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AQA GCSE Physics revision notes

Waves

Section 4.6
13 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 8463 section 4.6

Checked against AQA 8463 section 4.6. Review basis: the qualification registry sourced from the AQA GCSE Physics (8463) specification; registry verification recorded 17 July 2026.

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In the exam: Equation sheet provided · calculator allowed in every paper

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(physics only) means this content belongs to AQA GCSE Physics (8463), the separate-science qualification, but not AQA Combined Science: Trilogy (8464). It is not an exam tier: (HT only) separately marks Higher-tier content.

Transverse and longitudinal waves

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • In a transverse wave, oscillations are perpendicular to the direction of energy transfer; in a longitudinal wave, oscillations are parallel to it.
  • Use the direction the particles or points oscillate, not the direction the whole wave travels, to classify a wave.
  • A floating marker can bob up and down while a ripple moves horizontally, showing that the disturbance and energy travel without the water moving along with the wave.
  • Do not call every mechanical wave longitudinal: water-surface ripples are treated as transverse, while sound waves in air are longitudinal and contain compressions and rarefactions.
Transverse oscillations and longitudinal compressions shown relative to energy transfer.
Worked example

A wave travels from left to right. The particles of the medium vibrate up and down. State the wave type and explain your choice.

  1. 1.Compare the two directions. The wave travels horizontally but the particles vibrate vertically, so the oscillations are perpendicular to energy transfer. The wave is transverse.

Answer: transverse; the particle vibrations are perpendicular to the direction in which the wave travels

Common mistakes

  • Don't call every mechanical wave longitudinal: water-surface ripples are treated as transverse, while sound waves in air are longitudinal and contain compressions and rarefactions.
  • Don't fall into the trap of drawing particle motion along the direction of travel for a transverse wave.

Exam tip

For ‘compare transverse and longitudinal’, state the oscillation direction relative to energy transfer and give one example of each.

Tier 1 · Easy

ORIGINAL

Compare transverse and longitudinal waves by stating the direction of their oscillations relative to the direction of energy transfer.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A cork is floating on still water. A single ripple passes the cork and reaches the far side of the tank. Describe what happens to the cork and explain what this shows about wave motion.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A compression pulse travels along a horizontal spring. Explain how the motion of one coil differs from the motion of the pulse, and identify two features that show the pulse is longitudinal.

[4 marks]

Total for this question: 4

Your progress and exam materials

Properties of waves

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

The amplitude of a wave is the maximum displacement of a point on the wave away from its undisturbed position. The wavelength is the distance from a point on one wave to the equivalent point on the adjacent wave. The frequency is the number of waves passing a point each second, and the period is T=1fT=\dfrac{1}{f}. All waves obey the wave equation: wave speed = frequency × wavelength, v=fλv=f\lambda, with vv in m/s, ff in Hz and λ\lambda in m.

Explanation

  • Amplitude is the maximum displacement from the undisturbed position; wavelength is the distance between equivalent points on adjacent waves; frequency is waves per second and period is time per wave.
  • Measure several complete wavelengths or periods and divide by their number to reduce percentage uncertainty; for sound, two microphones and a measured separation can provide a travel time.
  • Use T=1/fT=1/f and v=fλv=f\lambda.
  • For example, a 5.0 Hz5.0\,\text{Hz} wave has period 0.20 s0.20\,\text{s}.
  • A common error is to use crest-to-trough distance as one wavelength; it is only half a wavelength, and amplitude must be measured from the undisturbed line rather than crest to trough.
Amplitude measured from the rest position and wavelength measured between adjacent crests.
Worked example

A student uses a ripple tank. She times how long it takes for 1010 complete waves to pass a fixed point and records 7.9 s7.9\,\text{s}, 8.3 s8.3\,\text{s} and 8.1 s8.1\,\text{s}. The wavelength of the waves is 2.5 cm2.5\,\text{cm}. Calculate the speed of the waves. Give your answer to 2 significant figures.

[5 marks]

Total for this question: 5

  1. 1.Mean time for 10 waves =7.9+8.3+8.13=8.1 s=\dfrac{7.9+8.3+8.1}{3}=8.1\,\text{s}, so the period T=0.81 sT=0.81\,\text{s}.
  2. 2.Frequency f=1T=10.81=1.2345… Hzf=\dfrac{1}{T}=\dfrac{1}{0.81}=1.2345\ldots\,\text{Hz}.
  3. 3.Convert the wavelength: λ=2.5 cm=0.025 m\lambda=2.5\,\text{cm}=0.025\,\text{m}.
  4. 4.v=fλ=1.2345×0.025=0.0309… m/sv=f\lambda=1.2345\times0.025=0.0309\ldots\,\text{m/s}.
  5. 5.To 2 significant figures, v=0.031 m/sv=0.031\,\text{m/s}.

Answer: v=0.031 m/sv=0.031\,\text{m/s} (2 significant figures)

Required practical 8

Measuring the frequency, wavelength and speed of waves in a ripple tank and waves in a solid (a stretched string)

Apparatus

  • Ripple tank with a lamp above and a white screen or paper below, a dipper driven by a signal generator, and a stroboscope or camera
  • Metre rule and stopwatch
  • For the string: vibration generator, string over a pulley with hanging masses, and a metre rule

Method

  1. Ripple tank: switch on the dipper so straight waves cross the tank; freeze the pattern with the stroboscope (or photograph it) and measure the distance across ten wavefronts with the rule, then divide by ten to get the wavelength.
  2. Read the frequency from the signal generator, or count the waves passing a point in ten seconds and divide by the time.
  3. Calculate wave speed from v=fλv=f\lambda; check by timing one wavefront across a measured length of the tank and dividing distance by time.
  4. String: switch on the vibration generator and adjust the frequency (or move the pulley/bridge) until a clear stationary pattern appears; measure the length of several loops and divide — each loop is half a wavelength.
  5. Calculate the wave speed on the string from v=fλv=f\lambda using the generator's frequency; repeat for a different tension by changing the hanging mass.

Variables

Independent:
frequency of the source (or, on the string, the tension)
Dependent:
wavelength, and hence the wave speed
Control:
depth of water in the ripple tank; the same string and its length between generator and pulley; amplitude of the dipper

Hazards and precautions

  • Water next to the mains-powered lamp and signal generator is an electric-shock hazard: keep leads and plugs dry and mop spills immediately.
  • A stroboscope can trigger photosensitive epilepsy: keep the flash rate low and warn the class; hanging masses can fall — keep feet clear and use a low-mass hanger.

Common mistakes

  • Don't use crest-to-trough distance as one wavelength; it is only half a wavelength, and amplitude must be measured from the undisturbed line rather than crest to trough.
  • Don't fall into the trap of using frequency in kilohertz without converting it to hertz.

Mark-scheme wording to use

  • “(count) number of waves that pass a point in a given time or (count) number of waves that are produced in a given time (1) and divide by that time in seconds (1)” — AQA 8463, June 2018 Paper 2F, Q03.2
  • measure across a practical number such as 10 waves with ruler … divide by that number — paraphrase of AQA 8463, Specimen Paper 2F, Q05.3
  • measure distance with metre rule … measure travel time with timer … divide distance by time — paraphrase of AQA 8463, November 2021 Paper 2H, Q06.3
  • “80cm = 0.8m … v = 55 × 0.8 … v = 44 (m/s)” — AQA 8463, June 2019 Paper 2F, Q09.3
  • mean 8.1 s; period 0.81 s; f = 1/0.81; 1.2345…; 1.2 Hz to 2 significant figures — paraphrase of AQA 8463, November 2021 Paper 2H, Q06.2

What the examiner wants

Measure several wavelengths or several periods and divide — one wave is too short to time or measure accurately. Name the instrument properly ('metre rule' or 'tape measure'; AQA does not accept 'ruler' for an 80 cm80\,\text{cm} wavelength) and convert centimetres to metres before v=fλv=f\lambda: 4400 m/s4400\,\text{m/s} instead of 44 m/s44\,\text{m/s} loses the final mark. Frequency is set by the source, so when the speed changes only the wavelength changes.

Exam tip

For a wave calculation, convert units first and write v=fλv=f\lambda before substitution.

Tier 1 · Easy

ORIGINAL

On a wave diagram, a crest is 3.0 cm3.0\,\text{cm} above the undisturbed line and the next crest is 0.40 m0.40\,\text{m} away. State the amplitude and wavelength.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

In a ripple tank, nine complete crest-to-crest intervals span 0.36 m0.36\,\text{m}. Five complete waves pass a marker in 2.0 s2.0\,\text{s}. Calculate the wavelength, frequency and wave speed.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A sound wave of frequency 2.40 kHz2.40\,\text{kHz} travels from air, where its speed is 336 m s−1336\,\text{m s}^{-1}, into water, where its speed is 1440 m s−11440\,\text{m s}^{-1}. The frequency does not change. Calculate the wavelength in each medium and explain the change.

[5 marks]

Total for this question: 5

Reflection of waves (physics only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • At a boundary between materials, some incident wave energy may be reflected, some transmitted and some absorbed.
  • For a reflection ray diagram, draw a normal perpendicular to the surface at the point of incidence and measure angles from the normal.
  • The angle of incidence equals the angle of reflection.
  • Energy accounting can test a boundary model: if 65%65\% is transmitted and 20%20\% reflected, the remaining 15%15\% is absorbed.
  • Do not measure the angles from the surface or assume that every boundary reflects all of the incident wave.
Reflection at a surface with equal angles measured from the normal.
Worked example

A light ray strikes a plane mirror at 35∘35^\circ to the normal. State the angle of reflection and name the line from which both angles are measured.

  1. 1.For reflection, the angle of reflection equals the angle of incidence. The incident angle is already measured from the normal, so the reflected ray is also at 35∘35^\circ to the normal.

Answer: angle of reflection =35∘=35^\circ; both angles are measured from the normal

Common mistakes

  • Don't measure the angles from the surface or assume that every boundary reflects all of the incident wave.
  • Don't fall into the trap of measuring the angle from the surface instead of from the normal.

Exam tip

On a reflection diagram, draw the normal and measure both angles from it.

Tier 1 · Easy

ORIGINAL

When a wave reaches a boundary between two materials, state the three possible outcomes for its energy.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

At a boundary, 68%68\% of the incident wave energy is transmitted and 17%17\% is reflected. Calculate the percentage absorbed and describe the three outcomes at the boundary.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Describe an investigation that compares reflection of light from a smooth white tile and a rough white card. Include the measurements, control variables and how the results would be compared.

[5 marks]

Total for this question: 5

Sound waves (physics only) (HT only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Sound waves can make solids vibrate; in the ear, sound makes the eardrum and other structures vibrate, producing the sensation of sound.
  • Trace a conversion by naming the incoming wave, the vibrating solid and any outgoing wave or electrical signal.
  • Normal human hearing extends from about 20 Hz20\,\text{Hz} to 20 kHz20\,\text{kHz}, so a 25 kHz25\,\text{kHz} vibration is above the usual audible range.
  • Do not assume that every vibrating system responds equally at all frequencies: conversion between sound and solid vibration works only over a limited frequency range.
Worked example

State the approximate lower and upper frequency limits of normal human hearing.

  1. 1.Recall the standard range: the lower limit is about 20 Hz20\,\text{Hz} and the upper limit is about 20 000 Hz20\,000\,\text{Hz}, which is 20 kHz20\,\text{kHz}.

Answer: 20 Hz20\,\text{Hz} to 20 kHz20\,\text{kHz}

Common mistakes

  • Don't assume that every vibrating system responds equally at all frequencies: conversion between sound and solid vibration works only over a limited frequency range.
  • Don't fall into the trap of saying sound travels through a vacuum even though it needs particles.

Exam tip

For sound, link vibration frequency to pitch and amplitude to loudness.

Tier 1 · Easy

ORIGINAL

Explain how a sound wave travelling through air can produce the sensation of sound in a listener.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A loudspeaker receives an alternating electrical signal. Describe the sequence of energy transfers that produces sound in the room and then makes a listener's eardrum vibrate.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A hearing test shows that a person detects tones from 40 Hz40\,\text{Hz} to 13 kHz13\,\text{kHz} but not tones outside this interval. Explain why sound-to-vibration conversion in the ear produces this result and compare it with normal human hearing.

[4 marks]

Total for this question: 4

Waves for detection and exploration (physics only) (HT only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Ultrasound is sound above 20 kHz20\,\text{kHz}; partial reflections at boundaries and their return times allow hidden interfaces to be located.
  • For echo measurements use the total out-and-back distance, so a boundary distance is d=vt/2d=vt/2.
  • P-waves are longitudinal and travel through solids and liquids, whereas transverse S-waves do not travel through liquids; their paths provide evidence about Earth's internal structure.
  • A common error is to omit the factor of 22 in echo sounding or to claim that an absent S-wave proves there are no waves rather than indicating a liquid region along its path.
Worked example

Define ultrasound and state what happens when an ultrasound pulse reaches a boundary between two different tissues.

  1. 1.The upper limit of normal human hearing is about 20 kHz20\,\text{kHz}, so ultrasound lies above it. A change of medium forms an interface, where some of the wave is reflected and can return to a detector.

Answer: Ultrasound has a frequency above 20 kHz20\,\text{kHz}; the pulse is partially reflected at the boundary.

Common mistakes

  • Don't omit the factor of 22 in echo sounding or to claim that an absent S-wave proves there are no waves rather than indicating a liquid region along its path.
  • Don't fall into the trap of using ultrasound for a calculation without using its echo travel time.

Exam tip

For echo ranging, account for the outward and return journey before finding depth.

Tier 1 · Easy

ORIGINAL

A seismic wave travels through both solid rock and a liquid layer inside Earth. Identify the wave as a P-wave or S-wave and state whether it is longitudinal or transverse.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

An echo sounder sends a pulse vertically down through seawater. The echo returns after 0.084 s0.084\,\text{s}. The speed of sound in seawater is 1500 m s−11500\,\text{m s}^{-1}. Calculate the water depth.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Seismic detectors on one side of Earth receive P-waves from an earthquake but receive no direct S-waves. Explain how the properties of P-waves and S-waves allow scientists to infer a liquid layer and locate boundaries inside Earth.

[5 marks]

Total for this question: 5

Types of electromagnetic waves

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Electromagnetic waves are transverse waves that transfer energy from a source to an absorber and form one continuous spectrum.
  • Order the spectrum from long wavelength and low frequency to short wavelength and high frequency: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma.
  • All electromagnetic waves travel at the same speed in a vacuum, about 3.00×108 m s−13.00\times10^8\,\text{m s}^{-1}; for example, f=5.0×1010 Hzf=5.0\times10^{10}\,\text{Hz} gives λ=6.0×10−3 m\lambda=6.0\times10^{-3}\,\text{m}.
  • Do not say that higher-frequency electromagnetic waves travel faster in a vacuum; frequency and wavelength change across the spectrum, but the vacuum speed is the same.
Worked example

Name the electromagnetic wave immediately below visible light in frequency and the wave immediately above visible light in frequency.

  1. 1.Use the frequency order radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. The neighbours of visible light are therefore infrared on the lower-frequency side and ultraviolet on the higher-frequency side.

Answer: infrared is below; ultraviolet is above

Common mistakes

  • Don't say that higher-frequency electromagnetic waves travel faster in a vacuum; frequency and wavelength change across the spectrum, but the vacuum speed is the same.
  • Don't fall into the trap of putting ultraviolet between microwaves and infrared in the spectrum order.

Exam tip

Memorise the spectrum order in increasing frequency and decreasing wavelength.

Tier 1 · Easy

ORIGINAL

State which region of the electromagnetic spectrum has the longest wavelength, and compare its speed in a vacuum with the speed of gamma rays.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

An electromagnetic wave has frequency 5.0×1010 Hz5.0\times10^{10}\,\text{Hz}. Calculate its wavelength in a vacuum and identify its region of the spectrum. Use c=3.00×108 m s−1c=3.00\times10^8\,\text{m s}^{-1}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A radio signal has frequency 75 MHz75\,\text{MHz} and a microwave signal has frequency 3.0 GHz3.0\,\text{GHz}. Calculate both wavelengths in air using 3.00×108 m s−13.00\times10^8\,\text{m s}^{-1}, then compare their speeds and wavelengths.

[5 marks]

Total for this question: 5

Properties of electromagnetic waves 1

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Refraction is the change in direction of a wave as it crosses the boundary between two media, caused by the difference in the velocity of the wave in the two substances (Higher tier). In a ray diagram, the angles of incidence and refraction are measured from the normal — a line at right angles to the boundary at the point where the ray enters — and a ray entering a denser medium bends towards the normal. (Higher tier) Different substances absorb, transmit, refract or reflect electromagnetic waves in ways that vary with wavelength.

Explanation

  • Construct a refraction ray diagram using a normal at the boundary. Higher tier: AQA restricts "use wave front diagrams to explain refraction in terms of the change of speed that happens when a wave travels from one medium to a different medium" to Higher —
  • tier, along with the statements that substances absorb, transmit, refract or reflect electromagnetic waves in ways that vary with wavelength and that refraction is due to the difference in wave velocity between substances.
  • To compare infrared emission or absorption by surfaces, keep area, temperature, distance and detector geometry fixed, repeat readings and change only the surface finish.
  • Higher only: on crossing into a slower medium, frequency stays constant, so wavelength decreases; closer wavefronts on the slower side show the speed change that causes refraction.
  • Higher only: do not say refraction is caused by a frequency change; speed and wavelength change at the boundary, while frequency is fixed by the source.
Refraction ray diagram for light entering and leaving a glass block: the angles of incidence ii and refraction rr are measured from the normal, the ray bends towards the normal in the glass, and the emergent ray is parallel to the incident ray.
Worked example

Water waves travel from deep water into shallow water. The wavefronts meet the boundary at an angle. The waves travel more slowly in the shallow water. Explain, in terms of wavefronts, why the direction of the waves changes at the boundary.

[3 marks]

Total for this question: 3

  1. 1.The waves are slower in the shallow water, so the end of each wavefront that reaches the boundary first slows down first.
  2. 2.The rest of that wavefront is still in deep water and is still travelling at the faster speed, so it gets ahead of the part that has slowed.
  3. 3.This swings the wavefront round — the direction of travel changes (refraction). The frequency is unchanged, so the wavelength is shorter in the shallow water and the wavefronts are closer together.

Answer: One end of each wavefront slows first while the other end keeps its higher speed, so the wavefront turns towards the normal; the wavelength decreases and the frequency stays the same.

Required practical 10

Investigating how the amount of infrared radiation absorbed or radiated by a surface depends on the nature of that surface

Apparatus

  • Leslie cube (a metal cube with matt black, shiny black, matt white and shiny silver faces) or four identical flasks with different surface finishes
  • Kettle or hot water, thermometer, and an infrared detector (or thermometer with a blackened bulb) on a clamp
  • Metre rule to fix the detector distance, and a stopwatch

Method

  1. Emission: fill the Leslie cube with hot water and put the lid on; wait a minute for the faces to reach the same temperature.
  2. Point the infrared detector at each face in turn from the same distance, and record the reading; the matt black face gives the highest reading, the shiny silver the lowest.
  3. Absorption: fill the four flasks with equal volumes of cold water at the same starting temperature and place them at equal distances from an infrared heater.
  4. Switch on the heater and start the clock; after ten minutes record each flask's temperature and calculate the temperature increase.
  5. Repeat and take means; conclude that matt black surfaces are the best absorbers and emitters, shiny silver the worst.

Variables

Independent:
the nature of the surface (colour and matt or shiny finish)
Dependent:
infrared detector reading (emission) or temperature increase in a fixed time (absorption)
Control:
distance from the surface to the detector or heater; starting temperature and volume of water; time allowed for heating

Hazards and precautions

  • Boiling water in the cube or flasks can scald: fill them on a tray, use a funnel, and let them cool before emptying.
  • The infrared heater and the cube's faces are hot enough to burn: do not touch them, and keep the heater away from paper and clothing.

Common mistakes

  • Don't say refraction is caused by a frequency change; speed and wavelength change at the boundary, while frequency is fixed by the source (Higher only).
  • Don't fall into the trap of saying electromagnetic waves have different speeds in a vacuum.

Mark-scheme wording to use

  • “ray refracts away from the normal … both rays in the air should be parallel” — AQA 8463, June 2019 Paper 2F, Q05.3
  • “velocity / speed is slower in shallow water … so edge of wave (front) entering shallow water slows down … but the part of the wave (front) in deeper water continues at a higher speed (leading to a change in direction of the wave fronts)” — AQA 8463, June 2018 Paper 2H, Q09.4
  • speed/velocity of light decreases — paraphrase of AQA 8463, Specimen Paper 2H, Q05.2
  • “keep the detector the same distance from each surface” — AQA 8463, June 2019 Paper 2F, Q04.6
  • black absorbs more than white … matt absorbs more than shiny of same colour — paraphrase of AQA 8463, June 2023 Paper 2F, Q09.5
  • harder to judge the centre of the beam — paraphrase of AQA 8463, June 2023 Paper 2F, Q06.1

What the examiner wants

Mark the normal and measure every angle from it; a narrow ray beats a wide one because with a wide beam it is harder to judge the centre. For the wavefront explanation, the marks are for what happens to one END of the wavefront first, not for a restatement that 'it refracts'. AQA's schemes for the infrared practical want the same distance from detector to each surface, matched starting temperatures and a comparison of the temperature CHANGE — and note that single readings can never show precision.

Exam tip

A comparison should state that all electromagnetic waves are transverse and share the same vacuum speed.

Tier 1 · Easy

ORIGINAL

A student compares infrared emission from a matt-black metal plate and a shiny metal plate. State two variables that must be kept the same for a fair comparison.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Higher only: parallel wavefronts are 2.4 cm2.4\,\text{cm} apart in medium A and 1.5 cm1.5\,\text{cm} apart in medium B. The frequency is unchanged. Calculate vB/vAv_B/v_A and explain what the wavefront spacing shows.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Plan an investigation to compare the rate of infrared emission from identical matt-black and shiny metal cans containing hot water. Include measurements, controls and a method of improving reliability.

[6 marks]

Total for this question: 6

Properties of electromagnetic waves 2

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Higher tier: AQA restricts "Radio waves can be produced by oscillations in electrical circuits" and the statement that absorbed radio waves "may create an alternating current with the same frequency as the radio wave itself" to Higher tier.
  • Use dose data by converting units consistently: 1000 mSv=1 Sv1000\,\text{mSv}=1\,\text{Sv}, then compare total doses rather than single exposures.
  • Electromagnetic waves can be emitted or absorbed when atoms or nuclei change; gamma rays specifically originate from changes in an atomic nucleus.
  • Do not treat ultraviolet, X-rays and gamma rays as equally hazardous: effects depend on radiation type and dose; ultraviolet can damage skin, while X-rays and gamma rays are ionising and can cause mutations and cancer.
Worked example

A radiation dose is 180 mSv180\,\text{mSv}. Convert this dose to sieverts.

  1. 1.Use 1000 mSv=1 Sv1000\,\text{mSv}=1\,\text{Sv}. Divide by 10001000: 180 mSv=180/1000=0.180 Sv180\,\text{mSv}=180/1000=0.180\,\text{Sv}.

Answer: 0.180 Sv0.180\,\text{Sv}

Common mistakes

  • Don't treat ultraviolet, X-rays and gamma rays as equally hazardous: effects depend on radiation type and dose; ultraviolet can damage skin, while X-rays and gamma rays are ionising and can cause mutations and cancer.
  • Don't fall into the trap of describing ionising radiation as making an object radioactive in every exposure.

Exam tip

For risk questions, identify the radiation, the tissue effect and how exposure is reduced.

Tier 1 · Easy

ORIGINAL

State one harmful effect of ultraviolet radiation and one harmful effect of X-rays or gamma rays.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Procedure A gives a dose of 4.5 mSv4.5\,\text{mSv} on each of six visits. Procedure B gives one dose of 18 mSv18\,\text{mSv}. Calculate the total dose for A and use the data to compare the radiation risk.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Higher only: a transmitter circuit oscillates at 92 MHz92\,\text{MHz}. Explain how it produces a radio wave and how a tuned receiving circuit can produce a signal at 92 MHz92\,\text{MHz}. Contrast this origin with the origin of gamma rays.

[5 marks]

Total for this question: 5

Uses and applications of electromagnetic waves

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Typical uses are radio for broadcasting; microwaves for satellite communication and cooking; infrared for heaters, cooking and thermal cameras; and visible light for fibre-optic communication.
  • Higher tier: AQA restricts "give brief explanations why each type of electromagnetic wave is suitable for the practical application" to Higher tier; explain suitability by linking the application to whether the wave is transmitted, absorbed, detected or able to penetrate the relevant material.
  • Ultraviolet is used in energy-efficient lamps and tanning, while X-rays and gamma rays are used for medical imaging and treatment.
  • A common error is to name a use without explaining suitability; at Higher tier, link the wave's penetration, absorption or effect on matter to the application.
Worked example

Name one electromagnetic wave used for each application: satellite communication, a thermal camera and medical imaging of bones.

  1. 1.Recall the standard application pairs. Satellite links use microwaves, thermal cameras detect infrared radiation, and bone imaging uses X-rays.

Answer: microwaves; infrared; X-rays

Common mistakes

  • Don't name a use without explaining suitability; at Higher tier, link the wave's penetration, absorption or effect on matter to the application.
  • Don't fall into the trap of naming a use without linking it to the wave's relevant property.

Exam tip

For ‘explain the use’, connect one wave property directly to why the application works.

Tier 1 · Easy

ORIGINAL

Put these applications in order of increasing wave frequency: thermal imaging, satellite communication and radio broadcasting. Name the wave used for each.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

Higher only: explain why an infrared camera can show warmer parts of a building and why visible light is unsuitable for measuring the same temperature pattern in darkness.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Higher only: a food manufacturer can heat a meal using microwaves or infrared radiation. Compare how the two waves heat the meal and explain why using both can improve the result.

[5 marks]

Total for this question: 5

Lenses (physics only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

A lens forms an image by refracting light. In a convex lens, parallel rays of light are brought to a focus at the principal focus; the distance from the lens to the principal focus is the focal length. The image produced by a convex lens can be real or virtual; the image produced by a concave lens is always virtual. Magnification = image height ÷ object height — a ratio, so it has no units, and both heights must be measured in the same unit (mm or cm).

Explanation

  • A convex lens refracts parallel rays towards its principal focus and may form real or virtual images; a concave lens spreads rays and always forms a virtual image.
  • Construct a ray diagram with at least two standard rays from the same point on the object; their intersection, or the intersection of backward extensions, locates the image.
  • Magnification is image height/object height\text{image height}/\text{object height} and has no unit; an image 4.5 cm4.5\,\text{cm} high from an object 1.5 cm1.5\,\text{cm} high has magnification 3.03.0.
  • Do not attach units to magnification or use mismatched units for the two heights; a virtual image cannot be projected onto a screen.
Two principal rays through a convex lens forming a real inverted image.
Worked example

A convex lens is used as a magnifying glass to look at a printed letter that is 3.0 mm3.0\,\text{mm} high. The image of the letter is 12 mm12\,\text{mm} high. Calculate the magnification produced by the lens, and give two properties of the image formed by a magnifying glass.

[4 marks]

Total for this question: 4

  1. 1.Substitute into magnification =image heightobject height=\dfrac{\text{image height}}{\text{object height}}: magnification =123.0=\dfrac{12}{3.0}.
  2. 2.Magnification =4.0=4.0 (a ratio, so no unit).
  3. 3.The image is virtual (it cannot be projected onto a screen; the rays only appear to come from it).
  4. 4.The image is upright (and magnified) because the object is inside the focal length of the lens.

Answer: Magnification =4.0=4.0 (no unit); the image is virtual and upright.

Common mistakes

  • Don't attach units to magnification or use mismatched units for the two heights; a virtual image cannot be projected onto a screen.
  • Don't fall into the trap of drawing a ray through the principal focus before it reaches a converging lens.

Mark-scheme wording to use

  • “6.0 = 9.0/object height … object height = 9.0/6.0 … object height = 1.5 (mm)” — AQA 8463, June 2019 Paper 2H, Q01.4
  • “(both are) diminished … concave is virtual and convex is real or concave is upright and convex is inverted … allow smaller for diminished” — AQA 8463, June 2019 Paper 2H, Q01.3
  • “upright; virtual” — AQA 8463, June 2018 Paper 2F, Q02.2
  • “magnification is a ratio” — AQA 8463, June 2025 Paper 2F, Q05.6
  • virtual and diminished — paraphrase of AQA 8463, November 2021 Paper 2H, Q04.1

What the examiner wants

Describe an image with the exam words — real or virtual, upright or inverted, magnified or diminished — and compare like with like when two lenses are involved (a concave lens gives a virtual, upright, diminished image; a convex lens with the object beyond the focal length gives a real, inverted one). Keep both heights in the same unit: AQA's 2019 scheme drops a mark for an unnecessary unit conversion, and 'focal length' is the only accepted name for the lens-to-focus distance.

Exam tip

Use two principal rays from the top of the object and mark their intersection as the image.

Tier 1 · Easy

ORIGINAL

A lens forms an image that can be projected onto a screen. State whether the image is real or virtual and identify whether the lens can be concave.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Describe how to construct a ray diagram for the image of an object formed by a concave lens, and state three properties of the image.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A lens produces a sharp image on a screen. The image is 7.2 cm7.2\,\text{cm} high and the magnification is 3.03.0. Calculate the object height, identify the lens as convex or concave, and justify your choice.

[4 marks]

Total for this question: 4

Visible light (physics only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Each visible colour occupies a narrow range of wavelengths and frequencies within the electromagnetic spectrum.
  • A smooth surface gives specular reflection mainly in one direction; a rough surface gives diffuse reflection by scattering light, while transparent and translucent materials both transmit light but differ in image clarity.
  • A colour filter absorbs some wavelength ranges and transmits others, so predict the emerging light by finding the wavelengths that both arrive and pass through the filter.
  • An opaque object appears the colour it reflects most strongly; it appears white if it reflects all visible wavelengths similarly and black if it absorbs them all.
  • Do not say an ordinary coloured object produces its own light.
Worked example

Under white light, one opaque card reflects all visible wavelengths equally and another absorbs all visible wavelengths. State the colour of each card.

  1. 1.White light contains the visible wavelength range. Equal reflection of all those wavelengths makes the first card look white. With no visible wavelengths reflected to the eye, the second card looks black.

Answer: The reflecting card appears white; the absorbing card appears black.

Common mistakes

  • Don't make this mistake: An opaque object appears the colour it reflects most strongly; it appears white if it reflects all visible wavelengths similarly and black if it absorbs them all. Do not say an ordinary coloured object produces its own light.
  • Don't fall into the trap of saying a blue object reflects every colour of white light.

Exam tip

For colour questions, state which wavelengths are absorbed, reflected and transmitted.

Tier 1 · Easy

ORIGINAL

Compare the reflection of a narrow beam of light from a smooth mirror and from rough paper.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A red book is viewed in white light through a blue filter. Explain why the book appears very dark.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A surface strongly reflects green light, weakly reflects red light and absorbs blue light. Predict and explain its appearance under white light, through a green filter and through a red filter.

[5 marks]

Total for this question: 5

Emission and absorption of infrared radiation (physics only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Every object emits and absorbs infrared radiation, and a hotter object emits more infrared energy in a given time.
  • To compare surfaces fairly, use equal areas at the same temperature and keep detector distance, angle and surroundings constant; repeat readings before comparing means.
  • A perfect black body absorbs all incident radiation, reflecting and transmitting none; because a good absorber is also a good emitter, it is the best possible emitter.
  • Do not confuse visible colour alone with the experimental variable: surface finish matters, and a shiny surface is generally a poorer absorber and emitter than a matt black surface.
Worked example

Two identical matt-black objects are at 35 ∘C35\,{}^\circ\text{C} and 75 ∘C75\,{}^\circ\text{C}. State which emits more infrared radiation each second.

  1. 1.The surfaces are identical, so temperature is the relevant difference. A hotter object radiates more infrared energy in a given time, so the 75 ∘C75\,{}^\circ\text{C} object emits more.

Answer: The object at 75 ∘C75\,{}^\circ\text{C} emits more infrared radiation each second.

Common mistakes

  • Don't confuse visible colour alone with the experimental variable: surface finish matters, and a shiny surface is generally a poorer absorber and emitter than a matt black surface.
  • Don't fall into the trap of assuming a shiny light surface is the best infrared emitter.

Exam tip

Compare surfaces using both emission and absorption: dull black is best; shiny light is poor.

Tier 1 · Easy

ORIGINAL

A student says, 'Only objects that are hotter than their surroundings emit radiation.' Correct the statement and give one way temperature affects the emitted radiation.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Identical hot-water cans have matt-black and polished-silver outer surfaces. Predict which can cools faster and explain the prediction in terms of infrared radiation.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A student uses an infrared lamp, four metal plates with different surface finishes and contact thermometers to compare absorption. Describe a valid method and explain how the data identify the best absorber.

[5 marks]

Total for this question: 5

Perfect black bodies and radiation (physics only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • All objects emit radiation, and both the intensity and wavelength distribution of the emitted radiation depend on temperature.
  • Higher tier: AQA restricts "explain how the temperature of a body is related to the balance between incoming radiation absorbed and radiation emitted, using everyday examples to illustrate this balance,
  • and the example of the factors which determine the temperature of the Earth" to Higher tier, together with drawing or interpreting diagrams of how radiation affects the temperature of Earth's surface and atmosphere;
  • equal rates of absorption and emission mean constant temperature. Higher only: if a body absorbs 480 J480\,\text{J} each second but emits 530 J530\,\text{J} each second, it has a net energy loss of 50 J s−150\,\text{J s}^{-1} and cools.
  • Higher only: do not infer constant temperature from a constant incoming rate alone; reflection, absorption and emission all affect the balance, including for Earth's surface and atmosphere.
Worked example

State two features of the radiation emitted by an object that depend on the object's temperature.

  1. 1.All objects emit radiation. The specification identifies two temperature-dependent properties of that emission: how intense it is and how the emitted energy is distributed across wavelengths.

Answer: the intensity; the wavelength distribution

Common mistakes

  • Don't infer constant temperature from a constant incoming rate alone; reflection, absorption and emission all affect the balance, including for Earth's surface and atmosphere (Higher only).
  • Don't fall into the trap of saying a constant-temperature body stops emitting infrared radiation.

Exam tip

For equilibrium, state that emission rate equals absorption rate, not that both rates are zero.

Tier 1 · Easy

ORIGINAL

State what a perfect black body does to radiation incident on it, including what it reflects and transmits.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

Higher only: a body absorbs radiation at 480 W480\,\text{W} and emits radiation at 530 W530\,\text{W}. Calculate the net rate of energy change and state what happens to its temperature.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Higher only: Earth receives an average solar power of 340 W m−2340\,\text{W m}^{-2}. Initially 102 W m−2102\,\text{W m}^{-2} is reflected and 238 W m−2238\,\text{W m}^{-2} is emitted to space. The reflected power then decreases to 90 W m−290\,\text{W m}^{-2} while the emitted power is initially unchanged. Calculate both initial and new net power, and explain the resulting temperature change.

[5 marks]

Total for this question: 5

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