4.6 Waves — revision question pack

13 specification points · notes, questions, answers and worked methods

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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(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.

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4.6.1.1 · Transverse and longitudinal waves

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.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

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
A sound wave travels to the right through air. Air particles are observed moving backwards and forwards along the same line, producing alternating high-pressure and low-pressure regions. Identify the wave type and name the two types of region.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
A camera tracks a marked point on a rope while a pulse moves past it. The point's horizontal position stays fixed, but its vertical position first rises and then falls. Use this evidence to classify the pulse and distinguish movement of the rope from movement of the wave.
[3 marks]
Total for this question: 3
A learner is given a single photograph of a stretched rope, showing only its shape. The learner says this is enough to decide whether the wave is transverse or longitudinal. Explain why this evidence is insufficient and state the observation needed to classify the wave if its energy travels east.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Two wave models are annotated. Model P has particle arrows parallel to the energy-transfer arrow but shows evenly spaced particles everywhere. Model Q has particle arrows perpendicular to energy transfer and shows crests and troughs. Critique both models and give the correct wave type for each.
[4 marks]
Total for this question: 4
Two waves transfer energy east. For wave R, a marked particle keeps the same east-west coordinate while its north-south displacement follows $0,+3,0,-3,0\,\text{cm}$ . For wave S, the north-south coordinate stays fixed while its east-west displacement follows $0,+2,0,-2,0\,\text{cm}$ . Classify both waves and explain why neither particle is carried east with the energy.
[5 marks]
Total for this question: 5
A repeating wave travels along a spring. A distance–displacement snapshot has adjacent matching points $0.72\,\text{m}$ apart. A displacement–time trace for one marked coil repeats every $0.30\,\text{s}$ and records displacement along the spring. State which record determines wavelength, which determines period and which observation fixes the wave type. Name the regions where the coils are closer together.
[5 marks]
Total for this question: 5
Describe a method and observations that would provide evidence that a water ripple and a sound pulse transfer energy without transferring the water or air across the apparatus. You have a ripple tank, a floating marker, a pulse loudspeaker, two microphones and an air-movement detector.
[6 marks]
Total for this question: 6

4.6.1.2 · Properties of waves

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 $T=1/f$ and $v=f\lambda$ .
For example, a $5.0\,\text{Hz}$ wave has period $0.20\,\text{s}$ .
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 vibrating source produces $8.0$ complete waves each second. Calculate the period of the wave.

1.The frequency is $f=8.0\,\text{Hz}$ . Use $T=1/f$ , giving $T=1/8.0=0.125\,\text{s}$ .

Answer: $T=0.125\,\text{s}$

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.

Exam tip

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

Tier 1 · Easy

On a wave diagram, a crest is $3.0\,\text{cm}$ above the undisturbed line and the next crest is $0.40\,\text{m}$ away. State the amplitude and wavelength.
[2 marks]
Total for this question: 2
A wave sketch labels the vertical distance from a trough to the next crest as the amplitude and the horizontal distance from a crest to the next trough as the wavelength. Correct both labels.
[2 marks]
Total for this question: 2

Tier 2 · Standard

In a ripple tank, nine complete crest-to-crest intervals span $0.36\,\text{m}$ . Five complete waves pass a marker in $2.0\,\text{s}$ . Calculate the wavelength, frequency and wave speed.
[4 marks]
Total for this question: 4
In RP₈, the times for 20 ripples are $8.0$ , $7.8$ and $8.2\,\text{s}$ . Repeated measurements of the span of five wavelengths are $0.30$ , $0.31$ and $0.29\,\text{m}$ . Use the mean measurements to determine frequency, wavelength and wave speed.
[5 marks]
Total for this question: 5
A wave crosses one region of a tank. Three rows list frequency, wavelength and claimed speed: A, $4.0\,\text{Hz}$ , $0.15\,\text{m}$ , $0.60\,\text{m s}^{-1}$ ; B, $5.0\,\text{Hz}$ , $0.12\,\text{m}$ , $0.60\,\text{m s}^{-1}$ ; C, $6.0\,\text{Hz}$ , $0.10\,\text{m}$ , $0.50\,\text{m s}^{-1}$ . Identify the inconsistent value, correct it and state what the corrected table shows.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A sound wave of frequency $2.40\,\text{kHz}$ travels from air, where its speed is $336\,\text{m s}^{-1}$ , into water, where its speed is $1440\,\text{m s}^{-1}$ . The frequency does not change. Calculate the wavelength in each medium and explain the change.
[5 marks]
Total for this question: 5
A student doing RP₈ measures one wavelength of a ripple as $6\,\text{cm}$ and tries to time one oscillation with a handheld stopwatch. Explain why the percentage uncertainties are large and redesign the measurements to obtain a more repeatable wave speed.
[6 marks]
Total for this question: 6
A wave-measuring grid is used to measure ripples in a tank. The smallest wavelength the grid can measure reliably is $8.0\,\text{cm}$ . Waves travel across the grid at $0.60\,\text{m s}^{-1}$ . The wave generator settings are $2.0\,\text{Hz}$ , $5.0\,\text{Hz}$ and $15\,\text{Hz}$ . Use the Physics Equations Sheet to calculate the wavelength at each setting and decide which settings the grid can resolve.
[5 marks]
Total for this question: 5
Two microphones are fixed $1.80\,\text{m}$ apart in line with a loudspeaker. Repeated sound pulses give delays of $5.2$ , $5.4$ and $5.3\,\text{ms}$ between the microphone traces. Calculate the mean speed of sound and explain two features of the method that reduce uncertainty.
[5 marks]
Total for this question: 5
A vibration generator attached to a taut cord produces steady transverse waves. A signal generator displays the frequency. Describe how to use the cord, a metre ruler and a video camera to determine wavelength and wave speed accurately over a range of frequencies. Use the Physics Equations Sheet.
[6 marks]
Total for this question: 6

4.6.1.3 · Reflection of waves (physics only)

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 $65\%$ is transmitted and $20\%$ reflected, the remaining $15\%$ 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 $35^\circ$ to the normal. State the angle of reflection and name the line from which both angles are measured.

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 $35^\circ$ to the normal.

Answer: angle of reflection $=35^\circ$ ; 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

When a wave reaches a boundary between two materials, state the three possible outcomes for its energy.
[3 marks]
Total for this question: 3
A reflection diagram gives the incident ray an angle of $28^\circ$ to the normal but draws the reflected ray at $28^\circ$ to the surface. State the correction needed and the correct angle between the reflected ray and the surface.
[2 marks]
Total for this question: 2

Tier 2 · Standard

At a boundary, $68\%$ of the incident wave energy is transmitted and $17\%$ is reflected. Calculate the percentage absorbed and describe the three outcomes at the boundary.
[3 marks]
Total for this question: 3
A narrow light beam strikes a surface at $35^\circ$ to the normal. A detector records reflected intensities of $2$ , $5$ , $21$ , $6$ and $2$ units at angles of $15^\circ$ , $25^\circ$ , $35^\circ$ , $45^\circ$ and $55^\circ$ to the normal. Interpret the pattern and state what it shows about the surface and the reflection law.
[3 marks]
Total for this question: 3
At a boundary, $42\,\text{J}$ of wave energy is transmitted and $18\,\text{J}$ is absorbed. The reflected energy is $20\%$ of the incident energy. Calculate the incident energy and the reflected energy.
[4 marks]
Total for this question: 4

Tier 3 · Hard

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
In RP₉, a student compares reflection from two surfaces by drawing the normal parallel to each surface, measuring angles from the surface, moving the lamp to a new distance for the second material and taking one detector reading. Identify four changes needed for a valid comparison.
[4 marks]
Total for this question: 4
Panel A reflects $8\%$ and transmits $82\%$ of incident wave energy. Panel B reflects $20\%$ and transmits $5\%$ . A student concludes that A is the better absorber because it reflects less. Evaluate the conclusion using a complete energy account.
[4 marks]
Total for this question: 4
Two parallel plane mirrors face each other. A ray travelling downwards and to the right strikes the lower mirror at $35^\circ$ to its normal, reflects to the upper mirror and reflects again. Describe how to construct the two reflected rays, state the angle of reflection at each mirror and compare the final ray direction with the incident direction.
[4 marks]
Total for this question: 4
A $150\,\text{J}$ wave pulse reaches the first of two boundaries. At the first boundary, $18\%$ is reflected and $62\%$ is transmitted. The energy transmitted reaches a second boundary, where half is absorbed and $20\%$ is reflected. Calculate the energy absorbed at the first boundary and the energy absorbed, reflected and transmitted at the second boundary.
[6 marks]
Total for this question: 6

4.6.1.4 · Sound waves (physics only) (HT only)

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 $20\,\text{Hz}$ to $20\,\text{kHz}$ , so a $25\,\text{kHz}$ 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.Recall the standard range: the lower limit is about $20\,\text{Hz}$ and the upper limit is about $20\,000\,\text{Hz}$ , which is $20\,\text{kHz}$ .

Answer: $20\,\text{Hz}$ to $20\,\text{kHz}$

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

Explain how a sound wave travelling through air can produce the sensation of sound in a listener.
[2 marks]
Total for this question: 2
A membrane is exposed to tones at $10\,\text{Hz}$ , $100\,\text{Hz}$ , $5.0\,\text{kHz}$ and $28\,\text{kHz}$ . It vibrates detectably only for the middle two tones. Explain what these results show about sound-to-solid conversion and normal human hearing.
[3 marks]
Total for this question: 3

Tier 2 · Standard

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
A microphone trace produced by a vibrating tuning fork contains six complete cycles in $1.5\,\text{ms}$ . Determine the sound frequency, decide whether it is within normal human hearing and describe the conversion from the fork to the trace.
[5 marks]
Total for this question: 5
A solid source is driven at $26\,\text{kHz}$ and a sensor confirms it is vibrating. A nearby listener reports no sound. A student claims this proves that the source failed to convert its vibration into a sound wave. Correct the claim.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A hearing test shows that a person detects tones from $40\,\text{Hz}$ to $13\,\text{kHz}$ 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
Sensor A converts sound to solid vibration from $100\,\text{Hz}$ to $5.0\,\text{kHz}$ ; sensor B works from $1.0\,\text{kHz}$ to $18\,\text{kHz}$ . Calculate the width of each sensor's frequency range and how many times wider B's range is. Compare their ability to reproduce low frequencies and a broad range of frequencies.
[5 marks]
Total for this question: 5
A signal must be converted by a source that works from $0.50$ to $18\,\text{kHz}$ , detected by a diaphragm that works from $2.0$ to $22\,\text{kHz}$ , and heard by a person with the normal $20\,\text{Hz}$ to $20\,\text{kHz}$ range. Candidate frequencies are $15\,\text{Hz}$ , $0.80\,\text{kHz}$ , $15\,\text{kHz}$ and $24\,\text{kHz}$ . Select the only candidate that satisfies all three constraints and explain why the others fail.
[5 marks]
Total for this question: 5
A sound system sends a signal from a loudspeaker to a microphone through these five stages: an alternating current in the loudspeaker wires, motion of the loudspeaker cone, the disturbance crossing the air, motion of the microphone diaphragm and the microphone's alternating output. Put the stages in order and classify every stage as an electrical signal, a sound wave or a solid vibration.
[4 marks]
Total for this question: 4
Describe a method to determine a person's upper hearing limit. You have a signal generator, a loudspeaker, a sound-level meter, a ruler and a quiet room. Include how frequency is changed, how equal loudness is maintained at each frequency, how responses are checked and how the result is compared with $20\,\text{kHz}$ .
[6 marks]
Total for this question: 6

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

Explanation

Ultrasound is sound above $20\,\text{kHz}$ ; 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 $d=vt/2$ .
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 $2$ 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.The upper limit of normal human hearing is about $20\,\text{kHz}$ , 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 $20\,\text{kHz}$ ; the pulse is partially reflected at the boundary.

Common mistakes

Don't omit the factor of $2$ 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

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
An ultrasound detector trace has echo peaks at $18\,\mu\text{s}$ and $42\,\mu\text{s}$ after the transmitted pulse. Without calculating distances, identify which echo comes from the nearer boundary and explain your choice.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An echo sounder sends a pulse vertically down through seawater. The echo returns after $0.084\,\text{s}$ . The speed of sound in seawater is $1500\,\text{m s}^{-1}$ . Calculate the water depth.
[3 marks]
Total for this question: 3
An ultrasound pulse produces echoes from the front and back surfaces of a wall at $14\,\mu\text{s}$ and $26\,\mu\text{s}$ . The wave speed in the wall material is $3200\,\text{m s}^{-1}$ . Calculate the wall thickness to 2 significant figures.
[4 marks]
Total for this question: 4
Identical ultrasound pulses produce echoes after $12\,\mu\text{s}$ , $34\,\mu\text{s}$ and $60\,\mu\text{s}$ . The echoes contain $4\%$ , $11\%$ and $6\%$ of the transmitted energy respectively. Identify the nearest boundary, the farthest boundary and the boundary that returns the greatest reflected fraction.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Two probes are tested on an opaque solid casing. For probe A, $70\%$ of the wave enters the casing and a separate sharp echo returns from an internal air gap. For probe B, only $5\%$ enters and no separate echo is detected. Select the better probe for locating the hidden gap and explain how its trace and known wave speed could reveal the gap position.
[5 marks]
Total for this question: 5
An ultrasound pulse passes through a $12\,\text{mm}$ coating at $2400\,\text{m s}^{-1}$ and then through an unnamed solid at $3600\,\text{m s}^{-1}$ . An echo from a flaw returns $30\,\mu\text{s}$ after transmission. Calculate the flaw's depth below the coating surface.
[5 marks]
Total for this question: 5
A station is $600\,\text{km}$ from an earthquake along a route through solid rock. The P-wave speed is $8.0\,\text{km s}^{-1}$ and the S-wave speed is $4.8\,\text{km s}^{-1}$ . Calculate the S–P arrival delay. A second station records the P-wave but no direct S-wave. Explain what the missing arrival adds to the evidence.
[5 marks]
Total for this question: 5
A ship is above water of known depth $36\,\text{m}$ . Its echo sounder records $48\,\text{ms}$ between sending a pulse and receiving the seabed echo. Calculate the speed of sound in the water and explain the factor of two in the calculation.
[5 marks]
Total for this question: 5

4.6.2.1 · Types of electromagnetic waves

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 $3.00\times10^8\,\text{m s}^{-1}$ ; for example, $f=5.0\times10^{10}\,\text{Hz}$ gives $\lambda=6.0\times10^{-3}\,\text{m}$ .
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.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

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
A spectrum diagram places X-rays below ultraviolet in frequency and labels gamma rays as longitudinal waves. Correct both errors.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An electromagnetic wave has frequency $5.0\times10^{10}\,\text{Hz}$ . Calculate its wavelength in a vacuum and identify its region of the spectrum. Use $c=3.00\times10^8\,\text{m s}^{-1}$ .
[3 marks]
Total for this question: 3
Four electromagnetic signals have wavelengths $2.0\,\text{m}$ , $4.0\,\text{mm}$ , $600\,\text{nm}$ and $0.020\,\text{nm}$ . Match them to radio, microwave, visible light and X-rays, then put the four signals in increasing frequency order.
[4 marks]
Total for this question: 4
Signals U and V cross the same vacuum distance. U has wavelength $700\,\text{nm}$ and V has wavelength $350\,\text{nm}$ . Compare their frequencies and arrival times, and correct the claim that V arrives first because its wavelength is shorter.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A radio signal has frequency $75\,\text{MHz}$ and a microwave signal has frequency $3.0\,\text{GHz}$ . Calculate both wavelengths in air using $3.00\times10^8\,\text{m s}^{-1}$ , then compare their speeds and wavelengths.
[5 marks]
Total for this question: 5
Three electromagnetic signals of frequencies $1.0\times10^8\,\text{Hz}$ , $3.0\times10^{13}\,\text{Hz}$ and $1.0\times10^{15}\,\text{Hz}$ cross $6.0\times10^7\,\text{m}$ of vacuum in $0.20\,\text{s}$ . Calculate their common speed, identify the three regions and explain why their arrival times match.
[6 marks]
Total for this question: 6
A table for three electromagnetic waves gives P: $f=1.0\times10^8\,\text{Hz}$ , $\lambda=3.0\,\text{m}$ , radio; Q: $f=6.0\times10^{14}\,\text{Hz}$ , $\lambda=5.0\times10^{-7}\,\text{m}$ , visible; R: $f=6.0\times10^{18}\,\text{Hz}$ , $\lambda=5.0\times10^{-10}\,\text{m}$ , X-ray. For this audit, their shared speed in vacuum is $3.00\times10^8\,\text{m s}^{-1}$ ; identify the incorrect wavelength and calculate its correction.
[5 marks]
Total for this question: 5
An infrared source and a detector plate are separated inside an evacuated chamber. When the source is switched on, the plate's temperature rises from $18\,{}^\circ\text{C}$ to $27\,{}^\circ\text{C}$ without contact between them. Explain what the observation shows about electromagnetic waves, state their wave type and compare the infrared speed in the chamber with the speed of gamma rays there.
[4 marks]
Total for this question: 4
Three electromagnetic waves have frequencies: radio, $1.0\times10^8\,\text{Hz}$ ; visible light, $5.0\times10^{14}\,\text{Hz}$ ; and X-rays, $1.0\times10^{18}\,\text{Hz}$ . An ultraviolet wave has frequency $1.0\times10^{16}\,\text{Hz}$ . Place ultraviolet between the correct two named waves, justify the placement, explain how all four belong to one continuous spectrum and state which the human eye detects directly.
[4 marks]
Total for this question: 4

4.6.2.2 · Properties of electromagnetic waves 1

Explanation

Construct a refraction ray diagram using a normal at the boundary.
Higher only: the amounts absorbed, transmitted, reflected or refracted depend on the material and wavelength.
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.

Worked example

A light ray enters glass from air at an angle to the normal. State how the ray changes direction and name the line from which the angles are measured.

1.Apply the refraction ray rule for light entering glass from air: draw the refracted ray closer to the normal. The normal is perpendicular to the boundary at the point where the ray enters, and both angles are referenced to it.

Answer: The ray bends towards the normal; angles are measured from the normal.

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.

Exam tip

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

Tier 1 · Easy

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
A ray enters a glass block from air. The drawing shows it bending away from the normal and labels both angles from the block surface. Identify the two diagram errors and describe the corrected ray.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Higher only: parallel wavefronts are $2.4\,\text{cm}$ apart in medium A and $1.5\,\text{cm}$ apart in medium B. The frequency is unchanged. Calculate $v_B/v_A$ and explain what the wavefront spacing shows.
[4 marks]
Total for this question: 4
In RP₉, a student shines a ray through a transparent block but marks only the entry and exit points, removes the block before tracing its outline, and measures one angle from the block edge. Repair the procedure so refraction by two substances can be compared.
[5 marks]
Total for this question: 5
A light ray enters one face of a parallel-sided transparent block from air and later leaves through the opposite face. Describe where to draw the two normals, how the ray bends at each face and how the emerging ray is directed relative to the incident ray.
[4 marks]
Total for this question: 4

Tier 3 · Hard

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
Higher only: an oblique wavefront enters a slower transparent material. A student's diagram draws closer wavefronts in the new material but keeps their direction unchanged and claims the frequency fell. Critique the diagram and explain the corrected refraction process.
[5 marks]
Total for this question: 5
Higher only: the same material receives $200\,\text{mJ}$ pulses at two wavelengths. For red light, $30\,\text{mJ}$ is reflected and $140\,\text{mJ}$ is transmitted. For infrared, $20\,\text{mJ}$ is reflected and $80\,\text{mJ}$ is transmitted. Calculate the absorbed percentage for each wavelength and state what the comparison demonstrates.
[5 marks]
Total for this question: 5
The same light ray enters transparent materials P and Q from air at $50^\circ$ to the normal. Its refracted angles are $29^\circ$ in P and $35^\circ$ in Q. Calculate the change in angle for each material, identify which material refracts the ray more and describe the ray-diagram evidence for your choice.
[4 marks]
Total for this question: 4
Higher only: a material proposed as a window coating has absorbed fractions $0.04$ at $450\,\text{nm}$ , $0.06$ at $550\,\text{nm}$ , $0.08$ at $650\,\text{nm}$ and $0.78$ at $10\,\mu\text{m}$ . Reflection is negligible. Recommend whether this coating should be used for a window that must transmit visible light but absorb infrared. Calculate the transmitted fraction at each wavelength and justify the recommendation from the data.
[6 marks]
Total for this question: 6

4.6.2.3 · Properties of electromagnetic waves 2

Explanation

Higher only: oscillations in electrical circuits can produce radio waves, and absorbed radio waves can induce an alternating current of the same frequency in a receiving circuit.
Use dose data by converting units consistently: $1000\,\text{mSv}=1\,\text{Sv}$ , 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 $180\,\text{mSv}$ . Convert this dose to sieverts.

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

Answer: $0.180\,\text{Sv}$

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

State one harmful effect of ultraviolet radiation and one harmful effect of X-rays or gamma rays.
[2 marks]
Total for this question: 2
Source P emits after a change in an atomic nucleus, while source R emits after a change in an atom. Which source must produce gamma rays?
[1 mark]
Total for this question: 1

Tier 2 · Standard

Procedure A gives a dose of $4.5\,\text{mSv}$ on each of six visits. Procedure B gives one dose of $18\,\text{mSv}$ . Calculate the total dose for A and use the data to compare the radiation risk.
[3 marks]
Total for this question: 3
A patient's dose limit for an imaging procedure is $0.0016\,\text{Sv}$ . Each scan image gives a dose of $0.60\,\text{mSv}$ . Calculate the greatest whole number of images that can be taken without exceeding the limit, and state one assumption in your calculation.
[4 marks]
Total for this question: 4
A technician works beside an ultraviolet curing lamp and an X-ray imaging unit. For each source, state the main biological hazard and explain one suitable way to reduce the technician's exposure.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Higher only: a transmitter circuit oscillates at $92\,\text{MHz}$ . Explain how it produces a radio wave and how a tuned receiving circuit can produce a signal at $92\,\text{MHz}$ . Contrast this origin with the origin of gamma rays.
[5 marks]
Total for this question: 5
Higher only: a receiving aerial produces an alternating-current trace with one complete cycle every $0.25\,\mu\text{s}$ . Determine the current frequency and explain how the transmitter and receiver produce matching-frequency oscillations.
[5 marks]
Total for this question: 5
Higher only: design an investigation to test the claim that a radio wave absorbed by a receiving circuit induces an alternating current at the same frequency as the transmitter. Include the measurements, controls, range of readings and expected result.
[5 marks]
Total for this question: 5
Procedure A uses X-rays and gives a dose of $0.0018\,\text{Sv}$ . Procedure B uses gamma rays and gives a dose of $0.45\,\text{mSv}$ . Convert the dose from A to millisieverts, calculate the total dose from both procedures and compare their contributions to the radiation risk.
[6 marks]
Total for this question: 6
Higher only: a transmitter circuit oscillates at $8.0\,\text{MHz}$ , but a receiver is observed producing an alternating-current trace at $4.0\,\text{MHz}$ , half the transmitter frequency. Explain why this contradicts the specified transmitter–receiver process and identify one measurement error that would explain the trace.
[5 marks]
Total for this question: 5

4.6.2.4 · Uses and applications of electromagnetic waves

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 only: 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.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

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
Name the electromagnetic wave used in each case: an energy-efficient lamp, communication through an optical fibre and an image of a broken bone.
[3 marks]
Total for this question: 3

Tier 2 · Standard

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
Higher only: a rescue system must send a signal to a satellite, locate warm people in darkness and produce an image of bones. Select a different electromagnetic wave for each job and link one relevant property to its suitability.
[6 marks]
Total for this question: 6
Higher only: three unlabelled waves have these measured interactions. Wave A passes through the atmosphere with little absorption. Wave B is emitted more strongly by warmer objects. Wave C passes through soft tissue but is absorbed more strongly by bone. Match A, B and C to satellite communication, thermal imaging and bone imaging, naming each wave. Explain each choice.
[6 marks]
Total for this question: 6

Tier 3 · Hard

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
Higher only: a designer proposes radio waves for a thermal camera, infrared for a satellite link and gamma rays for browning bread. Replace each choice with a suitable wave and explain the physical reason for every correction.
[6 marks]
Total for this question: 6
Higher only: a lighting and medical-equipment plan makes two claims: visible light should excite the coating in an energy-efficient lamp, and X-rays should sterilise instruments inside sealed packs. Evaluate both claims. Name the more suitable wave for each job, explain the relevant interaction, and state a medical use for which X-rays are suitable.
[5 marks]
Total for this question: 5
Higher only: choose a suitable electromagnetic wave for each of these jobs: carrying a television broadcast, heating the surface of an object in an electric heater and producing a suntan. Explain the relevant interaction or property for every choice and state the hazard associated with the tanning wave.
[6 marks]
Total for this question: 6
Higher only: gamma rays are used both to treat a tumour inside a patient and to sterilise food. Explain why the same electromagnetic wave is suitable for these very different jobs, and state how the dose, target and exposure time must differ between the two applications.
[6 marks]
Total for this question: 6

4.6.2.5 · Lenses (physics only)

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 $\text{image height}/\text{object height}$ and has no unit; an image $4.5\,\text{cm}$ high from an object $1.5\,\text{cm}$ high has magnification $3.0$ .
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 lens forms an image $3.6\,\text{cm}$ high from an object $1.2\,\text{cm}$ high. Calculate the magnification.

1.Use magnification $=\text{image height}/\text{object height}$ . Therefore magnification $=3.6/1.2=3.0$ . It is a ratio, so it has no unit.

Answer: magnification $=3.0$

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.

Exam tip

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

Tier 1 · Easy

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
A concave-lens diagram shows a parallel incident ray leaving the lens towards the far principal focus. Identify the error and describe the correct construction for that ray.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
A $1.8\,\text{cm}$ object gives repeated image heights of $4.5$ , $4.6$ , $7.9$ and $4.4\,\text{cm}$ without the apparatus being moved. Identify the anomalous result and use the other readings to calculate the mean magnification.
[4 marks]
Total for this question: 4
Lens P forms an upright image that cannot be caught on a screen. Lens Q forms an inverted image that is caught on a screen. Identify which lens must be convex, and explain why the observation for P does not uniquely identify its lens type.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A lens produces a sharp image on a screen. The image is $7.2\,\text{cm}$ high and the magnification is $3.0$ . Calculate the object height, identify the lens as convex or concave, and justify your choice.
[4 marks]
Total for this question: 4
An object is placed between a convex lens and its principal focus. A student draws an inverted real image on the far side of the lens and says it can be caught on a screen. Critique the ray diagram and state the correct image properties and location.
[5 marks]
Total for this question: 5
Audit three lens records. A: object $12\,\text{mm}$ , image $36\,\text{mm}$ , magnification $3.0$ , screen used, convex lens. B: object $20\,\text{mm}$ , image $10\,\text{mm}$ , magnification $0.50$ , screen used, concave lens. C: object $1.5\,\text{cm}$ , image $30\,\text{mm}$ , magnification $20$ . Identify the errors in B and C and give the corrections.
[5 marks]
Total for this question: 5
Parallel rays are brought to a focus $12\,\text{cm}$ from a lens. An object is then placed $30\,\text{cm}$ from the lens. State the lens type and focal length, show that the object is beyond twice the focal length, describe two construction rays and give all four image properties including whether it can be projected.
[5 marks]
Total for this question: 5
A display designer needs one lens setup to form an enlarged image on a screen and another to give an upright, diminished view that cannot be projected. Choose a convex or concave lens for each setup, state how the object must be positioned for the enlarged projected image, and explain the image properties using ray behaviour.
[6 marks]
Total for this question: 6

4.6.2.6 · Visible light (physics only)

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.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

Compare the reflection of a narrow beam of light from a smooth mirror and from rough paper.
[2 marks]
Total for this question: 2
Material A lets light through and a sharp image can be seen through it. Material B lets light through but the image is blurred. Classify A and B and distinguish both from an opaque material.
[3 marks]
Total for this question: 3

Tier 2 · Standard

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
Equal red, green and blue light shines on an opaque surface. A detector measures reflected signals of $12$ , $78$ and $9$ units for red, green and blue respectively. Predict the surface colour and explain what happens to most of the other incident wavelengths.
[3 marks]
Total for this question: 3
White light passes through an ideal red filter and then an ideal blue filter. Predict the light emerging from the second filter and explain the result using absorption and transmission.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Panel P transmits $88\%$ of incident light and preserves $95\%$ of a test grid's contrast. Panel Q transmits $55\%$ but preserves only $8\%$ contrast. Panel R transmits no light. Classify the panels, then select one for a clear window, one for a privacy screen and one for a blackout panel, justifying each choice.
[6 marks]
Total for this question: 6
Light that has passed through a filter carries red, green and blue signals of $60$ , $30$ and $10$ units and reaches an opaque surface. The surface reflects $20\%$ of incident red, $70\%$ of incident green and $40\%$ of incident blue. Calculate the three reflected signals, predict the dominant observed colour, and then predict the view through an ideal red filter.
[5 marks]
Total for this question: 5
A narrow white beam and a movable light detector are used with a glossy red card and a rough red card. Predict how the detector readings vary with angle for each card, explain why both cards still look red in white light, and predict their appearance when viewed through an ideal blue filter.
[5 marks]
Total for this question: 5
White light passes through a clear blue filter, then through a frosted colourless pane and finally reaches an opaque green card. Explain the wavelength and image changes at each stage. Predict the appearance of the card and what an observer would see if the card were removed.
[5 marks]
Total for this question: 5

4.6.3.1 · Emission and absorption of infrared radiation (physics only)

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 $35\,{}^\circ\text{C}$ and $75\,{}^\circ\text{C}$ . State which emits more infrared radiation each second.

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

Answer: The object at $75\,{}^\circ\text{C}$ 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

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
A matt-black reference gives an infrared detector reading of $9.2$ units and a polished-silver reference gives $2.1$ units. An unlabelled test surface gives $6.4$ units. The two candidate surfaces are the same colour: one is matt and one is highly polished. State which candidate the test surface is more likely to be. Justify the choice and state what must be held constant for this inference to be valid.
[3 marks]
Total for this question: 3

Tier 2 · Standard

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
In RP₁₀, a student tests infrared absorption using plates of different areas, places each at a different distance from the lamp, starts them at different temperatures and records one final temperature. Give four corrections that make the surface comparison valid and repeatable.
[4 marks]
Total for this question: 4
Two unlabelled surfaces are tested under identical conditions. Surface A absorbs $75\%$ of incident infrared and gives an emission reading of $7.5$ units at a fixed temperature. Surface B absorbs $25\%$ and gives $2.5$ units at the same temperature. Calculate the A:B ratio for absorption and for emission, then state the relationship supported by the data.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Plates that are identical apart from their surface finish are placed under the same infrared lamp. Their temperature rises are shown below. Matt-black trials: $12.1$ , $12.4$ , $12.3\,{}^\circ\text{C}$ . Polished-white trials: $7.4$ , $7.6$ , $12.0\,{}^\circ\text{C}$ . Identify any anomaly, calculate representative mean rises and use them to conclude which surface is the better absorber.
[5 marks]
Total for this question: 5
Three unlabelled panels at the same temperature have different emitting areas. A detector's total signal is proportional to emitted infrared power. Panel P: $20\,\text{cm}^2$ and $90$ units; Q: $50\,\text{cm}^2$ and $175$ units; R: $30\,\text{cm}^2$ and $60$ units. Calculate the signal per square centimetre, rank their finishes as emitters and predict the absorber ranking if finish is the only other difference.
[5 marks]
Total for this question: 5
Two identical metal plates have matt-black and polished surfaces. Under the same infrared lamp, both start at $20\,{}^\circ\text{C}$ ; after five minutes they are at $42\,{}^\circ\text{C}$ and $31\,{}^\circ\text{C}$ respectively. The plates are then both set to $60\,{}^\circ\text{C}$ with the lamp off; after ten minutes they are at $38\,{}^\circ\text{C}$ and $50\,{}^\circ\text{C}$ . Calculate the temperature change in each stage and explain the linked conclusions about the surfaces.
[5 marks]
Total for this question: 5
Describe an investigation that compares infrared emission from three identical metal plates with matt-black, matt-white and polished-silver finishes. You have a heater, a temperature sensor, an infrared detector, a ruler and a clamp stand. Keep the plates at one stated temperature, include repeats and explain why cooling-rate data alone would not isolate radiation.
[6 marks]
Total for this question: 6

4.6.3.2 · Perfect black bodies and radiation (physics only)

Explanation

All objects emit radiation, and both the intensity and wavelength distribution of the emitted radiation depend on temperature.
Higher only: for an energy-balance question, compare the incoming radiation absorbed each second with the radiation emitted each second; equal rates mean constant temperature.
Higher only: if a body absorbs $480\,\text{J}$ each second but emits $530\,\text{J}$ each second, it has a net energy loss of $50\,\text{J s}^{-1}$ 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.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

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
Two identical objects produce emission curves. Curve A has greater intensity than curve B across the measured wavelengths, but both curves are above zero. State what can be inferred about their temperatures and correct the claim that the cooler object emits no radiation.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Higher only: a body absorbs radiation at $480\,\text{W}$ and emits radiation at $530\,\text{W}$ . Calculate the net rate of energy change and state what happens to its temperature.
[3 marks]
Total for this question: 3
Higher only: three otherwise identical bodies have absorbed and emitted radiation powers as follows: A, $320\,\text{W}$ and $270\,\text{W}$ ; B, $415\,\text{W}$ and $415\,\text{W}$ ; C, $260\,\text{W}$ and $345\,\text{W}$ . Identify the body in thermal equilibrium. Calculate the net powers of the other two and compare their temperature changes.
[4 marks]
Total for this question: 4
Three otherwise identical bodies give these emission-curve data. P has a peak intensity of $18$ units at $4.0\,\mu\text{m}$ ; Q has $7$ units at $6.5\,\mu\text{m}$ ; R has $12$ units at $5.0\,\mu\text{m}$ . Order their temperatures from highest to lowest and state one conclusion about their temperatures that the data cannot support.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Higher only: Earth receives an average solar power of $340\,\text{W m}^{-2}$ . Initially $102\,\text{W m}^{-2}$ is reflected and $238\,\text{W m}^{-2}$ is emitted to space. The reflected power then decreases to $90\,\text{W m}^{-2}$ 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
Higher only: for $15$ minutes a roof absorbs radiation at $720\,\text{W}$ and emits at $640\,\text{W}$ . Cloud then reduces absorption to $590\,\text{W}$ for $10$ minutes while emission remains $640\,\text{W}$ . Assuming only these radiation transfers affect it, calculate the energy change in each interval and the overall change, then describe the temperature trend.
[6 marks]
Total for this question: 6
Higher only: a body absorbs radiation at a constant $420\,\text{W}$ . Its emitted power is $350\,\text{W}$ at $20\,{}^\circ\text{C}$ , $420\,\text{W}$ at $30\,{}^\circ\text{C}$ and $510\,\text{W}$ at $40\,{}^\circ\text{C}$ . Calculate the net power at $20\,{}^\circ\text{C}$ and $40\,{}^\circ\text{C}$ , identify the equilibrium temperature and explain why the data describe a stable equilibrium.
[5 marks]
Total for this question: 5
A matt-black ceramic block and a polished copper block are both held at $80\,{}^\circ\text{C}$ . At the same distance, an infrared detector gives a larger reading for the ceramic block. A student concludes that the ceramic block must be hotter. Explain why this hotter-or-cooler inference fails and state what would be needed to compare temperatures using infrared emission.
[4 marks]
Total for this question: 4
Higher only: a dark paving slab in sunlight warms rapidly and later reaches a constant temperature although sunlight continues. A reflective cover is then placed over it, and the slab eventually settles at a lower constant temperature. Explain both temperature changes using absorption, emission and reflection of radiation, including why the emitted power changes as the slab's temperature changes.
[5 marks]
Total for this question: 5

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.6.1.1 · Transverse and longitudinal waves

Tier 1 · Easy

Mark scheme for 4.6.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	In a transverse wave, the oscillations are perpendicular to the direction of energy transfer. In a longitudinal wave, the oscillations are parallel to the direction of energy transfer.	Use the energy-transfer direction as the reference. Transverse means across that direction; longitudinal means along the same direction.	2
Total Question 1			2
02.1	It is a longitudinal wave; the high-pressure and low-pressure regions are compressions and rarefactions.	The air particles oscillate parallel to the direction of energy transfer, so the wave is longitudinal. Regions where particles are closer together are compressions, while regions where they are farther apart are rarefactions.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.6.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The cork oscillates mainly up and down and returns close to its starting position; the ripple and energy travel across the tank, but the water does not travel across with them.	Track the marker rather than the crest. The cork follows the local water motion, so it oscillates as the disturbance passes. Because it has no sustained motion to the far side, the observation shows that the wave transfers energy without a net transfer of the water.	3
Total Question 1			3
02.1	The pulse is transverse because the marked point oscillates vertically while the pulse and energy travel horizontally; the rope point returns about its fixed position rather than travelling with the pulse.	Compare the two measured directions. Vertical displacement is perpendicular to horizontal energy transfer, which identifies a transverse wave. The fixed horizontal position is evidence that material in the rope is not transported along with the pulse.	3
Total Question 2			3
03.1	The photograph shows only the rope's shape at one instant, not the direction in which a point on the rope oscillates, so the wave type cannot be identified from this evidence. Track a marked point over time: north-south or vertical oscillation is perpendicular to eastward energy transfer and therefore transverse, whereas east-west oscillation is parallel and therefore longitudinal.	Do not infer the wave type from the rope's photographed shape. Classification requires the direction of oscillation as well as the energy-transfer direction. Establish the eastward energy transfer, track one marked point through a cycle and compare its oscillation direction with east.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	A coil oscillates backwards and forwards about a fixed position while the pulse and energy move along the spring; the oscillations are parallel to the direction of travel and the pulse contains compressed and less-compressed regions.	Separate local motion from wave motion. Each coil moves to and fro along the spring rather than being carried with the pulse. This particle motion is parallel to the travelling disturbance, and the changing coil spacing forms compression and rarefaction regions. Those are the defining features of a longitudinal wave.	4
Total Question 1			4
02.1	P has longitudinal particle motion but should show compressions and rarefactions, so it is an incomplete longitudinal-wave model. Q correctly represents a transverse wave because its oscillations are perpendicular to energy transfer and it has crests and troughs.	Test direction and structure separately. Parallel oscillations identify P as longitudinal, but changing particle spacing is needed to represent its compressions and rarefactions. Perpendicular oscillations identify Q as transverse, and its crest-trough shape is consistent with that classification.	4
Total Question 2			4
03.1	R is transverse because its particle oscillates north-south, perpendicular to the eastward energy transfer. S is longitudinal because its particle oscillates east-west, parallel to the energy transfer. Both displacement sequences return to zero, so each particle oscillates about a fixed position rather than travelling with the wave.	Compare each changing coordinate with the stated energy direction. R changes only in the perpendicular coordinate; S changes only in the parallel coordinate. The repeated return to the starting coordinate distinguishes local oscillation from the onward transfer of energy.	5
Total Question 3			5
04.1	The distance–displacement snapshot gives the wavelength, $0.72\,\text{m}$ , because it shows the spatial repeat. The displacement–time trace gives the period, $0.30\,\text{s}$ , because it shows the time for one coil to repeat its motion. Only the time trace states that the coil oscillates along the spring, parallel to the wave travel, so it fixes the wave as longitudinal; a snapshot alone does not show the direction of oscillation over time. Regions where the coils are closer together are compressions.	Read spatial repetition from the distance axis and temporal repetition from the time axis. Do not use the snapshot to infer the direction in which one coil moves. The trace explicitly records motion along the spring, so the oscillation is parallel to energy transfer and the wave is longitudinal.	5
Total Question 4			5
05.1	For the ripple, film the floating marker as a pulse passes: it should oscillate and return near its starting position while the ripple reaches the far side. For sound, place the two microphones at measured positions along the pulse path: the nearer microphone should detect the pulse first and the farther microphone later, showing onward energy transfer. The air-movement detector should show no sustained air movement in the pulse direction, so the air is not transported with the sound. Repeat the observations to check that the pattern is consistent.	Use the marker to distinguish local water motion from ripple travel. Use the microphones to show that the sound disturbance reaches successive positions, and use the air-movement detector to test for net transport of air. The combined observations separate energy transfer from bulk movement of either medium.	6
Total Question 5			6

4.6.1.2 · Properties of waves

Tier 1 · Easy

Mark scheme for 4.6.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Amplitude $=3.0\,\text{cm}$ , or equivalently $0.030\,\text{m}$ Wavelength $=0.40\,\text{m}$ , or equivalently $40\,\text{cm}$	Amplitude is the maximum displacement from the undisturbed line. Wavelength is the distance between equivalent adjacent points, such as one crest and the next. Either unit is acceptable provided it is stated.	2
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02.1	Amplitude is the vertical distance from the undisturbed line to a crest or trough; wavelength is the horizontal distance between equivalent adjacent points, such as crest to crest.	The labelled vertical distance is twice the amplitude, and the labelled horizontal distance is half a wavelength. Both quantities must use the undisturbed or equivalent-point definitions.	2
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Tier 2 · Standard

Mark scheme for 4.6.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$\lambda=0.040\,\text{m}$ $f=2.5\,\text{Hz}$ $v=0.10\,\text{m s}^{-1}$	Nine complete intervals give $\lambda=0.36/9=0.040\,\text{m}$ . The frequency is $f=5/2.0=2.5\,\text{Hz}$ . Therefore $v=f\lambda=(2.5)(0.040)=0.10\,\text{m s}^{-1}$ .	4
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02.1	mean time $=8.0\,\text{s}$ and mean five-wavelength span $=0.30\,\text{m}$ $f=2.5\,\text{Hz}$ , $\lambda=0.060\,\text{m}$ and $v=0.15\,\text{m s}^{-1}$	The mean time is $(8.0+7.8+8.2)/3=8.0\,\text{s}$ , so $f=20/8.0=2.5\,\text{Hz}$ . The mean span is $(0.30+0.31+0.29)/3=0.30\,\text{m}$ , so $\lambda=0.30/5=0.060\,\text{m}$ . Apply $v=f\lambda$ : $v=(2.5)(0.060)=0.15\,\text{m s}^{-1}$ .	5
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03.1	Row C's claimed speed is inconsistent. Its speed is $(6.0)(0.10)=0.60\,\text{m s}^{-1}$ . All three corrected rows therefore show the same wave speed in this region.	Check each row using $v=f\lambda$ . A gives $0.60\,\text{m s}^{-1}$ and B gives $0.60\,\text{m s}^{-1}$ . C must also give $(6.0)(0.10)=0.60\,\text{m s}^{-1}$ , so only its printed speed is wrong.	4
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Tier 3 · Hard

Mark scheme for 4.6.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	wavelength in air $=0.140\,\text{m}$ wavelength in water $=0.600\,\text{m}$ The wavelength increases because the speed increases while the frequency remains constant.	Convert $2.40\,\text{kHz}$ to $2400\,\text{Hz}$ . Rearrange $v=f\lambda$ to $\lambda=v/f$ . In air, $\lambda=336/2400=0.140\,\text{m}$ . In water, $\lambda=1440/2400=0.600\,\text{m}$ . The source fixes the frequency, so the higher speed in water requires a proportionally longer wavelength.	5
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02.1	One short distance and one short period give large fractional uncertainties. Measure several wavelengths and divide by their number; time many oscillations and divide by their number; repeat and average, then calculate frequency and use wave speed equals frequency times wavelength.	Use a ruler across several clear crest-to-crest intervals, perpendicular to the wavefronts, and divide the total span by the number of wavelengths. Count many oscillations at a fixed point while timing them, or use slow-motion video or a strobe, then divide the total time by the cycle count. Repeat both measurements, reject justified anomalies and use means. Convert centimetres to metres before applying $f=1/T$ and $v=f\lambda$ .	6
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03.1	The wavelengths are $0.30\,\text{m}$ at $2.0\,\text{Hz}$ , $0.12\,\text{m}$ at $5.0\,\text{Hz}$ and $0.040\,\text{m}$ at $15\,\text{Hz}$ . The grid can resolve the $2.0\,\text{Hz}$ and $5.0\,\text{Hz}$ settings, but not the $15\,\text{Hz}$ setting.	Rearrange $v=f\lambda$ to $\lambda=v/f$ . The three calculations are $0.60/2.0=0.30\,\text{m}$ , $0.60/5.0=0.12\,\text{m}$ and $0.60/15=0.040\,\text{m}$ . Convert the resolution threshold explicitly: $8.0\,\text{cm}=0.080\,\text{m}$ .	5
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04.1	The mean delay is $5.3\,\text{ms}$ , so the mean speed is $1.80/(5.3\times10^{-3})=3.4\times10^2\,\text{m s}^{-1}$ to 2 significant figures. Repeating and averaging reduces the effect of random timing variation. A large measured microphone separation makes the delay larger, reducing the percentage uncertainty in the delay and distance measurements.	Average the three delays: $(5.2+5.4+5.3)/3=5.3\,\text{ms}=5.3\times10^{-3}\,\text{s}$ . Use speed $=$ distance divided by time to obtain $339.6\,\text{m s}^{-1}$ , which rounds to $3.4\times10^2\,\text{m s}^{-1}$ . Then link repeats and a longer timing interval to reduced uncertainty.	5
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05.1	Set one frequency and wait for a steady pattern. Record the cord from directly above with the metre ruler in the same plane. Measure the distance covering several complete wavelengths and divide by their number. Read the frequency from the signal generator and calculate $v=f\lambda$ . Repeat the wavelength measurement and calculate a mean, then repeat for several frequencies while keeping the cord tension and length fixed. Compare the calculated speeds and identify any anomalous result.	Introduce a measurable scale in the plane of the moving cord and reduce percentage uncertainty by measuring several wavelengths rather than one. Pair each mean wavelength with its displayed frequency, calculate $v=f\lambda$ , and keep the properties of the cord constant so the calculated speeds can be compared validly.	6
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4.6.1.3 · Reflection of waves (physics only)

Tier 1 · Easy

Mark scheme for 4.6.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Some energy may be reflected. Some energy may be transmitted. Some energy may be absorbed.	Account for all incident wave energy at the boundary. It can return into the first material, pass into the second material or transfer to the materials.	3
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02.1	The reflected ray must be $28^\circ$ to the normal, so it is $62^\circ$ to the surface.	The law of reflection compares angles measured from the normal. Correct the reflected angle to $28^\circ$ from the normal, then use the right angle between normal and surface: $90^\circ-28^\circ=62^\circ$ .	2
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Tier 2 · Standard

Mark scheme for 4.6.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$15\%$ is absorbed; energy travels into the second material, returns into the first material and is transferred to the materials at the boundary.	The incident energy represents $100\%$ . The absorbed fraction is $100-68-17=15\%$ . Transmission carries energy onward through the boundary, reflection carries energy back, and absorption transfers wave energy to the material.	3
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02.1	The sharp maximum at $35^\circ$ shows that a smooth surface reflects most of the beam in one direction, and the reflected angle equals the $35^\circ$ incident angle.	Locate the detector angle with the strongest signal. Its match to the incident angle supports angle of incidence equals angle of reflection. Concentration around one direction, rather than similar readings across many angles, indicates that the surface is smooth and sends most reflected light along one path.	3
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03.1	The incident energy is $75\,\text{J}$ and the reflected energy is $15\,\text{J}$ .	Let the incident energy be $E$ . Energy conservation gives $E=42+18+0.20E$ . Therefore $0.80E=60$ and $E=75\,\text{J}$ . The reflected energy is $0.20(75)=15\,\text{J}$ ; the check is $42+18+15=75\,\text{J}$ .	4
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Tier 3 · Hard

Mark scheme for 4.6.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Direct the same light beam at each surface at the same angle; measure reflected light intensity with a light sensor at equal angles and distances; control source brightness, ambient light and illuminated area; repeat and compare mean reflected intensities or angular distributions.	Fix a lamp or ray box and mark one incidence angle. Put the sensor the same distance from the point of incidence and take readings at matching reflection angles for each surface, shielding the setup from ambient light. Keep the source, brightness, colour, illuminated area and geometry unchanged. Repeat each reading, calculate means and compare either the peak reflected intensity or readings across several detector angles. The smooth tile should give a more concentrated reflected beam, whereas the rough card scatters light over more directions.	5
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02.1	Draw the normal perpendicular to the surface; measure incident and reflected angles from the normal; keep lamp and detector distances and source intensity fixed; repeat readings at matching angles and compare means.	Construct the normal at $90^\circ$ to the surface through the point of incidence. Reference every ray angle to that line. Mark fixed positions so source brightness, distances, incidence angle and detector geometry stay the same when the surface changes. Repeat each matched reading and calculate a mean so random variation does not decide the comparison.	4
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03.1	A absorbs $100-8-82=10\%$ , while B absorbs $100-20-5=75\%$ . The conclusion is false: low reflection alone does not prove high absorption because most energy may be transmitted. Panel B is the better absorber.	For each panel, subtract both measured outgoing fractions from $100\%$ . Comparing reflected fractions alone omits transmission, so the conclusion must be based on the calculated absorbed fractions.	4
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04.1	Draw a normal perpendicular to each mirror at the point where the ray strikes. At each mirror draw the reflected ray on the opposite side of the normal with angle of reflection $35^\circ$ . After the second reflection, the ray travels downwards and to the right, parallel to its original direction.	The mirrors are parallel, so their normals are parallel. Apply angle of incidence equals angle of reflection at the lower mirror and again at the upper mirror. The first reflection reverses the vertical component of direction and the second reverses it again, leaving the final ray parallel to the incident ray.	4
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05.1	At the first boundary, $27\,\text{J}$ is reflected, $93\,\text{J}$ is transmitted and $30\,\text{J}$ is absorbed. At the second boundary, $46.5\,\text{J}$ is absorbed, $18.6\,\text{J}$ is reflected and $27.9\,\text{J}$ is transmitted.	At the first boundary, reflected energy is $0.18(150)=27\,\text{J}$ and transmitted energy is $0.62(150)=93\,\text{J}$ , leaving $150-27-93=30\,\text{J}$ absorbed. Use $93\,\text{J}$ as the incident energy at the second boundary: absorption is $0.50(93)=46.5\,\text{J}$ and reflection is $0.20(93)=18.6\,\text{J}$ . The remainder is $93-46.5-18.6=27.9\,\text{J}$ transmitted.	6
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4.6.1.4 · Sound waves (physics only) (HT only)

Tier 1 · Easy

Mark scheme for 4.6.1.4 Tier 1 · Easy
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01.1	The sound wave makes the eardrum and other structures in the ear vibrate. These vibrations are converted into signals that produce the sensation of sound.	Trace the conversion from the incoming sound wave to vibration of a solid structure in the ear, then to the signal interpreted as sound.	2
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02.1	The membrane converts sound to vibration only over a limited frequency range. The detected $100\,\text{Hz}$ and $5.0\,\text{kHz}$ tones lie within normal hearing, while $10\,\text{Hz}$ is below and $28\,\text{kHz}$ is above the normal $20\,\text{Hz}$ to $20\,\text{kHz}$ range.	Use the response evidence before comparing with the known hearing limits. A missing vibration at low and high frequency demonstrates that the conversion system does not respond equally to every tone.	3
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Tier 2 · Standard

Mark scheme for 4.6.1.4 Tier 2 · Standard
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01.1	The electrical signal makes the loudspeaker cone vibrate; the cone produces longitudinal sound waves in the air; the sound waves make the eardrum vibrate.	Follow each conversion in order. The alternating signal drives the solid cone backwards and forwards. The cone creates compressions and rarefactions in the air, transferring sound energy across the room. At the listener, the pressure changes exert forces on the eardrum and make that solid membrane vibrate.	3
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02.1	$f=4.0\,\text{kHz}$ , which is within normal human hearing; the vibrating fork produces a sound wave, the wave vibrates the microphone diaphragm, and the microphone converts this vibration into an electrical signal.	Convert $1.5\,\text{ms}$ to $0.0015\,\text{s}$ . Frequency is cycles divided by time, so $f=6/0.0015=4000\,\text{Hz}=4.0\,\text{kHz}$ . This lies between $20\,\text{Hz}$ and $20\,\text{kHz}$ . Trace the disturbance through solid fork vibration, air pressure oscillations, solid diaphragm vibration and the electrical trace.	5
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03.1	The absence of a human sensation does not show that no sound wave was produced. A frequency of $26\,\text{kHz}$ is above the normal upper hearing limit of about $20\,\text{kHz}$ , so the wave can make nearby matter vibrate without being audible to the listener.	Separate production from detection. The observed solid vibration can produce a sound wave, but the listener's ear converts sound to a sensation only over a limited frequency range. Compare $26\,\text{kHz}$ with the normal hearing limit.	3
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Tier 3 · Hard

Mark scheme for 4.6.1.4 Tier 3 · Hard
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01.1	Ear structures respond effectively only over a limited frequency range; this person's detectable interval is narrower than the normal range of about $20\,\text{Hz}$ to $20\,\text{kHz}$ , with both low-frequency and high-frequency loss.	The eardrum and other ear structures must be driven into vibration for a tone to be detected. Their mechanical response is limited, so frequencies outside the effective range do not produce a sufficient vibration or sensation. Compared with $20\,\text{Hz}$ to $20\,\text{kHz}$ , the measured lower limit is higher and the upper limit is lower, showing reduced hearing at both ends.	4
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02.1	A has a frequency-range width of $4900\,\text{Hz}$ ; B has a frequency-range width of $17\,000\,\text{Hz}$ , about $3.5$ times A's range width. A reaches lower frequencies, but B reproduces a broader range and extends to higher frequencies.	Convert the upper limits to hertz. For A, range width $=5000-100=4900\,\text{Hz}$ . For B, range width $=18\,000-1000=17\,000\,\text{Hz}$ . The ratio is $17\,000/4900=3.47\ldots$ , or about $3.5$ . A's lower cutoff is only $100\,\text{Hz}$ , so it reaches lower frequencies; B has the wider range and higher upper cutoff.	5
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03.1	$15\,\text{kHz}$ satisfies all three ranges. $15\,\text{Hz}$ is below every lower limit; $0.80\,\text{kHz}$ is below the diaphragm's $2.0\,\text{kHz}$ limit; $24\,\text{kHz}$ is above the source limit, diaphragm limit and normal hearing limit.	Find the overlap of the three working ranges: from $2.0\,\text{kHz}$ to $18\,\text{kHz}$ . Test each candidate against that common interval, converting $15\,\text{Hz}$ to $0.015\,\text{kHz}$ if needed.	5
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04.1	The five-stage chain is: alternating current in the loudspeaker wires — electrical signal; motion of the loudspeaker cone — solid vibration; disturbance crossing the air — sound wave; motion of the microphone diaphragm — solid vibration; alternating microphone output — electrical signal.	Follow the energy pathway without adding an extra stage. Current is an electrical signal, motion of each solid component is a solid vibration, and the travelling pressure disturbance in air is a sound wave.	4
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05.1	Seat the person at a fixed measured distance from the loudspeaker in quiet surroundings. Use the signal generator to present tones over a safe range, increasing frequency in small steps near the point where hearing stops. At every frequency use the sound-level meter at the person's position to set the same safe sound level, maintaining equal loudness so higher-frequency tones are not simply quieter. Ask the person to report whether each randomly presented tone is heard, include silent trials, and repeat each frequency. The highest frequency detected consistently is the estimated upper hearing limit; compare it with the normal upper limit of about $20\,\text{kHz}$ .	Change frequency while controlling the sound level at the ear, loudspeaker distance and background noise. Random order, silent trials and repeats make the person's yes-or-no responses more reliable. Use consistent detection, rather than one isolated response, to locate the threshold.	6
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4.6.1.5 · Waves for detection and exploration (physics only) (HT only)

Tier 1 · Easy

Mark scheme for 4.6.1.5 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	It is a P-wave. P-waves are longitudinal.	S-waves cannot travel through liquids, whereas P-waves travel through solids and liquids. P-waves have oscillations parallel to their direction of travel, so they are longitudinal.	2
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02.1	The $18\,\mu\text{s}$ echo is from the nearer boundary because it has the shorter out-and-back travel time.	Both echoes travel through the same material, so wave speed is common. A shorter return time means a shorter total path and therefore a closer reflecting boundary.	2
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Tier 2 · Standard

Mark scheme for 4.6.1.5 Tier 2 · Standard
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01.1	depth $=63\,\text{m}$	The pulse travels down and back, so its total distance is twice the depth. Calculate $vt=(1500)(0.084)=126\,\text{m}$ , then divide by $2$ : depth $=126/2=63\,\text{m}$ .	3
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02.1	wall thickness $=0.019\,\text{m}$ , or $19\,\text{mm}$ , to 2 significant figures	The extra round-trip time through the wall is $(26-14)\,\mu\text{s}=1.2\times10^{-5}\,\text{s}$ . The extra path is $vt=(3200)(1.2\times10^{-5})=0.0384\,\text{m}$ . Divide by $2$ for the one-way thickness: $0.0384/2=0.0192\,\text{m}$ , which is $0.019\,\text{m}$ or $19\,\text{mm}$ to 2 significant figures.	4
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03.1	The $12\,\mu\text{s}$ echo is from the nearest boundary, the $60\,\mu\text{s}$ echo is from the farthest boundary, and the $34\,\mu\text{s}$ echo returns the greatest reflected fraction at $11\%$ .	With a common wave speed, echo delay orders the out-and-back path lengths. Echo energy is a separate measurement, so compare the percentages rather than assuming that the nearest boundary gives the strongest echo.	3
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Tier 3 · Hard

Mark scheme for 4.6.1.5 Tier 3 · Hard
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01.1	P-waves are longitudinal and can pass through solids and liquids, while transverse S-waves cannot pass through liquids; missing S-waves therefore indicate a liquid layer, and changes in P-wave speed, direction and arrival time reveal the positions and sizes of boundaries.	Use the different transmission properties as the test. Both wave types can cross solid rock, but only P-waves continue through a liquid. If a detector geometry should receive both in an all-solid Earth but receives only P-waves, the S-wave path has been blocked by liquid. P-wave speed and direction change at material boundaries, so arrival-time patterns from many earthquakes and detectors can be used to map those interfaces and estimate the size of the liquid region.	5
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02.1	Probe A is better because much more of its wave enters the casing and the air-gap boundary gives a distinct reflection. The echo delay gives the out-and-back travel time, so the gap depth is wave speed multiplied by delay and divided by two. Probe B is mostly reflected or absorbed at the outer surface and provides no internal-boundary signal.	Compare transmission before using the echo. Probe A transfers enough energy into the solid to reach the hidden structure, and the change from solid to air partially reflects it as a distinct peak. Measure the time from the transmitted pulse to that peak; the wave covers twice the gap depth, so $d=vt/2$ . Probe B's small transmitted fraction and absent internal peak make its record unsuitable for locating the gap.	5
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03.1	The coating contributes $10\,\mu\text{s}$ to the round trip. The remaining $20\,\mu\text{s}$ gives a depth of $36\,\text{mm}$ in the solid, so the flaw is $48\,\text{mm}$ or $0.048\,\text{m}$ below the coating surface.	Convert $12\,\text{mm}$ to $0.012\,\text{m}$ . The coating round-trip time is $2(0.012)/2400=1.0\times10^{-5}\,\text{s}=10\,\mu\text{s}$ . The solid round-trip time is therefore $20\,\mu\text{s}$ . Its one-way depth is $(3600)(20\times10^{-6})/2=0.036\,\text{m}$ . Add the coating: $0.036+0.012=0.048\,\text{m}$ .	5
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04.1	The P-wave travel time is $600/8.0=75\,\text{s}$ and the S-wave travel time is $600/4.8=125\,\text{s}$ , so the S–P arrival delay is exactly $125-75=50\,\text{s}$ . The first station shows that the earthquake produced both waves. At the second station, the P-wave arrival but missing direct S-wave adds evidence that the route crosses a liquid layer, because P-waves travel through solids and liquids whereas S-waves do not travel through liquids.	Use $t=d/v$ for each wave with the supplied kilometres and kilometres per second, then subtract the earlier P arrival time from the later S arrival time. Interpret the second record only after using the first station to establish that an S-wave was produced.	5
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05.1	The pulse travels $2\times36=72\,\text{m}$ in $48\,\text{ms}=0.048\,\text{s}$ , so the speed is $v=72/0.048=1500\,\text{m s}^{-1}$ . The factor of two is needed because the recorded time includes the journey from the ship to the seabed and the return journey from the seabed to the ship.	Double the one-way depth to obtain the pulse's total distance and convert milliseconds to seconds before using $v=d/t$ : $v=2(36)/(48\times10^{-3})=1500\,\text{m s}^{-1}$ .	5
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4.6.2.1 · Types of electromagnetic waves

Tier 1 · Easy

Mark scheme for 4.6.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Radio waves have the longest wavelength. Radio waves and gamma rays travel at the same speed in a vacuum.	Radio waves are at the long-wavelength end of the spectrum. Frequency and wavelength vary across the spectrum, but every electromagnetic wave has the same vacuum speed.	2
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02.1	X-rays have a higher frequency than ultraviolet; gamma rays, like all electromagnetic waves, are transverse.	In increasing frequency, ultraviolet comes before X-rays. Every electromagnetic wave has oscillations perpendicular to its direction of energy transfer, so gamma rays are transverse rather than longitudinal.	2
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Tier 2 · Standard

Mark scheme for 4.6.2.1 Tier 2 · Standard
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01.1	$\lambda=6.0\times10^{-3}\,\text{m}$ ; microwave	Rearrange $c=f\lambda$ to $\lambda=c/f$ . Then $\lambda=(3.00\times10^8)/(5.0\times10^{10})=6.0\times10^{-3}\,\text{m}$ . A wavelength of $6.0\,\text{mm}$ is in the microwave region.	3
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02.1	$2.0\,\text{m}$ is radio, $4.0\,\text{mm}$ is microwave, $600\,\text{nm}$ is visible light and $0.020\,\text{nm}$ is X-ray; this is also the increasing-frequency order.	Classify by wavelength scale: metre wavelengths are radio, millimetre wavelengths are microwave, hundreds of nanometres are visible and hundredths of a nanometre are X-rays. Frequency increases as wavelength decreases, so ordering from $2.0\,\text{m}$ down to $0.020\,\text{nm}$ gives radio, microwave, visible, X-ray.	4
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03.1	V has twice the frequency of U because its wavelength is half as large. They arrive together because all electromagnetic waves have the same speed in a vacuum, so the claim about V arriving first is false.	For a common vacuum speed, frequency is inversely proportional to wavelength. Halving wavelength doubles frequency, but changing frequency does not change an electromagnetic wave's vacuum speed.	3
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Tier 3 · Hard

Mark scheme for 4.6.2.1 Tier 3 · Hard
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01.1	radio wavelength $=4.0\,\text{m}$ microwave wavelength $=0.10\,\text{m}$ They travel at the same speed in air, but the higher-frequency microwave has the shorter wavelength.	Convert the frequencies: $75\,\text{MHz}=75\times10^6\,\text{Hz}$ and $3.0\,\text{GHz}=3.0\times10^9\,\text{Hz}$ . Using $\lambda=v/f$ , the radio wavelength is $(3.00\times10^8)/(75\times10^6)=4.0\,\text{m}$ and the microwave wavelength is $(3.00\times10^8)/(3.0\times10^9)=0.10\,\text{m}$ . Both are electromagnetic waves, so they have the same speed in air; the frequency-wavelength product stays constant.	5
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02.1	speed $=3.0\times10^8\,\text{m s}^{-1}$ ; the signals are radio, infrared and ultraviolet respectively; their arrival times match because all electromagnetic waves have the same speed in a vacuum.	Calculate $v=d/t=(6.0\times10^7)/0.20=3.0\times10^8\,\text{m s}^{-1}$ . Locate each frequency in the continuous spectrum: $10^8\,\text{Hz}$ is radio, $3.0\times10^{13}\,\text{Hz}$ is infrared and $10^{15}\,\text{Hz}$ is ultraviolet. Different frequencies imply different wavelengths, not different vacuum speeds.	6
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03.1	P and Q are consistent because each gives $f\lambda=3.0\times10^8\,\text{m s}^{-1}$ . R's wavelength is incorrect; it should be $\lambda=c/f=(3.00\times10^8)/(6.0\times10^{18})=5.0\times10^{-11}\,\text{m}$ . The corrected scale remains consistent with an X-ray.	Multiply $f\lambda$ in each row. P gives $(1.0\times10^8)(3.0)=3.0\times10^8$ and Q gives $(6.0\times10^{14})(5.0\times10^{-7})=3.0\times10^8$ . R's printed values give $3.0\times10^9$ , so rearrange $c=f\lambda$ to obtain $5.0\times10^{-11}\,\text{m}$ .	5
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04.1	The plate gains energy when the infrared radiation reaches and is absorbed by it, showing that electromagnetic waves transfer energy. The transfer occurs across the evacuated gap, so electromagnetic waves do not require a material medium. They are transverse waves. Infrared and gamma rays travel at the same speed through the vacuum in the chamber.	Use the temperature rise as observable evidence that energy reached the plate. The evacuated gap rules out transfer by a material wave across the gap. Then apply the shared properties of the electromagnetic spectrum: transverse waves and a common speed in vacuum.	4
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05.1	Ultraviolet belongs between visible light and X-rays because $5.0\times10^{14}<1.0\times10^{16}<1.0\times10^{18}\,\text{Hz}$ . Radio, visible, ultraviolet and X-rays are regions of the same continuous electromagnetic spectrum, ordered by continuously increasing frequency rather than separated into different kinds of wave. The human eye detects the visible wave directly, but not the radio, ultraviolet or X-ray waves.	Compare the stated frequency with the two neighbouring values, then distinguish named spectrum regions from gaps: the electromagnetic spectrum is continuous. Apply the eye's limited response to the visible region only.	4
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4.6.2.2 · Properties of electromagnetic waves 1

Tier 1 · Easy

Mark scheme for 4.6.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Keep both plates at the same temperature — for example the same hot water, poured at the same time. Keep the emitting surface area, the material and the detector's position and angle the same.	Only the surface finish should change. Keep the emitting area, material and temperature equal — same volume and starting temperature of hot water, timed the same way — and hold the detector in the same position, angle and surroundings for both readings.	2
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02.1	The angles must be measured from the normal, and the ray should bend towards the normal on entering glass from air.	First construct a normal perpendicular to the boundary at the entry point. Reference the incident and refracted angles to that line. For the air-to-glass change, draw the refracted ray closer to the normal than the incident ray.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.6.2.2 Tier 2 · Standard
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01.1	$v_B/v_A=0.625$ The closer wavefronts show that the wavelength and speed are smaller in medium B.	Since $v=f\lambda$ and $f$ is unchanged, the speed ratio equals the wavelength ratio: $v_B/v_A=\lambda_B/\lambda_A=1.5/2.4=0.625$ . The wavefronts are closer together in B because each cycle travels a shorter distance there, so B is the slower medium.	4
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02.1	Trace each block before use; mark enough points along incident and emergent rays to reconstruct them; draw normals perpendicular to the boundaries; measure angles from the normals; keep the incident angle and geometry fixed, repeat, and compare results for the two substances.	Place each block on paper and trace its outline. Use a narrow ray and mark at least two points on each external ray before removing the block, then join the points and continue the path through the outline. Draw a normal at each boundary and measure incidence and refraction angles from it. Use the same incident angle and block orientation for both substances, repeat readings and compare mean refracted angles.	5
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03.1	Draw a normal perpendicular to each face at the point where the ray crosses it. The ray bends towards the normal on entering the block and away from the normal on returning to air. For parallel faces, the emerging ray is parallel to the incident ray but displaced sideways.	Treat the two boundaries separately. Use a perpendicular reference line at each crossing. The first change is air to the transparent material; the second reverses that material change. Equal and opposite angular changes at parallel faces restore the original ray direction.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.6.2.2 Tier 3 · Hard
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01.1	Use identical cans with equal water volumes at the same starting temperature; measure temperature at regular times or infrared intensity at a fixed distance; control surroundings, exposed area and detector position; repeat and compare mean cooling rates or intensities.	Fill identical cans to the same level with equal volumes of water at the same starting temperature. Put them side by side away from draughts. Either log each temperature at equal time intervals and find the gradient of temperature-time data, or place the same infrared detector at a fixed distance and angle from each surface. Keep can dimensions, exposed area, water volume, starting temperature, room conditions and detector geometry constant. Repeat the experiment, swap positions to reduce location bias and compare mean cooling rates or mean detector readings. The matt-black surface should be the better emitter.	6
Total Question 1			6
02.1	Closer wavefronts correctly show a shorter wavelength in the slower material, but an oblique front should change direction. The side entering first slows first, rotating the front and ray towards the normal; frequency remains fixed by the source rather than falling.	At oblique incidence, one side crosses the boundary before the other. That side covers less distance per cycle as it slows, while the other side is still moving faster, so the wavefront pivots. The propagation direction, perpendicular to the wavefront, therefore bends towards the normal. Since the source continues at the same frequency, $v=f\lambda$ is satisfied by a reduced wavelength, not a reduced frequency.	5
Total Question 2			5
03.1	Red absorption is $30\,\text{mJ}$ , or $15\%$ . Infrared absorption is $100\,\text{mJ}$ , or $50\%$ . The same material absorbs different fractions at different wavelengths, demonstrating wavelength-dependent interaction.	For red, absorbed energy is $200-30-140=30\,\text{mJ}$ and $30/200\times100=15\%$ . For infrared, it is $200-20-80=100\,\text{mJ}$ and $100/200\times100=50\%$ . Incident energy and material are controlled, leaving wavelength as the stated difference.	5
Total Question 3			5
04.1	In P the angle changes by $50-29=21^\circ$ . In Q it changes by $50-35=15^\circ$ . P refracts the ray more because its refracted ray has turned through the greater angle and lies closer to the normal.	Measure every angle from the normal, not from the surface. Subtract each refracted angle from the common $50^\circ$ incident angle to obtain $21^\circ$ and $15^\circ$ . The larger change and smaller final angle to the normal provide two consistent diagram observations that identify P.	4
Total Question 4			4
05.1	Use the coating. The transmitted fractions are $1-0.04=0.96$ at $450\,\text{nm}$ , $1-0.06=0.94$ at $550\,\text{nm}$ , $1-0.08=0.92$ at $650\,\text{nm}$ and $1-0.78=0.22$ at $10\,\mu\text{m}$ . It therefore transmits $92$ – $96\%$ of the tested visible wavelengths but absorbs $78\%$ of the tested infrared, showing that absorption depends on wavelength and matches the window requirement.	With negligible reflection, transmitted fraction equals one minus absorbed fraction. Treat $450$ , $550$ and $650\,\text{nm}$ as visible data and $10\,\mu\text{m}$ as infrared, then compare the three high visible transmissions with the low infrared transmission.	6
Total Question 5			6

4.6.2.3 · Properties of electromagnetic waves 2

Tier 1 · Easy

Mark scheme for 4.6.2.3 Tier 1 · Easy
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01.1	Ultraviolet radiation can damage skin, for example causing sunburn or increasing skin-cancer risk. X-rays or gamma rays are ionising and can cause mutations or cancer.	Match each region to its tissue effect. Ultraviolet damages skin; the ionising action of X-rays and gamma rays can damage DNA and lead to mutation or cancer.	2
Total Question 1			2
02.1	P must produce gamma rays.	Gamma rays specifically originate from changes in an atomic nucleus. An unspecified change in an atom can emit or absorb electromagnetic radiation at other frequencies, so R is not uniquely gamma.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.6.2.3 Tier 2 · Standard
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01.1	Procedure A gives $27\,\text{mSv}$ in total; on dose alone it presents the greater risk because $27\,\text{mSv}>18\,\text{mSv}$ .	Add repeated exposures by multiplication: total for A is $6(4.5)=27\,\text{mSv}$ . The dose for B is $18\,\text{mSv}$ . Since radiation dose measures risk of harm, A has the greater indicated risk if the radiation type and other conditions are comparable.	3
Total Question 1			3
02.1	The limit is $1.6\,\text{mSv}$ , so $1.6/0.60=2.67$ and at most $2$ whole images can be taken. This assumes that no other radiation exposure is counted towards this procedure's dose limit.	Convert the limit first: $0.0016\,\text{Sv}=1.6\,\text{mSv}$ . Divide by the dose per image: $1.6/0.60=2.666\ldots$ . Only a whole number of images is possible, and a third image would give $3(0.60)=1.8\,\text{mSv}$ , so the maximum is $2$ . State an assumption such as no other exposure contributing to the stated limit or every image delivering the quoted dose.	4
Total Question 2			4
03.1	Ultraviolet can damage skin and increase skin-cancer risk; shield the lamp or cover exposed skin and eyes. X-rays are ionising and can cause mutations or cancer; use suitable shielding, maximise distance or minimise exposure time.	Match the radiation to its tissue effect, then connect the control directly to reduced dose. A physical enclosure or protective clothing blocks ultraviolet reaching skin and eyes. Dense shielding, shorter operating time or greater distance reduces X-ray exposure.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.6.2.3 Tier 3 · Hard
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01.1	Electrical oscillations produce the radio wave; absorption by the receiver induces an alternating current at the same frequency; gamma rays are produced by changes in an atomic nucleus.	The alternating charges and currents in the transmitter form an oscillating electrical circuit, which emits radio-frequency electromagnetic waves. When the $92\,\text{MHz}$ wave is absorbed by a suitable receiving circuit, it induces electrical oscillations and an alternating current at the same $92\,\text{MHz}$ frequency. Gamma radiation is also electromagnetic, but it is generated by a change in the nucleus rather than by a macroscopic electrical circuit.	5
Total Question 1			5
02.1	$f=4.0\times10^6\,\text{Hz}$ or $4.0\,\text{MHz}$ ; an oscillating transmitter circuit emits the radio wave, and absorption by the receiving circuit induces an alternating current at the same frequency.	Convert the period: $0.25\,\mu\text{s}=0.25\times10^{-6}\,\text{s}=2.5\times10^{-7}\,\text{s}$ . Apply $f=1/T$ to obtain $f=1/(2.5\times10^{-7})=4.0\times10^6\,\text{Hz}$ . Electrical oscillations in the transmitting circuit generate the radio wave. When the wave is absorbed by a suitable receiver, it induces electrical oscillations and alternating current with that same frequency.	5
Total Question 2			5
03.1	Set the transmitter to several known frequencies and use an oscilloscope to measure the period or frequency of the induced alternating current in the receiver. Keep separation, orientation, transmitter amplitude and the circuits unchanged apart from frequency; repeat readings. The received-current frequency should equal the transmitter frequency at every setting.	Place fixed transmitting and receiving circuits at a marked distance and orientation. Select several transmitter frequencies within both circuits' operating ranges. For each, record multiple cycles of receiver current on an oscilloscope, calculate $f=1/T$ if needed, repeat and compare with the set frequency. Equality across the range supports the claim; amplitude need not be equal because the claim concerns frequency.	5
Total Question 3			5
04.1	Procedure A uses X-rays and gives $0.0018\times1000=1.8\,\text{mSv}$ . Procedure B uses gamma rays and gives $0.45\,\text{mSv}$ , so the total is $1.8+0.45=2.25\,\text{mSv}$ , or $0.00225\,\text{Sv}$ . Both X-rays and gamma rays are ionising electromagnetic radiation. A contributes exactly four times the dose from B and therefore the larger contribution to the risk of cell mutation and cancer, although the total risk depends on the combined dose.	Use $1\,\text{Sv}=1000\,\text{mSv}$ before adding the doses. Compare like units, then connect greater ionising-radiation dose with greater risk rather than claiming that harm is certain.	6
Total Question 4			6
05.1	An $8.0\,\text{MHz}$ alternating current in the transmitter should produce an $8.0\,\text{MHz}$ radio wave, and absorption of that wave should induce an alternating current at the same $8.0\,\text{MHz}$ in the receiving aerial. A genuine $4.0\,\text{MHz}$ output therefore is not the directly induced current described by the specification. The oscilloscope time-base could have been set to twice the correct value, or the period could have been measured across two complete cycles instead of one; either error makes the frequency appear halved.	Carry the transmitter frequency unchanged through emission and absorption. Because $4.0/8.0=0.50$ , look for a measurement error that doubles the apparent period: with $f=1/T$ , doubling the measured period halves the calculated frequency.	5
Total Question 5			5

4.6.2.4 · Uses and applications of electromagnetic waves

Tier 1 · Easy

Mark scheme for 4.6.2.4 Tier 1 · Easy
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01.1	Radio broadcasting uses radio waves. Satellite communication uses microwaves. Thermal imaging uses infrared radiation.	First match each application to its wave. Then use the spectrum order: radio waves have lower frequency than microwaves, which have lower frequency than infrared.	3
Total Question 1			3
02.1	ultraviolet for the energy-efficient lamp; visible light for optical-fibre communication; X-rays for the bone image	Match each application to the specification list: ultraviolet is used in energy-efficient lamps, visible light carries information through fibre optics, and X-rays are used for medical imaging.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.6.2.4 Tier 2 · Standard
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01.1	All objects emit infrared and hotter regions emit more infrared in a given time; the camera detects differences in infrared intensity, whereas visible light from the building depends mainly on illumination and may be absent in darkness.	A warm surface emits infrared radiation, and increasing temperature increases the amount emitted per second. An infrared detector converts the differing intensities into a thermal image, so hotter areas can be distinguished. In darkness there may be no visible light incident on the building to be reflected, so an ordinary visible-light image does not directly reveal the emitted thermal pattern.	4
Total Question 1			4
02.1	Microwaves are transmitted through the atmosphere for satellite communication; infrared emitted by warm bodies can be detected in darkness; X-rays penetrate soft tissue more than bone, producing image contrast.	Choose microwaves for the satellite link because suitable microwave wavelengths pass through the atmosphere and carry information. Choose infrared for locating people because warmer bodies emit more infrared and a detector can map it without visible illumination. Choose X-rays for bones because their unequal transmission through soft tissue and bone creates a detectable image.	6
Total Question 2			6
03.1	A is microwave radiation, suitable for satellite communication because it passes through the atmosphere with little absorption. B is infrared radiation, suitable for thermal imaging because warmer objects emit it more strongly. C is X-radiation, suitable for bone imaging because it passes through soft tissue but is absorbed more strongly by bone.	Infer the application from the stated interaction rather than from a memorised list alone. Atmospheric transmission selects microwaves, temperature-dependent emission selects infrared, and contrasting transmission through soft tissue and bone selects X-rays.	6
Total Question 3			6

Tier 3 · Hard

Mark scheme for 4.6.2.4 Tier 3 · Hard
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01.1	Microwaves penetrate into the food and are absorbed within it, while infrared is absorbed mainly near the surface; microwaves heat deeper regions and infrared can brown or heat the outside, so combining them gives more even internal heating with a hot surface.	Link each wave to where its energy is absorbed. Microwaves can enter the food and transfer energy within a greater depth, so they heat material below the surface. Infrared is absorbed strongly at the exposed surface, transferring energy there. Microwave heating alone may leave less surface browning, while infrared alone heats inward more slowly. Using both provides internal heating and a hotter outer layer.	5
Total Question 1			5
02.1	Use infrared for the thermal camera because warm objects emit detectable infrared; use microwaves for the satellite link because they can pass through the atmosphere; use infrared for browning because it is absorbed near the food surface, whereas ionising gamma radiation would be hazardous.	Judge each job by the interaction required. Thermal imaging needs detection of radiation whose emission varies with temperature, which is infrared. Satellite communication needs a wave that carries information through the atmosphere, which is microwave. Surface cooking needs energy absorbed near the outer layer, which infrared provides; gamma rays are penetrating ionising radiation used in medical contexts, not a safe cooking source.	6
Total Question 2			6
03.1	Use ultraviolet in the energy-efficient lamp because the coating absorbs ultraviolet and emits visible light. Use gamma rays to sterilise the sealed instruments because penetrating ionising radiation can pass through the packaging and kill microorganisms. X-rays are suitable for medical imaging, for example imaging bones because soft tissue transmits more X-rays than bone.	Separate the radiation that produces an effect from the visible output or image. In the lamp, ultraviolet is absorbed by the fluorescent coating and converted to visible light, so visible light is the output rather than the radiation used to excite the coating. Sterilisation needs penetrating ionising radiation to reach packaged surfaces and destroy microorganisms, so use gamma rays. Keep X-rays for imaging, where unequal transmission through soft tissue and bone produces contrast.	5
Total Question 3			5
04.1	Use radio waves for the television broadcast because they can carry an information signal and induce matching-frequency oscillations in a receiving aerial. Use infrared for the heater because surfaces absorb infrared and gain thermal energy. Use ultraviolet for tanning because it interacts with skin to cause tanning, but it can also age the skin and increase the risk of skin cancer.	Match the required effect rather than frequency alone: information transfer and aerial reception for radio, absorption and heating for infrared, and a biological skin response for ultraviolet. Include the harmful effect because the same ultraviolet interaction that makes tanning possible creates a health risk.	6
Total Question 4			6
05.1	Gamma rays are penetrating ionising radiation, so they can reach material below a surface and damage or kill living cells. In treatment the target is the tumour: beams and dose are directed to destroy cancer cells while limiting the dose to surrounding healthy tissue. In food sterilisation the targets are microorganisms throughout the food or its packaging, while unnecessary changes to the food must be limited. The absorbed dose and exposure time must therefore be selected separately for each target: a prescribed therapeutic dose and treatment time for the tumour, but a sterilising dose and exposure time sufficient to kill microorganisms in the food.	Start with the two shared properties—penetration and ionisation—then separate the applications by what is meant to be killed. Dose and exposure time are controlled variables, not universal settings for gamma radiation.	6
Total Question 5			6

4.6.2.5 · Lenses (physics only)

Tier 1 · Easy

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01.1	The image is real. The lens cannot be concave; a concave lens forms only virtual images.	A projected image must be formed where rays actually meet, so it is real. A concave lens always gives a virtual image, which cannot be projected onto a screen.	2
Total Question 1			2
02.1	A concave lens should make the ray diverge; draw the emergent ray so its backward extension passes through the principal focus on the incident side.	A concave lens does not bring parallel light to a real focus. Refract the ray away from the principal axis, then use a dashed backward extension to the near focus to show where the diverging ray appears to originate.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.6.2.5 Tier 2 · Standard
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01.1	Draw a ray through the optical centre undeviated and a parallel ray refracted as if it came from the near focus; extend the diverging rays backwards; the image is virtual, upright and diminished.	Start both rays at the top of the object. Send one through the optical centre without deviation. Draw the other parallel to the principal axis, then refract it outwards so its backward extension passes through the principal focus on the object side. Extend the rays backwards until they meet. Their apparent intersection gives an upright, smaller virtual image on the same side of the lens as the object.	4
Total Question 1			4
02.1	$7.9\,\text{cm}$ is anomalous; mean image height $=4.5\,\text{cm}$ ; mean magnification $=2.5$	The $7.9\,\text{cm}$ value is far from the cluster and is excluded. The mean of the consistent readings is $(4.5+4.6+4.4)/3=4.5\,\text{cm}$ . Magnification is image height divided by object height, so $4.5/1.8=2.5$ . Magnification has no unit.	4
Total Question 2			4
03.1	Q must be convex because its projected image is real and a concave lens cannot form a real image. P's image is virtual, but it could come from a concave lens or from a convex lens with the object inside its focal length, so P cannot be identified uniquely.	Use screen evidence first: a screen receives only a real image. Then apply the image possibilities of each lens. Concave lenses always form virtual images, while convex lenses can form either real or virtual images depending on object position.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.6.2.5 Tier 3 · Hard
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01.1	object height $=2.4\,\text{cm}$ ; convex lens; the screen shows the image is real, and a concave lens produces only virtual images	Rearrange magnification $=\text{image height}/\text{object height}$ to object height $=7.2/3.0=2.4\,\text{cm}$ . A screen can intercept only a real image. A concave lens always produces a virtual image, so the lens must be convex.	4
Total Question 1			4
02.1	The refracted rays diverge and do not meet on the far side. Their backward extensions meet on the object's side, forming a virtual, upright and magnified image that cannot be projected onto a screen.	Draw one ray through the optical centre without deviation and one ray parallel to the axis then through the far focus. With the object inside the focal length, those emergent rays separate. Extend them backwards until they intersect on the incident side. Because only extensions meet, the image is virtual; its geometry makes it upright and enlarged, so the screen claim is false.	5
Total Question 2			5
03.1	B's numerical magnification is correct, but a concave lens produces a virtual image that cannot be caught on a screen; either the screen observation or lens label is wrong. In C, $30\,\text{mm}=3.0\,\text{cm}$ , so the magnification is $3.0/1.5=2.0$ , not $20$ . Magnification has no unit.	Check the ratio and the image claim separately. A is consistent. For B, $10/20=0.50$ , but the combination of concave lens and projected image is impossible. For C, convert both heights to the same unit before dividing: $30\,\text{mm}/15\,\text{mm}=2.0$ .	5
Total Question 3			5
04.1	The lens is convex and has focal length $12\,\text{cm}$ because parallel incident rays meet at its principal focus. Twice the focal length is $24\,\text{cm}$ , so the $30\,\text{cm}$ object is beyond $2f$ . Draw one ray from the object's top parallel to the principal axis and then through the far focus, and a second through the optical centre without changing direction. The rays meet on the far side between $f$ and $2f$ , forming a real, inverted, diminished image that can be projected onto a screen.	Use the parallel-ray focus as direct evidence for both a convex lens and $f=12\,\text{cm}$ . Compare $30\,\text{cm}$ with $2f=24\,\text{cm}$ , then use the central ray and the ray refracted through the far focus. Actual ray intersection makes the image real and projectable.	5
Total Question 4			5
05.1	Use a convex lens for the enlarged projected image and place the object between the principal focus and twice the focal length. Refracted rays then actually converge beyond twice the focal length on the far side, so the image is real, inverted, magnified and can be caught on a screen. Use a concave lens for the upright diminished view. Its rays diverge and their backward extensions meet on the object side, producing a virtual, upright, diminished image that cannot be projected.	Start with the screen requirement: only a real image formed by converging rays can be projected, so use a convex lens with the object outside its focal length. The always-virtual image from a concave lens supplies the second requirement; backward ray extensions explain why it appears upright and cannot reach a screen.	6
Total Question 5			6

4.6.2.6 · Visible light (physics only)

Tier 1 · Easy

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01.1	The smooth mirror gives specular reflection, with reflected rays mainly in one direction. The rough paper gives diffuse reflection, scattering the rays in many directions.	Surface orientation changes the direction of each reflected ray. A smooth surface has nearly parallel normals and keeps the reflected beam organised; a rough surface has varying normals and scatters it.	2
Total Question 1			2
02.1	A is transparent, B is translucent, and an opaque material does not transmit light through it.	Use both transmission and image clarity. Transparent materials transmit light with enough directional information for a clear image. Translucent materials transmit but scatter the light, blurring the image. Opaque materials transmit no light through the material.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.6.2.6 Tier 2 · Standard
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01.1	The blue filter transmits blue light and absorbs most other colours; the red book reflects red but absorbs most blue, so little light from the book passes through the filter to the eye.	Work from the object to the observer. Under white light the book reflects mainly red wavelengths. A blue filter does not transmit those red wavelengths. Any blue light reaching the book is mostly absorbed by the red surface, so very little visible light completes the path to the eye and the book appears dark.	3
Total Question 1			3
02.1	The surface appears green because it reflects green much more strongly; most red and blue light is absorbed.	For an opaque object, compare reflected wavelengths because none are transmitted through it. The dominant reflected signal is green. The much smaller red and blue signals show that most of those wavelengths are absorbed rather than returned to the eye.	3
Total Question 2			3
03.1	No light emerges ideally. The red filter transmits red and absorbs the other visible wavelengths; the blue filter then absorbs the red light because it transmits blue rather than red.	Track the available wavelengths after each stage. Only red remains after the first filter, so no blue reaches the second. An ideal blue filter removes the remaining red component.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.2.6 Tier 3 · Hard
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01.1	It appears mainly green in white light, bright green through the green filter, and dim red or very dark through the red filter; the filters transmit only their colour while the surface reflects green much more strongly than red and absorbs blue.	In white light, all three colour ranges arrive, but strong green reflection dominates, so the surface appears green. A green filter passes the strongly reflected green light and absorbs most other wavelengths, so the surface remains clearly green. A red filter blocks green and passes only the weak red reflection, so much less light reaches the eye and the surface looks dim red or nearly dark. Blue contributes nothing because the surface absorbs it.	5
Total Question 1			5
02.1	P is transparent and suits the clear window; Q is translucent and suits the privacy screen; R is opaque and suits the blackout panel.	Use both transmission and retained contrast. P passes most light while preserving a clear image, so it is transparent and appropriate for seeing through a window. Q still passes light but destroys most image contrast through scattering, so it is translucent and provides privacy. R passes no light, so it is opaque and blocks the view and illumination.	6
Total Question 2			6
03.1	The reflected signals are $12$ red, $21$ green and $4$ blue units. Green is dominant, so the unfiltered surface appears mainly green. An ideal red viewing filter removes the green and blue components and transmits the $12$ red units, so the surface appears dim red.	Apply each reflection percentage to the light of that colour reaching the surface: $0.20(60)=12$ , $0.70(30)=21$ and $0.40(10)=4$ . Compare the returned signals for the first appearance. Then keep only the red component for the final ideal filter.	5
Total Question 3			5
04.1	The glossy card gives a strong reading mainly at the specular reflection angle, whereas the rough card gives weaker readings spread over many angles because its reflection is diffuse. Both cards look red in white light because they reflect red wavelengths more strongly and absorb most other visible wavelengths. An ideal blue filter transmits blue and absorbs red, so both cards appear very dark through it.	Separate surface texture from colour. Texture controls whether reflected light is concentrated in one direction or scattered. The material's wavelength-dependent reflection produces the red appearance. Apply the blue filter last: it removes the reflected red light before it reaches the observer.	5
Total Question 4			5
05.1	The blue filter absorbs most non-blue wavelengths and transmits blue light. The frosted pane is translucent: it transmits the blue light but scatters it, so a sharp image cannot pass through. The green card reflects green strongly but receives mainly blue, which it mostly absorbs, so it appears very dark. If the card is removed, the observer sees diffuse blue light through the pane rather than a sharp view.	Track the surviving wavelengths and the direction information separately. The filter selects blue. The translucent pane preserves transmission but scatters directions. The opaque green surface receives no strong green component to reflect, leaving little light for the observer.	5
Total Question 5			5

4.6.3.1 · Emission and absorption of infrared radiation (physics only)

Tier 1 · Easy

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01.1	All objects emit radiation, whatever their temperature. Temperature affects the intensity or the wavelength distribution of the emitted radiation.	Emission does not switch off when an object is cool or at the same temperature as its surroundings. Changing temperature changes how intense the emission is and how it is distributed across wavelengths.	2
Total Question 1			2
02.1	The test surface is more likely to be the matt candidate because its $6.4$ -unit reading is nearer the stronger emission from matt black than the weak emission from polished silver. The surfaces must be at the same temperature and measured at the same distance with the same detector and geometry.	A greater detector reading indicates stronger infrared emission. The $6.4$ -unit result lies between the references but is closer to $9.2$ than to $2.1$ , which supports the matt finish. That comparison is valid only if temperature, detector distance, detector and angle are controlled so that surface finish is the relevant difference.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.6.3.1 Tier 2 · Standard
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01.1	The matt-black can cools faster because a matt-black surface is a better emitter of infrared radiation than a polished-silver surface, so it transfers energy to the surroundings at a greater rate.	Because the cans and starting conditions are identical, compare only their emitting surfaces. Matt black is a good infrared emitter, while polished silver is a poor emitter. The black can therefore loses internal energy by radiation more rapidly and its temperature falls faster.	3
Total Question 1			3
02.1	Use equal-area plates of the same material and mass; set the same starting temperature; keep lamp distance, angle, power and exposure time fixed; repeat each finish and compare mean temperature rises.	Change only surface finish. Match plate area, material, mass and thermometer contact; allow every plate to reach the same initial temperature. Mark one position and orientation relative to a constant-power infrared lamp and expose for one fixed time. Measure temperature change, not just final temperature, repeat the trials and compare means.	4
Total Question 2			4
03.1	The absorption ratio A:B is $75/25=3.00$ . The emission ratio A:B is $7.5/2.5=3.00$ . The matching ratios support that a good absorber is also a good emitter.	Compare like quantities between the surfaces. Divide A's absorbed percentage by B's, then divide A's emission reading by B's. Both calculations give $3.00$ , so A absorbs three times the fraction and emits three times the detector signal under the controlled conditions. This supports the generic absorber-emitter relationship.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.3.1 Tier 3 · Hard
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01.1	Expose equal plates at equal distance and angle for equal times from the same lamp; measure equal-mass plates' temperature rises with identical contact thermometers; control starting temperature and surroundings; repeat; the surface with the greatest mean temperature rise is the best absorber.	Use plates of the same material, mass and area, differing only in surface finish. Let them reach the same starting temperature, place each at the same distance and orientation to the lamp, and expose each for the same time at constant lamp power. Measure the temperature before and after with the same sensor. Shield the setup from draughts and other heat sources, repeat each finish and calculate its mean temperature rise. With the energy input controlled, the largest mean rise indicates the greatest infrared absorption.	5
Total Question 1			5
02.1	$12.0\,{}^\circ\text{C}$ is anomalous for polished white; matt-black mean $=12.3\,{}^\circ\text{C}$ and polished-white mean $=7.5\,{}^\circ\text{C}$ ; matt black is the better absorber.	The polished-white $12.0\,{}^\circ\text{C}$ result does not fit its other two trials, so exclude it with that justification. The matt-black mean is $(12.1+12.4+12.3)/3=12.266\ldots\,{}^\circ\text{C}$ , reported as $12.3\,{}^\circ\text{C}$ . The representative polished-white mean is $(7.4+7.6)/2=7.5\,{}^\circ\text{C}$ . Under controlled exposure, the larger temperature rise identifies greater absorption.	5
Total Question 2			5
03.1	P gives $4.5$ , Q gives $3.5$ and R gives $2.0$ units per $\text{cm}^2$ . The emitter ranking is P, Q, R from best to worst. The predicted absorber ranking is also P, Q, R because good emitters are good absorbers.	Normalise the total signal by emitting area: $90/20=4.5$ , $175/50=3.5$ and $60/30=2.0$ units per $\text{cm}^2$ . This removes area as the cause of the different totals. With temperature and geometry controlled, compare the normalised values and apply the absorber-emitter relationship.	5
Total Question 3			5
04.1	During heating, the matt-black plate rises by $22\,{}^\circ\text{C}$ and the polished plate by $11\,{}^\circ\text{C}$ , so the matt-black surface is the better absorber in this comparison. During cooling, the matt-black plate falls by $22\,{}^\circ\text{C}$ and the polished plate by $10\,{}^\circ\text{C}$ , so the matt-black surface emits infrared at a greater rate in this comparison.	For heating, subtract the common starting temperature from each final temperature. For cooling, subtract each final temperature from $60\,{}^\circ\text{C}$ . Because the plates are otherwise identical and each comparison uses equal times and conditions, the larger rise identifies stronger absorption and the larger fall identifies stronger emission.	5
Total Question 4			5
05.1	Use the heater and temperature sensor to bring each plate to the same stated temperature. Mount each plate in turn in the same position and use the clamp stand and ruler to keep the infrared detector at the same distance and angle from the same area of the plate. Record the detector reading, repeat at least three times for each finish and compare the mean readings; the finish with the greatest mean reading is the best emitter. Keep plate material, area, thickness, temperature, detector position and surroundings constant. Cooling-rate data alone would not isolate radiation because the plates also lose energy by conduction and convection.	Measure outgoing infrared directly at a controlled temperature and geometry. Comparing means across the three finishes isolates finish more clearly than comparing total cooling, which includes non-radiative transfers by conduction and convection.	6
Total Question 5			6

4.6.3.2 · Perfect black bodies and radiation (physics only)

Tier 1 · Easy

Mark scheme for 4.6.3.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	It absorbs all incident radiation. It reflects none of the incident radiation. It transmits none of the incident radiation.	A perfect black body is an ideal absorber. All incoming radiation is absorbed, leaving no incident radiation to be reflected or transmitted.	3
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02.1	Object A is hotter than B; B still emits radiation because every object emits radiation, and temperature affects the intensity and wavelength distribution rather than switching emission on or off.	The objects are identical, so the stronger emission pattern indicates the higher temperature. A non-zero B curve is also consistent with the rule that all bodies emit radiation whatever their temperature.	3
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Tier 2 · Standard

Mark scheme for 4.6.3.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	net energy change $=-50\,\text{W}$ ; the body cools	Take absorbed power minus emitted power: $480-530=-50\,\text{W}$ . The negative result means the body loses $50\,\text{J}$ of energy each second. Its internal energy and temperature therefore decrease.	3
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02.1	B is in thermal equilibrium and remains at constant temperature. A has net power $+50\,\text{W}$ and warms; C has net power $-85\,\text{W}$ and cools. C's temperature changes faster because its net power has the greater magnitude.	B absorbs and emits at equal rates, so it has no net energy transfer and is in thermal equilibrium. For A, net power $=320-270=+50\,\text{W}$ , so it warms. For C, net power $=260-345=-85\,\text{W}$ , so it cools. Because the bodies are otherwise identical, the larger magnitude $85\,\text{W}$ gives C the greater rate of temperature change.	4
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03.1	The temperature order is P, R, Q from highest to lowest. P has the greatest peak intensity and shortest peak wavelength, while Q has the smallest peak intensity and longest peak wavelength. The data cannot give the bodies' exact temperatures, or exact temperature differences, without a calibrated relationship.	Use both features consistently: higher peak intensity together with a peak at shorter wavelength indicates the hotter otherwise identical body. This orders the bodies P, R, Q. The table provides comparative emission data but no conversion from either measurement to temperature, so it supports an order rather than numerical temperatures.	3
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Tier 3 · Hard

Mark scheme for 4.6.3.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	initial net power $=0\,\text{W m}^{-2}$ new net power $=+12\,\text{W m}^{-2}$ Earth warms until increased emission restores the balance.	Initially the absorbed solar power is $340-102=238\,\text{W m}^{-2}$ , equal to the $238\,\text{W m}^{-2}$ emitted, so the net power is zero. After the reflection change, absorption is $340-90=250\,\text{W m}^{-2}$ . The initial new imbalance is $250-238=+12\,\text{W m}^{-2}$ , so Earth gains energy and warms. As temperature rises, emitted radiation increases until outgoing power again matches absorbed incoming power.	5
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02.1	first interval $=+72\,\text{kJ}$ ; second interval $=-30\,\text{kJ}$ ; overall change $=+42\,\text{kJ}$ ; the roof warms first, cools during the cloudy interval, and ends with a net energy gain	Before the cloud, net power is $720-640=80\,\text{W}$ for $15\times60=900\,\text{s}$ , giving $80\times900=72\,000\,\text{J}=72\,\text{kJ}$ . Under cloud, net power is $590-640=-50\,\text{W}$ for $10\times60=600\,\text{s}$ , giving $-50\times600=-30\,000\,\text{J}=-30\,\text{kJ}$ . The total is $72-30=+42\,\text{kJ}$ . Positive then negative net power gives warming then cooling, with an overall energy gain.	6
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03.1	The net power is $+70\,\text{W}$ at $20\,{}^\circ\text{C}$ and $-90\,\text{W}$ at $40\,{}^\circ\text{C}$ . Equilibrium is at $30\,{}^\circ\text{C}$ , where absorbed and emitted powers are equal. Below it the body gains energy and warms; above it the body loses energy and cools, so changes drive it back towards equilibrium.	Use net power equals absorbed minus emitted. At $20\,{}^\circ\text{C}$ , $420-350=+70\,\text{W}$ . At $40\,{}^\circ\text{C}$ , $420-510=-90\,\text{W}$ . The zero-net row is $30\,{}^\circ\text{C}$ . The signs on either side show restoring temperature changes.	5
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04.1	The inference fails because the blocks are different materials with different surface finishes, so they can have different abilities to emit infrared at the same temperature. The stated temperatures are both $80\,{}^\circ\text{C}$ ; the larger detector reading from the matt-black ceramic can be caused by stronger infrared emission from its material and finish, not a higher temperature. Temperature comparison requires otherwise identical bodies with the same finish and geometry, or a detector calibrated separately for each material and surface.	An infrared reading depends on temperature and on how good an emitter the surface is. Because two variables differ here, material and finish, the reading cannot isolate temperature even though distance is controlled.	4
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05.1	At first the slab absorbs radiation faster than it emits radiation, so its energy and temperature increase. As it becomes hotter, it emits more radiation until its emitted power equals its absorbed power; the net energy change is then zero and the temperature is constant. The reflective cover sends more incident radiation away, so the slab absorbs less than it emits at the old temperature and cools. As it cools its emitted power falls until it again equals the lower absorbed power, giving a new, lower constant temperature.	Follow the sign of absorbed power minus emitted power through each stage. Warming raises emission until the first balance is reached. Increased reflection lowers absorption, breaking that balance in the cooling direction; falling temperature then lowers emission until equality is restored.	5
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