (no particles) on the Moon’s surface to carry the longitudinal sound vibrations.

Part 2 · Echoes and echo sounding

Read the passage, then answer the questions.

An echo is a reflection of sound from a surface. If you know the speed of sound and time how long an echo takes to return, you can work out how far away the surface is. The sound travels there and back, so the distance is half of speed × time.

Bats use this to navigate in the dark: they emit high-frequency sound pulses and listen for the echoes from insects and obstacles. The time between pulse and echo tells the bat how far away an object is.

Echo sounding (sonar) uses the same idea at sea. A ship or submarine sends a pulse of high-frequency sound down through the water; the time for the echo to return from the seabed, a wreck or another vessel gives its distance. It is used to measure water depth and to detect objects in deep water.

Fig 9.1: Echo sounding from a submarine, and a bat using echoes
Fig 9.1: Echo sounding from a submarine, and a bat using echoes

A submarine’s sonar pulse returns from the seabed after 0.8 s. Sound travels at 1500 m/s in water. Find the depth.

Worked Example · Echo sounding
V
Variables speed = 1500 m/s   time = 0.8 s   depth = ?
E
Equation distance = speed × time
S
Substitute distance = 1500 × 0.8 = 1200 (there and back)
R
Rearrange depth = 12002
S
Solve depth = 600
U
Units m
Answer: depth = 600 m

Questions · Echoes and echo sounding

Q7
What is an echo? (1 mark)
Model AnswerA reflection of sound from a surface.
Q8
Explain why the measured time must be halved when using an echo to find a distance. (2 marks)
Model AnswerThe pulse travels to the surface and back, so the measured time is for twice the distance; halving gives the one-way distance.
Q9
Describe how a bat uses sound to find insects in the dark. (2 marks)
Model AnswerIt emits high-frequency sound pulses and listens for the echoes reflected from the insect; the time between pulse and echo gives the distance to the insect.
Q10
What is echo sounding used for? Give two uses. (2 marks)
Model AnswerMeasuring the depth of water and detecting objects (e.g. wrecks, shoals of fish or other vessels) in deep water.
Q11
A ship’s echo sounder receives an echo from the seabed 1.2 s after the pulse is sent. Sound travels at 1500 m/s in water. Calculate the depth of the water. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables speed = 1500 m/s   time = 1.2 s   depth = ?
E
Equation distance = speed × time
S
Substitute distance = 1500 × 1.2 = 1800 (there and back)
R
Rearrange depth = 18002
S
Solve depth = 900
U
Units m
Answer: depth = 900 m

Part 3 · Ultrasound

Read the passage, then answer the questions.

Ultrasound is partly reflected at a boundary between two different materials.

By timing the returning echo, the distance to the boundary can be calculated. This is used in prenatal scans and to find flaws in metal.

Ultrasound is used for baby scans because it is non-ionising (safer than X-rays).

Fig 9.2: Ultrasound reflecting at a boundary
Fig 9.2: Ultrasound reflecting at a boundary

Questions · Ultrasound

Q12
What happens to ultrasound at a boundary between two materials? (1 mark)
Model AnswerIt is partly reflected.
Q13
State two uses of ultrasound. (2 marks)
Model AnswerPrenatal (baby) scanning and detecting flaws in materials (non-destructive testing).
Q14
Why is ultrasound used instead of X-rays for scanning unborn babies? (2 marks)
Model AnswerUltrasound is non-ionising, so it does not damage the baby’s cells; X-rays are ionising and could be harmful.
Q15
Ultrasound travels at 1500 m/s in tissue and an echo returns after 0.0002 s. Calculate the depth of the boundary. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables speed = 1500 m/s   time = 0.0002 s   depth = ?
E
Equation distance = speed × time
S
Substitute distance = 1500 × 0.0002 = 0.3 (there and back)
R
Rearrange depth = 0.32
S
Solve depth = 0.15
U
Units m
Answer: depth = 0.15 m
Q16
Explain why ultrasound rather than audible sound is used for medical scans. (2 marks)
Model AnswerIts higher frequency means a shorter wavelength, so it can show much finer detail; it is also inaudible and does not harm the patient.

Part 4 · Seismic waves

Read the passage, then answer the questions.

Earthquakes produce two types of seismic wave: P waves (longitudinal, travel through solids and liquids, but at different speeds in each) and S waves (transverse, travel through solids only).

P stands for primary and S for secondary: P waves travel faster, so they are the first to arrive at a seismometer after an earthquake, and S waves arrive second.

P waves are refracted at the core boundary because their speed changes, and S waves cannot pass through the liquid outer core. Together, these give evidence for the structure and size of the Earth’s core.

Fig 9.3: P and S waves travelling through the Earth
Fig 9.3: P and S waves travelling through the Earth

Questions · Seismic waves

Q17
Name the two types of seismic wave. (2 marks)
Model AnswerP waves and S waves.
Q18
State whether P waves and S waves are longitudinal or transverse. (2 marks)
Model AnswerP waves are longitudinal; S waves are transverse.
Q19
Which type of seismic wave cannot travel through liquids? (1 mark)
Model AnswerS waves.
Q20
Explain how S waves give evidence for a liquid outer core. (3 marks)
Model AnswerS waves are not detected on the far side of the Earth; since transverse S waves cannot travel through liquids, this suggests part of the core is liquid.
Q21
P waves are detected all around the Earth but S waves are not. Explain what this tells scientists. (2 marks)
Model AnswerP waves travel through both solid and liquid layers, while S waves are blocked by the liquid outer core, which is evidence for the Earth’s layered structure.

Exam-style questions

Q1
Sound and Ultrasound(a)(i) Human ears can detect a range of sound frequencies. Use the correct answers from the box to complete the sentence. (2 marks)2 · 20 · 200 · 2000 · 20 000The range of human hearing is from about ______ Hz to ______ Hz.(ii) What is ultrasound? (1 mark)(iii) Ultrasound can be used to find the speed of blood flow in an artery. State one other medical use of ultrasound. (1 mark)(b) The speed of an ultrasound wave in soft tissue in the human body is 1.5 × 103 m/s and the frequency of the wave is 2.0 × 106 Hz. Calculate the wavelength of the ultrasound wave. (2 marks)(c) When ultrasound is used to find the speed of blood flow in an artery, an ultrasound transducer emits ultrasound which reflects from moving blood cells and is detected back at the transducer. Describe the differences between the emitted waves and the reflected waves detected. (2 marks)Total: 8 marks
Mark Scheme(a)(i) 20 Hz to 20 000 Hz.(ii) Sound with a frequency higher than the upper limit of human hearing (above 20 000 Hz).(iii) Any valid medical use, e.g. prenatal scanning of an unborn baby (or imaging soft tissue/organs).(b)
Worked Answer · VESSU
V
Variables v = 1.5 × 103 m/s   f = 2.0 × 106 Hz   λ = ?
E
Equation v = f λ
S
Substitute 1500 = 2 000 000 × λ
R
Rearrange λ = 15002 000 000
S
Solve λ = 7.5 × 10−4
U
Units m
(c) Because the blood cells are moving, the reflected wave returns with a different frequency (and wavelength) from the emitted wave (the Doppler effect); the size of this frequency shift is used to calculate the speed of the blood flow.
Q2
Seismic WavesP-waves and S-waves are two types of seismic wave caused by earthquakes.(a) Which one of the following statements is correct? Tick one box. (1 mark)☐ P-waves and S-waves are transverse.☐ P-waves and S-waves are longitudinal.☐ P-waves are transverse and S-waves are longitudinal.☐ P-waves are longitudinal and S-waves are transverse.Seismometers on the Earth’s surface record the vibrations caused by seismic waves. Figure 1 shows the vibration recorded by a seismometer for one P-wave.
exam l8 seismic
(b) Calculate the frequency of the P-wave shown in Figure 1. (1 mark)(c) Write down the equation which links frequency, wavelength and wave speed. (1 mark)(d) The P-wave shown in Figure 1 is travelling at 7200 m/s. Calculate its wavelength. (3 marks)(e) Explain why the study of seismic waves provides evidence for the structure of the Earth’s core. (2 marks)Total: 8 marks
Mark Scheme(a) P-waves are longitudinal and S-waves are transverse.(b)
Worked Answer · VESSU
V
Variables 4 complete waves in 10 s   f = ?
E
Equation frequency = number of wavestime
S
Substitute f = 410
S
Solve f = 0.4
U
Units Hz
(c) v = f × λ.(d)
Worked Answer · VESSU
V
Variables v = 7200 m/s   f = 0.4 Hz   λ = ?
E
Equation v = f λ
S
Substitute 7200 = 0.4 × λ
R
Rearrange λ = 72000.4
S
Solve λ = 18 000
U
Units m
(e) S-waves (transverse) cannot travel through liquids and are not detected on the far side of the Earth, while P-waves are detected everywhere but are refracted as they pass through the core; this pattern of shadow zones shows that part of the Earth’s core must be liquid.

Lesson 10 · Lenses Triple Only

Do Now

Q1
What does a magnifying glass do?
Model AnswerIt makes objects look bigger (a virtual, upright, magnified image).
Q2
What is the shape of a convex lens?
Model AnswerIt bulges outwards (thicker in the middle than at the edges).
Q3
Calculate 6 ÷ 2.
Model Answer6 ÷ 2 = 3.

Part 1 · Convex lenses

Read the passage, then answer the questions.

A convex (converging) lens bulges outwards (it is thicker in the middle than at the edges).

Rays of light parallel to the axis are brought together by a convex lens to a point called the principal focus, F. The distance from the centre of the lens to the principal focus is the focal length.

Lenses change the direction of light by refraction: the light bends as it enters and leaves the glass. A more curved (fatter) lens refracts the rays more, so it has a shorter focal length.

Fig 10.1: A convex lens bringing parallel rays to the principal focus
Fig 10.1: A convex lens bringing parallel rays to the principal focus

Questions · Convex lenses

Q1
What shape is a convex lens? (1 mark)
Model AnswerIt bulges outwards (thicker in the middle than at the edges).
Q2
What does a convex lens do to parallel rays of light? (1 mark)
Model AnswerIt brings them together (converges them) to a focus.
Q3
What is the name of the point where a convex lens focuses parallel rays? (1 mark)
Model AnswerThe principal focus (F).
Q4
Define the focal length of a lens. (1 mark)
Model AnswerThe distance from the (centre of the) lens to the principal focus.
Q5
By what process do lenses change the direction of light? (1 mark)
Model AnswerRefraction.
Q6
Parallel rays enter a convex lens with a focal length of 10 cm. How far from the lens do the rays meet? (1 mark)
Model Answer10 cm from the lens (at the principal focus).
Q7
Two convex lenses are made of the same glass. Lens X is more curved than lens Y. State which lens has the shorter focal length and explain why. (2 marks)
Model AnswerLens X: it is more curved so it refracts the rays through a larger angle, bringing them to a focus closer to the lens.
Q8
Sketch a diagram to show three parallel rays passing through a convex lens. Label the principal focus and the focal length. (2 marks)
Model AnswerAs Fig 10.1: parallel rays refracted to meet at F; the focal length marked from the lens to F.

Part 2 · How to draw ray diagrams (convex lenses)

Read the passage, then answer the questions.

A real image is formed where the rays actually meet, so it can be projected onto a screen. On a ray diagram the refracted rays cross on the opposite side of the lens to the object. Examples: the picture on a cinema screen; the image on the sensor of a camera.

A virtual image is formed where the rays only appear to come from, so it cannot be projected onto a screen. On a ray diagram the refracted rays spread apart, so they are extended backwards with dashed lines and cross on the same side of the lens as the object. Examples: your reflection in a plane mirror; the enlarged image seen through a magnifying glass.

A magnified image is larger than the object (a magnifying glass, a projector). A diminished image is smaller than the object (a camera, the eye).

An upright image is the same way up as the object (a mirror, a magnifying glass). An inverted image is upside down (the image in a camera or on the back of the eye).

The image formed by a convex lens depends on where the object is (Fig 10.2):

(a) object beyond 2F: real, inverted, diminished image (a camera)
(a) object beyond 2F: real, inverted, diminished image (a camera)
(b) object between F and 2F: real, inverted, magnified image (a projector)
(b) object between F and 2F: real, inverted, magnified image (a projector)
(c) object between F and the lens: virtual, upright, magnified image (a magnifying glass)
(c) object between F and the lens: virtual, upright, magnified image (a magnifying glass)
(d) object at F: the rays leave the lens parallel, so no image is formed (a searchlight or torch: the bulb is at F to give a parallel beam)
(d) object at F: the rays leave the lens parallel, so no image is formed (a searchlight or torch: the bulb is at F to give a parallel beam)
Fig 10.2: Ray diagrams for a convex lens: (a) object beyond 2F, (b) object between F and 2F, (c) object between F and the lens, (d) object at F

How to draw a ray diagram for a convex lens

  1. Ray 1: from the top of the object, parallel to the axis, to the lens; then through F on the far side.
  2. Ray 2: from the top of the object straight through the centre of the lens, with no change of direction.
  3. Where the two rays cross is the top of the image. Draw the image arrow from the axis to this point and label it.
  4. If the refracted rays spread apart, extend them backwards with dashed lines; where the dashed lines meet is the top of a virtual image.

Questions · How to draw ray diagrams (convex lenses)

Q9
For each of the ray diagrams in this question and the next two: (a) complete the ray diagram to show how the lens forms the image. Draw two rays from the top of the object and label the image. (b) Describe the image on the lines beside the diagram. Diagrams A1 to A4: the object is beyond 2F.
Diagram A1: convex lens, object beyond 2F
Diagram A1: convex lens, object beyond 2F
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
Diagram A2: convex lens, object beyond 2F
Diagram A2: convex lens, object beyond 2F
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
Diagram A3: convex lens, object beyond 2F
Diagram A3: convex lens, object beyond 2F
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
Diagram A4: convex lens, object beyond 2F
Diagram A4: convex lens, object beyond 2F
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
(16 marks)
Model AnswerCompleted diagrams and image descriptions are shown below. Ray 1 from the top of the object travels parallel to the axis and is refracted through F on the far side; ray 2 passes straight through the centre of the lens. The image is where the rays, or their dashed extensions, cross.
Diagram A1: convex lens, object beyond 2F
Diagram A1: convex lens, object beyond 2F
  • Position of imageBetween F and 2F, other side
  • Real or virtualReal
  • Upright or invertedInverted
  • Magnified or diminishedDiminished
Diagram A2: convex lens, object beyond 2F
Diagram A2: convex lens, object beyond 2F
  • Position of imageBetween F and 2F, other side
  • Real or virtualReal
  • Upright or invertedInverted
  • Magnified or diminishedDiminished
Diagram A3: convex lens, object beyond 2F
Diagram A3: convex lens, object beyond 2F
  • Position of imageBetween F and 2F, other side
  • Real or virtualReal
  • Upright or invertedInverted
  • Magnified or diminishedDiminished
Diagram A4: convex lens, object beyond 2F
Diagram A4: convex lens, object beyond 2F
  • Position of imageBetween F and 2F, other side
  • Real or virtualReal
  • Upright or invertedInverted
  • Magnified or diminishedDiminished
Q10
Diagrams B1 to B4: the object is between F and 2F. Complete each diagram and describe the image, as in the previous question.
Diagram B1: convex lens, object between F and 2F
Diagram B1: convex lens, object between F and 2F
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
Diagram B2: convex lens, object between F and 2F
Diagram B2: convex lens, object between F and 2F
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
Diagram B3: convex lens, object between F and 2F
Diagram B3: convex lens, object between F and 2F
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
Diagram B4: convex lens, object between F and 2F
Diagram B4: convex lens, object between F and 2F
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
(16 marks)
Model AnswerCompleted diagrams and image descriptions are shown below. Ray 1 from the top of the object travels parallel to the axis and is refracted through F on the far side; ray 2 passes straight through the centre of the lens. The image is where the rays, or their dashed extensions, cross.
Diagram B1: convex lens, object between F and 2F
Diagram B1: convex lens, object between F and 2F
  • Position of imageBeyond 2F, other side
  • Real or virtualReal
  • Upright or invertedInverted
  • Magnified or diminishedMagnified
Diagram B2: convex lens, object between F and 2F
Diagram B2: convex lens, object between F and 2F
  • Position of imageBeyond 2F, other side
  • Real or virtualReal
  • Upright or invertedInverted
  • Magnified or diminishedMagnified
Diagram B3: convex lens, object between F and 2F
Diagram B3: convex lens, object between F and 2F
  • Position of imageBeyond 2F, other side
  • Real or virtualReal
  • Upright or invertedInverted
  • Magnified or diminishedMagnified
Diagram B4: convex lens, object between F and 2F
Diagram B4: convex lens, object between F and 2F
  • Position of imageBeyond 2F, other side
  • Real or virtualReal
  • Upright or invertedInverted
  • Magnified or diminishedMagnified
Q11
Diagrams C1 to C4: the object is between F and the lens (a magnifying glass). Complete each diagram and describe the image, as before.
Diagram C1: convex lens, object between F and the lens
Diagram C1: convex lens, object between F and the lens
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
Diagram C2: convex lens, object between F and the lens
Diagram C2: convex lens, object between F and the lens
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
Diagram C3: convex lens, object between F and the lens
Diagram C3: convex lens, object between F and the lens
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
Diagram C4: convex lens, object between F and the lens
Diagram C4: convex lens, object between F and the lens
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
(16 marks)
Model AnswerCompleted diagrams and image descriptions are shown below. Ray 1 from the top of the object travels parallel to the axis and is refracted through F on the far side; ray 2 passes straight through the centre of the lens. The image is where the rays, or their dashed extensions, cross.
Diagram C1: convex lens, object between F and the lens
Diagram C1: convex lens, object between F and the lens
  • Position of imageSame side as object, further from the lens
  • Real or virtualVirtual
  • Upright or invertedUpright
  • Magnified or diminishedMagnified
Diagram C2: convex lens, object between F and the lens
Diagram C2: convex lens, object between F and the lens
  • Position of imageSame side as object, further from the lens
  • Real or virtualVirtual
  • Upright or invertedUpright
  • Magnified or diminishedMagnified
Diagram C3: convex lens, object between F and the lens
Diagram C3: convex lens, object between F and the lens
  • Position of imageSame side as object, further from the lens
  • Real or virtualVirtual
  • Upright or invertedUpright
  • Magnified or diminishedMagnified
Diagram C4: convex lens, object between F and the lens
Diagram C4: convex lens, object between F and the lens
  • Position of imageSame side as object, further from the lens
  • Real or virtualVirtual
  • Upright or invertedUpright
  • Magnified or diminishedMagnified

Part 3 · Concave lenses

Read the passage, then answer the questions.

A concave (diverging) lens caves inwards (it is thinner in the middle than at the edges).

Rays of light parallel to the axis are made to spread out (diverge) by a concave lens. The refracted rays appear to come from a point on the same side as the incoming light: this is the principal focus, F, of the concave lens. Because the rays never actually pass through it, it is a virtual focus.

The focal length is the distance from the centre of the lens to F, just as for a convex lens.

Fig 10.3: A concave lens spreading parallel rays out as if from the principal focus
Fig 10.3: A concave lens spreading parallel rays out as if from the principal focus

Questions · Concave lenses

Q12
What shape is a concave lens? (1 mark)
Model AnswerIt caves inwards (thinner in the middle than at the edges).
Q13
What does a concave lens do to parallel rays? (1 mark)
Model AnswerIt makes them spread out (diverge), as if they came from the principal focus on the object side.
Q14
State whether each lens converges or diverges light: (a) convex (b) concave. (2 marks)
Model Answera) converges  b) diverges.
Q15
Where is the principal focus of a concave lens, and why is it described as virtual? (2 marks)
Model AnswerOn the same side of the lens as the incoming (parallel) rays; the refracted rays only appear to come from it and never actually pass through it.
Q16
On Fig 10.3, what do the dashed lines show? (1 mark)
Model AnswerThe refracted rays extended backwards to the point they appear to come from (the principal focus).
Q17
Sketch a diagram to show three parallel rays passing through a concave lens. Mark the principal focus and the focal length. (2 marks)
Model AnswerAs Fig 10.3: rays spread out after the lens, dashed lines back to F on the incoming side, focal length marked from the lens to F.
Q18
A concave lens and a convex lens have the same focal length of 8 cm. Describe one similarity and one difference in what they do to parallel rays. (2 marks)
Model AnswerSimilarity: both refract the rays and have a focal length of 8 cm. Difference: the convex lens brings the rays together to a real focus 8 cm beyond the lens; the concave lens spreads them out as if from a virtual focus 8 cm in front of the lens.
Q19
Describe the image always formed by a concave lens. (3 marks)
Model AnswerVirtual, upright and diminished.
Q20
Explain why the image formed by a concave lens cannot be projected onto a screen. (2 marks)
Model AnswerThe refracted rays spread apart and never actually meet; the image is where they only appear to come from, so it is virtual and no light arrives at that point to form it on a screen.

Part 4 · How to draw ray diagrams (concave lenses)

Read the passage, then answer the questions.

A concave lens always produces a virtual, upright, diminished image, on the same side of the lens as the object, between the lens and F.

On the ray diagram the refracted rays spread apart and never meet, so the image is found where their dashed extensions cross; this is why the image is virtual and cannot be projected onto a screen.

Fig 10.4: A concave lens forming a virtual image
Fig 10.4: A concave lens forming a virtual image

How to draw a ray diagram for a concave lens

  1. Ray 1: from the top of the object, parallel to the axis, to the lens; it then spreads out as if it came from F on the same side as the object. Draw a dashed line from F to the lens to show this.
  2. Ray 2: from the top of the object straight through the centre of the lens, with no change of direction.
  3. The refracted rays spread apart, so trace ray 1 backwards along its dashed line: where it crosses ray 2 is the top of the virtual image. Draw and label the image arrow (upright, smaller, between the lens and F).

Questions · How to draw ray diagrams (concave lenses)

Q21
For each of Diagrams D1 to D3: (a) complete the ray diagram to show how the lens forms the image. Draw two rays from the top of the object and label the image. (b) Describe the image on the lines beside the diagram.
Diagram D1: concave lens
Diagram D1: concave lens
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
Diagram D2: concave lens
Diagram D2: concave lens
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
Diagram D3: concave lens
Diagram D3: concave lens
  • Position of image 
  • Real or virtual 
  • Upright or inverted 
  • Magnified or diminished 
(12 marks)
Model AnswerCompleted diagrams and image descriptions are shown below. Ray 1 from the top of the object travels parallel to the axis and then spreads out as if from F on the object side (dashed line back to F); ray 2 passes straight through the centre of the lens. The virtual image is where the dashed extension of ray 1 crosses ray 2.
Diagram D1: concave lens
Diagram D1: concave lens
  • Position of imageSame side as object, between the lens and F
  • Real or virtualVirtual
  • Upright or invertedUpright
  • Magnified or diminishedDiminished
Diagram D2: concave lens
Diagram D2: concave lens
  • Position of imageSame side as object, between the lens and F
  • Real or virtualVirtual
  • Upright or invertedUpright
  • Magnified or diminishedDiminished
Diagram D3: concave lens
Diagram D3: concave lens
  • Position of imageSame side as object, between the lens and F
  • Real or virtualVirtual
  • Upright or invertedUpright
  • Magnified or diminishedDiminished

Part 5 · Magnification

Read the passage, then answer the questions.

The magnification of a lens is found from the AQA equation:

magnification = image heightobject height

Magnification has no units (it is a ratio). A value greater than 1 means the image is enlarged.

An object 2 cm tall forms an image 6 cm tall. Find the magnification.

Worked Example · VESSU
V
Variables image height = 6 cm   object height = 2 cm   m = ?
E
Equation m = image heightobject height
S
Substitute m = 62
S
Solve m = 3
U
Units (no units)
Answer: m = 3

Questions · Magnification

Q22
Write the equation for magnification. (1 mark)
Model Answermagnification = image height ÷ object height.
Q23
Does magnification have a unit? Explain. (1 mark)
Model AnswerNo: it is a ratio of two heights, so the units cancel.
Q24
Calculate the magnification if:(a) image height 5 cm, object height 1 cm(b) image height 3 m, object height 0.2 m(c) image height 2 mm, object height 15 mm (3 marks)
Model Answer(a) 5(b) 15(c) about 0.13
Q25
An object 4 cm tall produces an image 1 cm tall. Calculate the magnification and state whether the image is enlarged or diminished. (3 marks)
Model AnswerSee the worked answer below.
Worked Answer · VESSU
V
Variables image height = 1 cm   object height = 4 cm   m = ?
E
Equation m = image heightobject height
S
Substitute m = 14
S
Solve m = 0.25 (diminished)
U
Units (no units)
Answer: m = 0.25 (diminished)
Q26
A lens gives a magnification of 4 for an object 3 cm tall. Calculate the image height. (2 marks)
Model Answerimage height = m × object height = 4 × 3 = 12 cm.

Exam-style questions

Q1
A camera was used to take a photograph of a distant object. The camera contains a convex (converging) lens.
exam l9 lens
(a) Complete the ray diagram to show how the lens produces an image of the object. Draw at least two rays from the top of the object to locate the top of the image. (4 marks)(b) State two words to describe the nature of the image produced by the lens in the camera. (2 marks)Total: 6 marks
Mark Scheme(a) Draw a ray from the top of the object parallel to the axis; it refracts through the lens and passes through the principal focus F on the far side. Draw a second ray from the top of the object straight through the centre of the lens, undeviated. The two rays cross beyond F on the far side (since the object is beyond 2F); this marks the top of the image; drop a line down to the axis to complete it.(b) Real and inverted (also diminished, since the object is beyond 2F).Completed ray diagram for part (a):
Completed ray diagram for part (a)
Completed ray diagram for part (a)

Lesson 11 · Colour and Visible Light Triple Only

Do Now

Q1
What colours make up white light?
Model AnswerAll the colours of the visible spectrum: red, orange, yellow, green, blue, indigo, violet.
Q2
Which colour of light bends most in a prism?
Model AnswerViolet (it has the shortest wavelength).
Q3
What does a red filter do?
Model AnswerIt transmits (lets through) red light and absorbs all the other colours.

Part 1 · The spectrum of white light

Read the passage, then answer the questions.

White light is made of a spectrum of colours: red, orange, yellow, green, blue, indigo, violet.

Each colour of the visible spectrum has its own narrow band of wavelength and frequency.

A prism splits white light into these colours because each colour is refracted by a different amount.

Violet has the shortest wavelength and is refracted the most; red has the longest wavelength and is refracted the least.

Fig 11.1: A prism splitting white light
Fig 11.1: A prism splitting white light
Key ideaTo remember which colour refracts the most: Blue bends best.

Questions · The spectrum of white light

Q1
What is white light made of? (1 mark)
Model AnswerA spectrum (mixture) of all the colours of visible light.
Q2
List the colours of the visible spectrum from longest to shortest wavelength. (2 marks)
Model AnswerRed, orange, yellow, green, blue, indigo, violet.
Q3
Which colour is refracted the most by a prism, and why? (2 marks)
Model AnswerViolet, because it has the shortest wavelength.
Q4
Which colour is refracted the least? (1 mark)
Model AnswerRed (longest wavelength).
Q5
Explain why a prism separates white light into colours. (3 marks)
Model AnswerEach colour travels at a slightly different speed in the glass, so each is refracted by a different amount; violet is refracted the most and red the least, spreading the colours into a spectrum.
Q6
State the colour with (a) the highest frequency (b) the lowest frequency. (2 marks)
Model Answera) violet  b) red.

Part 2 · Colour filters

Read the passage, then answer the questions.

A colour filter only lets through light of its own colour and absorbs the other colours.

A red filter transmits red light and absorbs the rest; a green filter transmits only green, and so on.

Key ideaA filter transmits its own colour and absorbs all the others.
Fig 11.2: Colour filters: each transmits its own colour and absorbs the rest
Fig 11.2: Colour filters: each transmits its own colour and absorbs the rest

Questions · Colour filters

Q7
What does a colour filter do? (2 marks)
Model AnswerIt transmits (lets through) light of its own colour and absorbs the other colours.
Q8
What colour passes through a red filter when white light shines on it? (1 mark)
Model AnswerRed light only.
Q9
White light passes through a green filter. Explain what is seen and why. (2 marks)
Model AnswerGreen light, because the filter transmits green and absorbs all the other colours.
Q10
White light passes through a red filter and then a blue filter. Explain what is seen. (3 marks)
Model AnswerNo light (it looks black): the red filter only lets red through, but the blue filter then absorbs the red, so nothing is transmitted.
Q11
Explain why a red object looks red in white light. (2 marks)
Model AnswerIt reflects red light and absorbs the other colours, so only red reaches our eyes.

Part 3 · The colour of objects

Read the passage, then answer the questions.

An opaque object’s colour is the colour of light it reflects; it absorbs the other colours.

A white object reflects all colours; a black object absorbs all colours.

If an object is lit by only one colour, it can look black if that colour is not one it reflects.

Reflection from a smooth surface in a single direction is called specular reflection (e.g. a mirror). Reflection from a rough surface causes scattering: this is called diffuse reflection (e.g. paper).

Objects that transmit light are either transparent (you can see clearly through them, e.g. glass) or translucent (light passes through but is scattered, so you cannot see clearly, e.g. frosted glass).

Fig 11.3: The colour of an opaque object, and specular and diffuse reflection
Fig 11.3: The colour of an opaque object, and specular and diffuse reflection

Questions · The colour of objects

Q12
What determines the colour of an opaque object in white light? (1 mark)
Model AnswerThe colour(s) of light it reflects (the rest are absorbed).
Q13
What colours does (a) a white object and (b) a black object reflect? (2 marks)
Model Answera) all colours  b) none (it absorbs all colours).
Q14
A white ball is lit with red light. Explain why it appears red. (2 marks)
Model AnswerA white surface reflects all colours, but only red light is shining on it, so only red is reflected and it looks red.
Q15
A blue ball is lit with red light. Explain why it appears black. (3 marks)
Model AnswerA blue object reflects only blue and absorbs other colours; with only red light shining on it there is no blue to reflect and the red is absorbed, so no light is reflected and it looks black.
Q16
Explain why a red jumper looks black in a room lit only by green light. (2 marks)
Model AnswerThe red jumper reflects only red and absorbs green; with only green light present, the green is absorbed and no light is reflected, so it appears black.
Q17
Explain the difference between specular and diffuse reflection. (2 marks)
Model AnswerSpecular reflection is from a smooth surface, where all the light is reflected in a single direction; diffuse reflection is from a rough surface, which scatters the light in many directions.
Q18
What is the difference between a transparent and a translucent object? (2 marks)
Model AnswerBoth transmit light; a transparent object lets light through without scattering so you can see clearly through it, while a translucent object scatters the light so you cannot see clearly through it.

Exam-style questions

Q1
Peter had two different-coloured tennis balls: one white and one green. He shone white light through a red filter onto each ball in turn.(a)(i) Experiment 1: the white ball appeared red. Explain why. (2 marks)(ii) Experiment 2: the green ball was used instead of the white ball. What colour did this ball appear? (1 mark)Explain your answer. (2 marks)(b) Peter then cut two holes in a piece of card and covered one hole with a red filter and the other with a blue filter. He placed a white screen behind the card and shone white light at it. What would Peter see on the screen behind each hole? (1 mark)Total: 6 marks
Mark Scheme(a)(i) The red filter only transmits red light and absorbs all other colours. The white ball reflects all colours that reach it, so it reflects the red light coming through the filter and appears red.(ii) Black. A green ball reflects only green light and absorbs other colours; the red filter lets through only red light, so there is no green light for the ball to reflect; the red light is absorbed and no light is reflected, so it looks black.(b) Red light through the red filter; blue light through the blue filter (each filter only transmits its own colour of light).
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