3.3 Waves — revision question pack

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

Checked against AQA 7408 section 3.3. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

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Answer all questions in the spaces provided.

3.3.1.1 · Progressive waves

Explanation

A progressive wave transfers energy while particles of a material medium oscillate about equilibrium without net travel with the wave.
Amplitude is maximum displacement, wavelength is the shortest distance between points in phase, frequency is cycles per second and period is time per cycle, so $f=1/T$ .
Wave speed obeys $v=f\lambda$ .
Phase locates a point within a cycle; two points separated by $x$ have phase difference $2\pi x/\lambda$ radians, $360x/\lambda$ degrees or $x/\lambda$ cycles.
A sound-speed experiment can use direct travel time or stationary waves followed by graphical analysis, with distance and timing uncertainties considered.

Amplitude is measured from equilibrium and wavelength between adjacent points in phase.

Worked example

A wave has wavelength $0.80\,\text{m}$ and period $2.5\,\text{ms}$ . Find its frequency and speed.

1.Convert $T=2.5\times10^{-3}\,\text{s}$ .
2. $f=1/T=400\,\text{Hz}$ .
3. $v=f\lambda=400(0.80)$ .

Answer: The frequency is 400 Hz and the speed is 3.2 × 10² m s⁻¹.

Common mistakes

Don't measure amplitude from crest to trough instead of from equilibrium.
Don't use the distance between a crest and adjacent trough as one wavelength.
Don't leave phase difference as a distance rather than degrees, radians or a fraction of a cycle.

Exam tip

For phase questions, first express separation as a fraction of one wavelength.

Tier 1 · Easy

Successive crests of a progressive wave are $0.80\,\text{m}$ apart and pass a point every $2.5\,\text{ms}$ . Calculate the wave speed.
[1 mark]
Total for this question: 1
A sound wave travels at $340\,\text{m s}^{-1}$ and has frequency $680\,\text{Hz}$ . Calculate its wavelength.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Two sensors on a ripple tank are $0.24\,\text{m}$ apart along the direction of a progressive wave of wavelength $0.64\,\text{m}$ . Determine the magnitude of their phase difference in both degrees and radians.
[3 marks]
Total for this question: 3
Two microphones are $2.50\,\text{m}$ apart. A sharp sound reaches the second microphone $7.35\,\text{ms}$ after the first. Determine the speed of sound and predict the delay for microphones $4.00\,\text{m}$ apart.
[3 marks]
Total for this question: 3
Two sensors are $0.540\,\text{m}$ apart along the direction of a progressive wave. The second sensor's trace is delayed by $1.50\,\text{ms}$ relative to the first. The period is $6.00\,\text{ms}$ and the sensor separation is less than one wavelength. Determine the wavelength and wave speed.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Water waves of frequency $2.4\,\text{Hz}$ and wavelength $0.35\,\text{m}$ travel from P towards Q. Q is $0.22\,\text{m}$ beyond P along the direction of travel. Determine the wave speed, the phase change from P to Q, and the delay between a crest passing P and that crest passing Q.
[5 marks]
Total for this question: 5
A loudspeaker and reflector produce a stationary sound wave in air. Moving a microphone between adjacent signal minima gives a separation of $8.55\,\text{cm}$ . An oscilloscope trace shows six complete cycles in $3.00\,\text{ms}$ . Determine the frequency, wavelength and speed of the sound.
[5 marks]
Total for this question: 5
A progressive wave of frequency $40.0\,\text{Hz}$ travels from sensor A to sensor B, which is $1.35\,\text{m}$ farther along the direction of travel. The trace at B lags that at A by $90.0^{\circ}$ within a cycle. A coarse timing measurement shows that a crest takes between $25.0\,\text{ms}$ and $40.0\,\text{ms}$ to travel from A to B. Determine the wavelength and wave speed.
[5 marks]
Total for this question: 5
A graph of phase difference against sensor separation for two sensors on a string is a straight line through the origin with gradient $7.80\,\text{rad m}^{-1}$ . The progressive wave has frequency $58.5\,\text{Hz}$ . Determine its wavelength and speed, and predict the phase difference for sensors separated by $0.275\,\text{m}$ .
[5 marks]
Total for this question: 5
A microphone beside a loudspeaker is $2.13\,\text{m}$ from a flat wall. The time between receiving a short direct pulse and its echo is $12.2\,\text{ms}$ . The loudspeaker is then driven with period $1.63\,\text{ms}$ . Determine the speed, frequency and wavelength of the sound, and the phase difference between points $0.310\,\text{m}$ apart along its direction of travel.
[6 marks]
Total for this question: 6

3.3.1.2 · Longitudinal and transverse waves

Explanation

In a transverse wave, particle or field displacement is perpendicular to energy propagation; electromagnetic waves and waves on a string are examples. In a longitudinal wave, displacement is parallel to propagation, producing compressions and rarefactions as in sound.
All electromagnetic waves travel at the same speed, $3.00\times10^8\,\text{m s}^{-1}$ , in vacuum. Polarisation restricts oscillations to one transverse direction and is evidence that electromagnetic waves are transverse.
Polaroid material selects an electric-field direction, while transmitting and receiving aerials give the strongest signal when aligned with that direction.
Malus’s law is not required.
Water-wave investigations can relate speed to experimental factors.

Transverse displacement is perpendicular to propagation; longitudinal displacement forms compressions and rarefactions along it.

Worked example

A vertically polarised radio wave reaches a straight receiving aerial. State the aerial orientation for maximum and minimum signal.

1.The radio wave’s electric field oscillates vertically.
2.Maximum response occurs when the aerial is parallel to that field.
3.A horizontal aerial is perpendicular to the field and gives the ideal minimum.

Answer: Maximum signal with a vertical aerial; minimum signal with a horizontal aerial.

Common mistakes

Don't define transverse and longitudinal waves using the drawn wave shape rather than displacement direction.
Don't claim sound in air can be polarised.
Don't use the speed of sound for a radio wave in vacuum.

Exam tip

State both displacement direction and energy-propagation direction when classifying a wave.

Tier 1 · Easy

State whether a sound wave travelling through air is longitudinal or transverse, and state the direction in which the air molecules oscillate.
[2 marks]
Total for this question: 2
Infrared radiation has frequency $3.00\times10^{14}\,\text{Hz}$ and ultraviolet radiation has frequency $1.20\times10^{15}\,\text{Hz}$ . Compare their speeds and wavelengths in vacuum.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A radio transmitter produces a vertically polarised wave. Explain why the signal received by a straight aerial decreases when the aerial is rotated from vertical towards horizontal.
[3 marks]
Total for this question: 3
Explain how Polaroid sunglasses reduce glare from a horizontal road and why rotating the glasses through $90^{\circ}$ changes the transmitted glare.
[3 marks]
Total for this question: 3
A vertically polarised microwave travels towards a vertical receiving aerial. A grille made from parallel metal rods is placed between the transmitter and receiver. Explain why the received signal is greater when the rods are horizontal than when they are vertical.
[3 marks]
Total for this question: 3

Tier 3 · Hard

At a distance of $1.20\,\text{km}$ , a detector receives a radio pulse and a sound pulse that were emitted simultaneously. Take the speed of sound as $340\,\text{m s}^{-1}$ . Calculate the arrival-time difference and explain why passing the radio wave through a correctly oriented polariser supports its classification as transverse.
[5 marks]
Total for this question: 5
Describe an experiment to test how the speed of water waves depends on water depth.
[5 marks]
Total for this question: 5
Two ideal Polaroid sheets have perpendicular transmission axes, one vertical and one horizontal, so no light passes through the pair. A third ideal Polaroid with its axis at $45^{\circ}$ to the vertical is inserted between them. Explain why light is now transmitted and how this observation supports the classification of light as a transverse wave.
[5 marks]
Total for this question: 5
Two seismic pulses P and S travel through solid rock at $8.20\,\text{km s}^{-1}$ and $4.70\,\text{km s}^{-1}$ respectively. Both have frequency $2.00\,\text{Hz}$ . P is detected at a far-side station along a path that crosses Earth's liquid outer core, whereas S is not. Calculate the wavelength of each pulse. Deduce which pulse is longitudinal and explain why an S-wave shadow zone is observed beyond the liquid outer core. The two pulses are emitted together and arrive at a station $1.15\times10^3\,\text{km}$ away. Calculate the difference in their arrival times.
[5 marks]
Total for this question: 5
Describe a sequence of observations using a microwave transmitter and a straight receiving aerial that would show that the radiation is plane polarised and support the conclusion that it is transverse. Your method must distinguish an orientation effect from a change in transmitter output.
[5 marks]
Total for this question: 5

3.3.1.3 · Principle of superposition of waves and formation of stationary waves

Explanation

Superposition makes resultant displacement the vector sum of individual displacements. Two progressive waves of equal frequency travelling in opposite directions form a stationary wave: fixed nodes have zero amplitude and antinodes have maximum amplitude, with adjacent nodes separated by $\lambda/2$ .
A string fixed at both ends has $n$ loops in its $n$ th harmonic; the first harmonic satisfies $f=(1/2l)\sqrt{T/\mu}$ .
A graphical explanation should show cancellation at nodes and reinforcement at antinodes.
Stationary patterns also occur with microwaves and sound.
Required practical 1 investigates how string frequency varies with length, tension and mass per unit length; “fundamental” and “overtone” terminology is not used.

A third-harmonic stationary wave on a string has fixed nodes and three antinodes.

Worked example

A $0.75\,\text{m}$ string has tension $45\,\text{N}$ and mass per unit length $1.8\times10^{-3}\,\text{kg m}^{-1}$ . Find its first-harmonic frequency.

1.Use $f=(1/2l)\sqrt{T/\mu}$ .
2. $f=[2(0.75)]^{-1}\sqrt{45/(1.8\times10^{-3})}$ .
3.Evaluate the square root before dividing by $1.50$ .

Answer: The first-harmonic frequency is 1.1 × 10² Hz.

Common mistakes

Don't set string length equal to one wavelength for the first harmonic.
Don't call an antinode a point of zero displacement at every instant.
Don't change more than one of length, tension and mass per unit length without controlling variables.

Exam tip

Count loops to identify the harmonic, then use half a wavelength between adjacent nodes.

Tier 1 · Easy

State the difference between a node and an antinode in a stationary wave.
[1 mark]
Total for this question: 1
State the phase relationship between points between the same pair of adjacent nodes in a stationary wave and between points in adjacent loops.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A string of length $0.75\,\text{m}$ is fixed at both ends. Its tension is $45\,\text{N}$ and its mass per unit length is $1.8\times10^{-3}\,\text{kg m}^{-1}$ . Determine the frequency of its first harmonic.
[3 marks]
Total for this question: 3
Successive detector maxima in a microwave stationary wave are $1.62\,\text{cm}$ apart. Determine the microwave frequency.
[3 marks]
Total for this question: 3
Four consecutive nodes on a stationary wave span $0.720\,\text{m}$ from the first node to the fourth. The wave frequency is $250\,\text{Hz}$ . Determine the wavelength and the speed of the progressive waves that form the pattern.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $0.90\,\text{m}$ string fixed at both ends has mass per unit length $2.5\times10^{-3}\,\text{kg m}^{-1}$ . Its third-harmonic stationary wave has frequency $120\,\text{Hz}$ . Determine the string tension and the numbers of nodes and antinodes in this pattern.
[5 marks]
Total for this question: 5
For a string of fixed length $0.800\,\text{m}$ , a graph of first-harmonic frequency against $\sqrt{T}$ has gradient $18.0\,\text{Hz N}^{-1/2}$ . Determine the string's mass per unit length and the mass of the vibrating length.
[5 marks]
Total for this question: 5
A string of length $0.900\,\text{m}$ and mass per unit length $1.50\times10^{-3}\,\text{kg m}^{-1}$ is driven at a fixed frequency of $180\,\text{Hz}$ . It produces a stationary wave with a node at each end. Its tension can be set only between $6.00\,\text{N}$ and $45.0\,\text{N}$ . Determine every tension in this range that produces a stationary wave, and identify the harmonic for each tension.
[5 marks]
Total for this question: 5
A string of vibrating length $1.35\,\text{m}$ is kept under tension by a hanging mass of $1.48\,\text{kg}$ . Resonances are observed at $112\,\text{Hz}$ and $168\,\text{Hz}$ , with no resonance between them. Take $g=9.81\,\text{N kg}^{-1}$ . Identify both harmonics and determine the mass per unit length of the string.
[5 marks]
Total for this question: 5
In a stationary-wave experiment, a $0.640\,\text{m}$ string is driven in its first harmonic. A graph of $f^2$ against the measured tension $T$ has gradient $2.85\times10^3\,\text{Hz}^2\,\text{N}^{-1}$ and intercept $51.3\,\text{Hz}^2$ . Calculate the string's mass per unit length and the measured tension required for a frequency of $180\,\text{Hz}$ . Explain what the intercept implies if the actual tension is $T+T_0$ .
[6 marks]
Total for this question: 6

3.3.2.1 · Interference

Explanation

Coherent sources have the same frequency and constant phase difference, producing a stable interference pattern. For in-phase sources, path difference $n\lambda$ gives constructive interference and $(n+\tfrac12)\lambda$ gives destructive interference.
Young’s double slits create two coherent sources from one source, with fringe spacing $w=\lambda D/s$ . White light gives a white central fringe and coloured side fringes because spacing depends on wavelength.
Interference must also be described for sound and other electromagnetic waves. Required practical 2 includes Young’s slits and a diffraction grating.
Laser light is monochromatic, but the beam must not be viewed directly.
Interference evidence contributed to changing models of electromagnetic radiation.

Two coherent slits separated by s form fringes on a screen a distance D away.

Worked example

Light of wavelength $600\,\text{nm}$ passes through slits $0.40\,\text{mm}$ apart onto a screen $2.4\,\text{m}$ away. Find fringe spacing.

1.Convert $\lambda=6.00\times10^{-7}\,\text{m}$ and $s=4.0\times10^{-4}\,\text{m}$ .
2.Use $w=\lambda D/s$ .
3. $w=(6.00\times10^{-7})(2.4)/(4.0\times10^{-4})$ .

Answer: The fringe spacing is 3.6 × 10⁻³ m, or 3.6 mm.

Common mistakes

Don't define coherent sources as merely having equal amplitude.
Don't count bright fringes rather than the intervals between them when finding mean spacing.
Don't look into a laser beam or omit laser safety from a practical method.

Exam tip

Measure across several fringe intervals and divide by the number of intervals to reduce percentage uncertainty.

Tier 1 · Easy

State the two conditions that two sources must satisfy to be coherent.
[2 marks]
Total for this question: 2
Two coherent sources emit in phase with wavelength $0.240\,\text{m}$ . State the smallest path difference for destructive interference.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Light of wavelength $600\,\text{nm}$ illuminates two slits separated by $0.40\,\text{mm}$ . A screen is $2.4\,\text{m}$ from the slits. Calculate the fringe spacing.
[3 marks]
Total for this question: 3
Two in-phase loudspeakers emit a $425\,\text{Hz}$ tone. Sound travels at $340\,\text{m s}^{-1}$ . A microphone is $3.40\,\text{m}$ from one speaker and $4.20\,\text{m}$ from the other. Determine whether constructive or destructive interference occurs.
[3 marks]
Total for this question: 3
Two coherent loudspeakers are driven in antiphase. The sound speed is $330\,\text{m s}^{-1}$ . At a detector, one path is $0.375\,\text{m}$ longer than the other. Determine the lowest frequency at which a minimum is detected.
[3 marks]
Total for this question: 3

Tier 3 · Hard

In a Young double-slit experiment, the distance from the first to the ninth bright fringe is $28.8\,\text{mm}$ . The screen is $1.80\,\text{m}$ from the slits and the wavelength is $520\,\text{nm}$ . Determine the slit separation. Explain the appearance near the centre when the laser is replaced by white light.
[5 marks]
Total for this question: 5
Two coherent microwave transmitters emit in phase. The path difference to a detector is $0.0900\,\text{m}$ . Determine the two lowest frequencies that produce destructive interference at the detector and explain their numerical relationship.
[5 marks]
Total for this question: 5
Coherent sources S₁ and S₂ have a constant phase difference. At detector A, the path from S₂ is $0.150\,\text{m}$ longer than the path from S₁ and a maximum is detected. At detector B, the corresponding path difference is $0.450\,\text{m}$ and a minimum is detected. The wavelength is known to lie between $0.400\,\text{m}$ and $0.800\,\text{m}$ . Determine the wavelength and the phase by which S₂ leads S₁.
[5 marks]
Total for this question: 5
In a Young double-slit experiment, the distance across $20$ fringe intervals is $(63.8\pm0.4)\,\text{mm}$ . The slit separation is $(0.285\pm0.003)\,\text{mm}$ and the screen distance is $(1.73\pm0.01)\,\text{m}$ . Determine the wavelength and its maximum percentage uncertainty.
[5 marks]
Total for this question: 5
A mixture of two monochromatic wavelengths, $462\,\text{nm}$ and $616\,\text{nm}$ , illuminates double slits separated by $0.320\,\text{mm}$ . A screen is $2.15\,\text{m}$ away. Determine the first non-central position where bright fringes from both wavelengths coincide, and identify the order of each fringe.
[5 marks]
Total for this question: 5

3.3.2.2 · Diffraction

Explanation

Diffraction is wave spreading at an aperture or obstacle and becomes more pronounced when aperture size is comparable with wavelength. A monochromatic single-slit pattern has a broad central bright maximum with narrower, weaker side maxima.
Increasing wavelength or decreasing slit width increases the central maximum’s angular width; an intensity-against-angle graph is not required. White light produces a white centre with coloured edges.
For a plane transmission grating at normal incidence, maxima satisfy $d\sin\theta=n\lambda$ ; this relationship should be derived from path difference.
Convert line density to spacing with $d=1/N$ .
Gratings separate wavelengths and are used in spectral analysis; spectrometer operation is not tested.

A narrow slit spreads a wave into a broad central diffraction maximum and weaker side maxima.

Worked example

A grating has $400$ lines per millimetre. Find the first-order angle for $589\,\text{nm}$ light at normal incidence.

1. $N=400\times10^3\,\text{m}^{-1}$ , so $d=1/N=2.50\times10^{-6}\,\text{m}$ .
2.Use $d\sin\theta=n\lambda$ with $n=1$ .
3. $\theta=\sin^{-1}[(589\times10^{-9})/(2.50\times10^{-6})]$ .

Answer: The first-order angle is 13.6°.

Common mistakes

Don't say diffraction is greatest when the aperture is much larger than the wavelength.
Don't use line density directly as grating spacing.
Don't accept an order for which the calculated sine of the angle exceeds one.

Exam tip

For the highest order, require order times wavelength not to exceed grating spacing, then choose the greatest integer order.

Tier 1 · Easy

State how the width of the central diffraction maximum changes when monochromatic light passes through a narrower single slit.
[1 mark]
Total for this question: 1
A narrow single slit is illuminated by monochromatic light. Compare the central bright maximum with the bright side maxima.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A diffraction grating has $400$ lines per $\text{mm}$ . Light of wavelength $589\,\text{nm}$ is incident normally. Calculate the angle of the first-order maximum.
[3 marks]
Total for this question: 3
White light passes through a narrow single slit. Describe and explain the appearance of the diffraction pattern near its centre.
[3 marks]
Total for this question: 3
A plane wave is incident normally on a transmission grating whose adjacent slits are separated by $d$ . Assume rays reaching a distant screen are parallel. Explain why a principal maximum at angle $\theta$ satisfies $d\sin\theta=n\lambda$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

A plane transmission grating has $600$ lines per $\text{mm}$ . It is illuminated normally with light of wavelength $480\,\text{nm}$ . Determine the highest observable order and the angle of this order. Explain why the next order cannot occur.
[5 marks]
Total for this question: 5
A plane transmission grating has $500$ lines per $\text{mm}$ . The second-order maximum from an unknown spectral line coincides with the third-order maximum from light of wavelength $450\,\text{nm}$ . Determine the unknown wavelength and the angular separation of the two first-order maxima.
[5 marks]
Total for this question: 5
White light of wavelengths $380\,\text{nm}$ to $780\,\text{nm}$ is incident normally on a grating with $800$ lines per $\text{mm}$ . Determine the angular interval over which the first-order and second-order spectra overlap on one side of the central maximum. State the first-order and second-order wavelength ranges present in this interval.
[5 marks]
Total for this question: 5
A grating is labelled $(725\pm5)$ lines per millimetre. Its first-order maximum is measured at $(24.6\pm0.2)^{\circ}$ for monochromatic light at normal incidence. Determine the wavelength and estimate its maximum uncertainty.
[5 marks]
Total for this question: 5
Light containing wavelengths $487.3\,\text{nm}$ and $488.1\,\text{nm}$ is incident normally on a grating with $640$ lines per millimetre. A spectrometer can resolve two maxima only if their angular separation is at least $0.240^{\circ}$ . Calculate the angular separation of the third-order maxima and determine whether the two wavelengths are resolved in this order. Do not round the intermediate angles.
[5 marks]
Total for this question: 5

3.3.2.3 · Refraction at a plane surface

Explanation

Refractive index is $n=c/v$ , with air approximately $1$ . At a boundary, $n_1\sin\theta_1=n_2\sin\theta_2$ , and angles are measured from the normal.
Total internal reflection requires travel from higher to lower refractive index and incidence greater than critical angle, where $\sin\theta_c=n_2/n_1$ . A step-index optical fibre uses a higher-index core and lower-index cladding to maintain a suitable boundary and guide light.
Modal dispersion arises from different ray path lengths; material dispersion arises from wavelength-dependent speed.
Both broaden pulses and limit data rate, while absorption reduces pulse energy and limits transmission distance.
Consequences should be linked to overlapping or weakened received pulses.

A step-index fibre guides light by total internal reflection at the core–cladding boundary.

Worked example

Glass of refractive index $1.52$ borders air. Find the critical angle and decide whether incidence at $44.0^{\circ}$ gives total internal reflection.

1. $\sin\theta_c=n_{\text{air}}/n_{\text{glass}}=1/1.52$ .
2. $\theta_c=41.1^{\circ}$ .
3.The ray travels from higher to lower $n$ and $44.0^{\circ}>41.1^{\circ}$ .

Answer: The critical angle is $41.1^{\circ}$ , so total internal reflection occurs.

Common mistakes

Don't measure incidence and refraction angles from the surface.
Don't check only the critical angle but not travel from higher to lower refractive index.
Don't swap material dispersion and modal dispersion.

Exam tip

A total-internal-reflection answer needs both the refractive-index direction and incidence greater than the critical angle.

Tier 1 · Easy

Light travels through a transparent material at $2.00\times10^8\,\text{m s}^{-1}$ . Calculate its refractive index.
[2 marks]
Total for this question: 2
A transparent glass has refractive index $1.62$ . Calculate the speed of light in the glass.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A ray in glass of refractive index $1.52$ meets a glass-air boundary at an incidence angle of $44.0^{\circ}$ . Determine whether total internal reflection occurs.
[3 marks]
Total for this question: 3
A ray travels from air into acrylic of refractive index $1.49$ at an incidence angle of $35.0^{\circ}$ . Calculate the angle of refraction and the angle through which the ray changes direction.
[3 marks]
Total for this question: 3
Two monochromatic rays follow different routes through a step-index fibre of refractive index $1.50$ . One route is $2.00\,\text{km}$ long and the other is $2.10\,\text{km}$ long. Determine the difference between their arrival times and name the dispersion responsible.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A step-index optical fibre has core refractive index $1.50$ and cladding refractive index $1.47$ . A ray in the core meets the boundary at $80.0^{\circ}$ to the normal. Show that the ray is guided. Explain how material dispersion and modal dispersion broaden a light pulse in this fibre.
[6 marks]
Total for this question: 6
A ray in air enters an oil at $52.0^{\circ}$ to the normal and refracts at $30.0^{\circ}$ . Determine the oil's refractive index, the critical angle at an oil-air boundary and the speed of light in the oil.
[5 marks]
Total for this question: 5
A monochromatic pulse of duration $5.00\,\text{ns}$ enters a $900\,\text{m}$ step-index fibre of core refractive index $1.50$ and cladding refractive index $1.45$ . Some light travels along the fibre axis and some follows a route at $10.0^{\circ}$ to the axis. Assume both routes remain in the core and that material dispersion is negligible. Determine the shortest possible output-pulse duration and the greatest pulse frequency that avoids overlap.
[5 marks]
Total for this question: 5
A ray enters a parallel-sided acrylic block at $48.0^{\circ}$ to the normal. The acrylic has refractive index $1.47$ and air has refractive index $1.00$ . Determine the angle inside the block, the angle at which the ray emerges into air, and the speed of light in the acrylic.
[4 marks]
Total for this question: 4
A transmitter sends $35.0\,\text{ns}$ light pulses at a repetition frequency of $4.00\,\text{MHz}$ through a fibre. The pulses contain two wavelengths whose refractive indices are $1.472$ and $1.478$ . Assuming this refractive-index difference is the only cause of broadening, determine the greatest fibre length for which adjacent received pulses do not overlap. Explain why using a fibre with a core narrow enough to support one path would not remove this limit.
[6 marks]
Total for this question: 6

Answer key

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

3.3.1.1 · Progressive waves

Tier 1 · Easy

Mark scheme for 3.3.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.2\times10^{2}\,\text{m s}^{-1}$	The period is $T=2.5\times10^{-3}\,\text{s}$ , so $f=1/T=400\,\text{Hz}$ . Hence $v=f\lambda=400\times0.80=3.2\times10^{2}\,\text{m s}^{-1}$ .	1
02.1	$0.500\,\text{m}$	Use $v=f\lambda$ , so $\lambda=v/f=340/680=0.500\,\text{m}$ to three significant figures.	1

Tier 2 · Standard

Mark scheme for 3.3.1.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$135^{\circ}$ , or $2.36\,\text{rad}$	The separation is $x/\lambda=0.24/0.64=0.375$ cycles. Therefore $\Delta\phi=360^{\circ}\times0.375=135^{\circ}$ . In radians, $\Delta\phi=2\pi\times0.375=0.75\pi=2.36\,\text{rad}$ .	3
02.1	$3.40\times10^2\,\text{m s}^{-1}$ and $11.8\,\text{ms}$	Convert the measured delay to seconds: $\Delta t=7.35\times10^{-3}\,\text{s}$ . The speed is $v=x/\Delta t=2.50/(7.35\times10^{-3})=340.136\,\text{m s}^{-1}=3.40\times10^2\,\text{m s}^{-1}$ . For a $4.00\,\text{m}$ separation, $\Delta t=x/v=4.00/340.136=1.176\times10^{-2}\,\text{s}=11.8\,\text{ms}$ to three significant figures.	3
03.1	$2.16\,\text{m}$ and $360\,\text{m s}^{-1}$	The delay is $1.50/6.00=0.250$ of a cycle. Since the separation is less than one wavelength, $0.540\,\text{m}=0.250\lambda$ , so $\lambda=2.16\,\text{m}$ . The wave speed is $v=x/\Delta t=0.540/(1.50\times10^{-3})=360\,\text{m s}^{-1}$ .	3

Tier 3 · Hard

Mark scheme for 3.3.1.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.84\,\text{m s}^{-1}$ ; $3.95\,\text{rad}$ (or $226^{\circ}$ ); $0.26\,\text{s}$	First, $v=f\lambda=2.4\times0.35=0.84\,\text{m s}^{-1}$ . The phase change is $\Delta\phi=2\pi x/\lambda=2\pi(0.22/0.35)=3.95\,\text{rad}$ , which is $226^{\circ}$ . The propagation delay is $t=x/v=0.22/0.84=0.262\,\text{s}$ , so to two significant figures it is $0.26\,\text{s}$ .	5
02.1	$2.00\,\text{kHz}$ ; $0.171\,\text{m}$ ; $342\,\text{m s}^{-1}$	The period is $T=(3.00\times10^{-3})/6=5.00\times10^{-4}\,\text{s}$ , so $f=1/T=2.00\times10^3\,\text{Hz}$ . Adjacent minima are separated by $\lambda/2$ , giving $\lambda=2(8.55\times10^{-2})=0.171\,\text{m}$ . Hence $v=f\lambda=(2.00\times10^3)(0.171)=342\,\text{m s}^{-1}$ .	5
03.1	$1.08\,\text{m}$ and $43.2\,\text{m s}^{-1}$	The period is $T=1/f=1/40.0=25.0\,\text{ms}$ . A $90.0^{\circ}$ lag corresponds to a travel time of $(n+0.250)T$ . The only value within the measured interval is obtained with $n=1$ , giving $\Delta t=1.250(25.0\,\text{ms})=31.25\,\text{ms}$ . Thus $v=1.35/(31.25\times10^{-3})=43.2\,\text{m s}^{-1}$ and $\lambda=v/f=43.2/40.0=1.08\,\text{m}$ .	5
04.1	$0.806\,\text{m}$ ; $47.1\,\text{m s}^{-1}$ ; $2.15\,\text{rad}$ (accept $2.14$ to $2.15\,\text{rad}$ )	Since $\Delta\phi=(2\pi/\lambda)x$ , the graph gradient is $2\pi/\lambda$ . Hence $\lambda=2\pi/7.80=0.805536\,\text{m}=0.806\,\text{m}$ . The speed is $v=f\lambda=58.5(0.805536)=47.1239\,\text{m s}^{-1}=47.1\,\text{m s}^{-1}$ . At $x=0.275\,\text{m}$ , $\Delta\phi=7.80(0.275)=2.145\,\text{rad}=2.15\,\text{rad}$ .	5
05.1	$349\,\text{m s}^{-1}$ ; $613\,\text{Hz}$ ; $0.569\,\text{m}$ ; $3.42\,\text{rad}$	The echo travels an extra distance $2d=2(2.13)=4.26\,\text{m}$ , so $v=4.26/(12.2\times10^{-3})=349.180\,\text{m s}^{-1}$ . The frequency is $f=1/T=1/(1.63\times10^{-3})=613.497\,\text{Hz}$ . Therefore $\lambda=v/f=349.180/613.497=0.569164\,\text{m}$ . The phase difference is $\Delta\phi=2\pi x/\lambda=2\pi(0.310/0.569164)=3.42178\,\text{rad}=3.42\,\text{rad}$ .	6

3.3.1.2 · Longitudinal and transverse waves

Tier 1 · Easy

Mark scheme for 3.3.1.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Longitudinal; the molecules oscillate parallel to the direction of energy propagation.	Sound in air consists of compressions and rarefactions, so it is longitudinal. The molecules move back and forth parallel to the direction in which the wave transfers energy.	2
02.1	They have the same speed; the infrared wavelength is four times the ultraviolet wavelength.	All electromagnetic waves travel at the same speed, $c$ , in vacuum. Since $\lambda=c/f$ , wavelength is inversely proportional to frequency. The ultraviolet frequency is four times the infrared frequency, so its wavelength is one quarter as large.	2

Tier 2 · Standard

Mark scheme for 3.3.1.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The electric field oscillates vertically, so it drives charges most effectively in a vertical aerial; rotation reduces the field component along the aerial and a horizontal aerial receives a minimum signal.	Vertical polarisation means that the electric field oscillates vertically. The field component parallel to the conducting aerial drives its charges and induces the received signal. Rotating the aerial reduces this parallel component, which becomes zero in the ideal horizontal orientation.	3
02.1	Reflected glare is partially horizontally plane polarised; a vertical Polaroid transmission axis rejects most of this component, whereas a $90^{\circ}$ rotation aligns the axis horizontally and transmits more glare.	Light reflected from a horizontal road is preferentially polarised with its electric field horizontal. Polaroid material absorbs the electric-field component perpendicular to its transmission axis and transmits the parallel component. A vertical axis therefore rejects most horizontal glare; rotating the glasses by $90^{\circ}$ makes the axis horizontal and increases the transmitted glare.	3
03.1	The vertical electric field drives currents in vertical rods, so they absorb or reflect much of the wave; horizontal rods are perpendicular to the electric-field oscillation, so little current is driven and more of the wave reaches the aerial.	The transmitting aerial produces a vertical electric-field oscillation. Vertical conducting rods let charges move along the field, producing currents that remove energy from the transmitted beam by absorption or reflection. With horizontal rods, the field has no component along the rods, so much less current is induced and the received signal is larger.	3

Tier 3 · Hard

Mark scheme for 3.3.1.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.53\,\text{s}$ ; polarisation selects one direction of field oscillation, which is possible only for a transverse wave.	For the radio pulse, $t_{\rm radio}=1200/(3.00\times10^8)=4.00\times10^{-6}\,\text{s}$ . For sound, $t_{\rm sound}=1200/340=3.529\,\text{s}$ . The difference is $3.529-0.000004=3.53\,\text{s}$ . A polariser transmits a selected direction of oscillation; a longitudinal wave has oscillations only along its travel direction and cannot show this effect, so the observation supports a transverse model.	5
02.1	Vary depth alone, keep the generator frequency fixed, measure the wavelength at each depth, calculate $v=f\lambda$ , repeat and plot speed against depth.	Use a ripple tank with a straight wave generator driven at a measured fixed frequency. Set a known water depth, use a strobe or frozen image to measure the distance across several wavefront intervals, and divide by the number of wavelengths. Calculate $v=f\lambda$ . Repeat the wavelength measurement and average it, then repeat for several depths while keeping generator frequency and tank geometry unchanged. Plot mean wave speed against depth.	5
03.1	The first sheet produces vertical plane-polarised light; this has a component along the $45^{\circ}$ axis, so the middle sheet transmits and repolarises it at $45^{\circ}$ ; that light has a horizontal component which the final sheet transmits. Selection of perpendicular oscillation directions is possible only for a transverse wave.	After the first sheet, the electric field oscillates vertically. It has a non-zero component along the middle sheet's $45^{\circ}$ axis, so some light passes and emerges polarised at $45^{\circ}$ . This new oscillation has a non-zero horizontal component, allowing some light through the final sheet. A wave can be restricted to selected directions perpendicular to its propagation only when its oscillations are transverse, so the effect is evidence that light is transverse.	5
04.1	$4.10\,\text{km}$ and $2.35\,\text{km}$ ; P is longitudinal; liquids do not transmit transverse S waves, so no direct S waves reach the shadow zone; difference in arrival times $104\,\text{s}$ (accept $1.0\times10^2\,\text{s}$ )	Using $v=f\lambda$ , $\lambda_{\rm P}=8.20/2.00=4.10\,\text{km}$ and $\lambda_{\rm S}=4.70/2.00=2.35\,\text{km}$ . A liquid can transmit compressions but has no shear rigidity, so the pulse that crosses the liquid outer core, P, is longitudinal. S is transverse and cannot propagate through the liquid outer core, leaving a region on the far side in which no direct S waves are detected. The speeds are $v_{\rm P}=2.00(4.10\times10^3)=8.20\times10^3\,\text{m s}^{-1}$ and $v_{\rm S}=2.00(2.35\times10^3)=4.70\times10^3\,\text{m s}^{-1}$ . Hence $t_{\rm S}-t_{\rm P}=1.15\times10^6/(4.70\times10^3)-1.15\times10^6/(8.20\times10^3)=244.68-140.24=104\,\text{s}$ .	5
05.1	Rotate the receiver about the beam axis through at least $180^{\circ}$ , record repeated maxima and minima at perpendicular orientations, restore the maximum orientation between readings to monitor source output, and show that rotating the transmitter rotates the receiver orientation for maximum signal by the same angle. Directional polarisation is possible only for transverse oscillations.	Keep transmitter–receiver distance and alignment fixed. Rotate the receiving aerial about the propagation axis and record signal strength at regular angles; a maximum when the aerials are parallel and a minimum when they are perpendicular demonstrates selection of one electric-field direction. Recheck the original maximum orientation during the run so drift in output cannot imitate the angular pattern. Rotate the transmitter by a known angle and repeat: the receiver maximum should rotate by that angle. A wave whose oscillation can be restricted to a direction perpendicular to propagation is transverse.	5

3.3.1.3 · Principle of superposition of waves and formation of stationary waves

Tier 1 · Easy

Mark scheme for 3.3.1.3 Tier 1 · Easy
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01.1	A node has zero amplitude, whereas an antinode has maximum amplitude.	At a node, the two component waves always cancel. At an antinode, their displacements reinforce to give the largest oscillation amplitude.	1
02.1	Points between the same pair of adjacent nodes oscillate in phase; points in adjacent loops oscillate in antiphase, so they are $180^{\circ}$ or $\pi\,\text{rad}$ out of phase.	All points between one pair of neighbouring nodes reach corresponding displacements together. Crossing a node reverses the displacement, giving a phase difference of $180^{\circ}$ between adjacent loops.	2

Tier 2 · Standard

Mark scheme for 3.3.1.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.1\times10^2\,\text{Hz}$	Use $f=(2l)^{-1}\sqrt{T/\mu}$ . Thus $f=[2(0.75)]^{-1}\sqrt{45/(1.8\times10^{-3})}=(1/1.50)\sqrt{25000}=105\,\text{Hz}$ , which is $1.1\times10^2\,\text{Hz}$ to two significant figures.	3
02.1	$9.26\times10^9\,\text{Hz}$	Successive maxima are adjacent antinodes, separated by $\lambda/2$ . Hence $\lambda=2(1.62\times10^{-2})=3.24\times10^{-2}\,\text{m}$ . Then $f=c/\lambda=(3.00\times10^8)/(3.24\times10^{-2})=9.26\times10^9\,\text{Hz}$ to three significant figures.	3
03.1	$0.480\,\text{m}$ and $120\,\text{m s}^{-1}$	Four consecutive nodes contain three node intervals, so one interval is $0.720/3=0.240\,\text{m}$ . Adjacent nodes are separated by $\lambda/2$ , giving $\lambda=0.480\,\text{m}$ . Hence $v=f\lambda=250(0.480)=120\,\text{m s}^{-1}$ .	3

Tier 3 · Hard

Mark scheme for 3.3.1.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$13\,\text{N}$ ; $4$ nodes and $3$ antinodes	For harmonic number $n$ , $f_n=n(2l)^{-1}\sqrt{T/\mu}$ . Rearranging gives $T=\mu(2lf_n/n)^2$ . Hence $T=2.5\times10^{-3}[2(0.90)(120)/3]^2=2.5\times10^{-3}(72)^2=12.96\,\text{N}$ , so $T=13\,\text{N}$ . The third harmonic has three loops, so it has three antinodes and four nodes including the fixed ends.	5
02.1	$1.21\times10^{-3}\,\text{kg m}^{-1}$ and $0.965\,\text{g}$ (accept $0.968\,\text{g}$ if the rounded value of $\mu$ is carried forward)	Since $f=(2l)^{-1}\sqrt{T/\mu}$ , the gradient of $f$ against $\sqrt{T}$ is $G=1/(2l\sqrt{\mu})$ . Thus $\mu=[1/(2lG)]^2=[1/(2(0.800)(18.0))]^2=1.2056\times10^{-3}\,\text{kg m}^{-1}=1.21\times10^{-3}\,\text{kg m}^{-1}$ . Using the unrounded value, the mass of the vibrating length is $m=\mu l=9.6451\times10^{-4}\,\text{kg}=0.965\,\text{g}$ to three significant figures.	5
03.1	$39.4\,\text{N}$ for the second harmonic; $17.5\,\text{N}$ for the third; $9.84\,\text{N}$ for the fourth; $6.30\,\text{N}$ for the fifth	For harmonic number $n$ , $f=n(2l)^{-1}\sqrt{T/\mu}$ , so $T=\mu(2lf/n)^2$ . Here $T=(1.50\times10^{-3})[2(0.900)(180)/n]^2=157.464/n^2\,\text{N}$ . The allowed range excludes $n=1$ , for which $T=157\,\text{N}$ , and $n\ge6$ , because $T\le4.37\,\text{N}$ . Evaluating $n=2,3,4,5$ gives $39.366\,\text{N}$ , $17.496\,\text{N}$ , $9.8415\,\text{N}$ and $6.29856\,\text{N}$ , respectively, which round to the stated values.	5
04.1	Second and third harmonics; $6.35\times10^{-4}\,\text{kg m}^{-1}$	Adjacent harmonics differ by the first-harmonic frequency, so $f_1=168-112=56.0\,\text{Hz}$ . Thus $112=2f_1$ is the second harmonic and $168=3f_1$ is the third. The tension is $T=mg=1.48(9.81)=14.5188\,\text{N}$ . From $f_1=(1/2l)\sqrt{T/\mu}$ , $\mu=T/(2lf_1)^2=14.5188/[2(1.35)(56.0)]^2=6.35078\times10^{-4}\,\text{kg m}^{-1}=6.35\times10^{-4}\,\text{kg m}^{-1}$ .	5
05.1	$2.14\times10^{-4}\,\text{kg m}^{-1}$ ; $11.4\,\text{N}$ ; an additional tension $T_0=0.0180\,\text{N}$ is present when the measured tension is zero	For the first harmonic, $f^2=T/(4l^2\mu)$ , so the gradient is $1/(4l^2\mu)$ . Hence $\mu=1/[4(0.640)^2(2.85\times10^3)]=2.14158\times10^{-4}\,\text{kg m}^{-1}=2.14\times10^{-4}\,\text{kg m}^{-1}$ . From the graph equation, the reading at $180\,\text{Hz}$ is $T=(180^2-51.3)/(2.85\times10^3)=11.3504\,\text{N}=11.4\,\text{N}$ . If $f^2=G(T+T_0)$ , the intercept is $GT_0$ . Therefore $T_0=51.3/(2.85\times10^3)=0.0180\,\text{N}$ , consistent with a small unaccounted initial tension rather than a line through the origin.	6

3.3.2.1 · Interference

Tier 1 · Easy

Mark scheme for 3.3.2.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	They have the same frequency and a constant phase difference.	Coherence requires equal frequency so the relative phase does not drift, and a constant phase difference so fixed maxima and minima can form.	2
02.1	$0.120\,\text{m}$	For in-phase coherent sources, destructive interference first occurs at a path difference of $\lambda/2$ . Hence the smallest value is $0.240/2=0.120\,\text{m}$ .	1

Tier 2 · Standard

Mark scheme for 3.3.2.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.6\,\text{mm}$	Convert to SI units: $\lambda=6.00\times10^{-7}\,\text{m}$ and $s=4.0\times10^{-4}\,\text{m}$ . Then $w=\lambda D/s=(6.00\times10^{-7})(2.4)/(4.0\times10^{-4})=3.6\times10^{-3}\,\text{m}=3.6\,\text{mm}$ .	3
02.1	Constructive interference occurs.	The wavelength is $\lambda=v/f=340/425=0.800\,\text{m}$ . The path difference is $4.20-3.40=0.800\,\text{m}=1\lambda$ . An integer number of wavelengths from in-phase sources gives constructive interference.	3
03.1	$880\,\text{Hz}$	The sources begin $180^{\circ}$ out of phase, so a minimum occurs when the path difference is a whole number of wavelengths: $\Delta x=n\lambda$ . Hence $f=nv/\Delta x=n(330)/0.375=880n\,\text{Hz}$ . The smallest allowed value is $n=1$ ( $n=0$ would require an infinite wavelength), so the required frequency is $880\,\text{Hz}$ .	3

Tier 3 · Hard

Mark scheme for 3.3.2.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.260\,\text{mm}$ ; a white central fringe with coloured fringes on either side, violet nearer the centre and red farther out.	From the first to the ninth bright fringe there are eight spacings, so $w=28.8/8=3.60\,\text{mm}=3.60\times10^{-3}\,\text{m}$ . Rearranging $w=\lambda D/s$ gives $s=\lambda D/w=(520\times10^{-9})(1.80)/(3.60\times10^{-3})=2.60\times10^{-4}\,\text{m}=0.260\,\text{mm}$ . At zero path difference every visible wavelength reinforces, giving a white central fringe. Since $w\propto\lambda$ , violet fringes are closer to the centre and red fringes are farther away.	5
02.1	$1.67\,\text{GHz}$ and $5.00\,\text{GHz}$ ; the second is three times the first because destructive frequencies are odd-number multiples of the lowest	For in-phase sources, destructive interference occurs when $\Delta x=(n+\tfrac12)\lambda$ . For $n=0$ , $\lambda=2(0.0900)=0.180\,\text{m}$ , so $f=c/\lambda=(3.00\times10^8)/0.180=1.67\times10^9\,\text{Hz}$ . For $n=1$ , $0.0900=1.5\lambda$ , so $\lambda=0.0600\,\text{m}$ and $f=5.00\times10^9\,\text{Hz}$ . In general $f=(2n+1)c/(2\Delta x)$ , so the allowed destructive frequencies are odd-number multiples of the lowest and the second is three times the first.	5
03.1	$0.600\,\text{m}$ and $90.0^{\circ}$ (or $\pi/2\,\text{rad}$ )	Changing from a maximum to a minimum requires an odd half-wavelength change in path difference. Here the change is $0.450-0.150=0.300\,\text{m}$ . The stated wavelength range permits only $0.300=\lambda/2$ , so $\lambda=0.600\,\text{m}$ . At A, the extra $0.150\,\text{m}$ path delays the wave from S₂ by $2\pi(0.150/0.600)=\pi/2$ . For the waves to arrive in phase at the maximum, S₂ must therefore lead S₁ by $\pi/2\,\text{rad}=90.0^{\circ}$ .	5
04.1	$(525\pm12)\,\text{nm}$ , with maximum percentage uncertainty $2.26\%$	The fringe spacing is $w=63.8/20=3.190\,\text{mm}$ . Hence $\lambda=ws/D=(3.190\times10^{-3})(0.285\times10^{-3})/1.73=5.25520\times10^{-7}\,\text{m}=525\,\text{nm}$ . For products and quotients, maximum fractional uncertainties add: $0.4/63.8+0.003/0.285+0.01/1.73=0.022576$ , or $2.26\%$ . The absolute uncertainty is $0.022576(525)=11.9\,\text{nm}$ , giving $(525\pm12)\,\text{nm}$ .	5
05.1	$12.4\,\text{mm}$ from the centre; fourth order for $462\,\text{nm}$ and third order for $616\,\text{nm}$	Coincident bright fringes require $n_1\lambda_1=n_2\lambda_2$ . Since $4(462)=3(616)=1848\,\text{nm}$ , the first coincidence is the fourth-order $462\,\text{nm}$ fringe with the third-order $616\,\text{nm}$ fringe. Its displacement is $y=n\lambda D/s=4(462\times10^{-9})(2.15)/(0.320\times10^{-3})=0.0124163\,\text{m}=12.4\,\text{mm}$ .	5

3.3.2.2 · Diffraction

Tier 1 · Easy

Mark scheme for 3.3.2.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	The central maximum becomes wider.	A smaller slit-width-to-wavelength ratio produces more diffraction, so the angular spread and hence the central maximum width increase.	1
02.1	The central maximum is brighter and about twice as wide as the side maxima.	The central maximum has the greatest intensity and is about twice the width of each dimmer side maximum.	2

Tier 2 · Standard

Mark scheme for 3.3.2.2 Tier 2 · Standard
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01.1	$13.6^{\circ}$	The line density is $400\times10^3\,\text{m}^{-1}$ , so $d=1/(400\times10^3)=2.50\times10^{-6}\,\text{m}$ . For $n=1$ , $\sin\theta=n\lambda/d=(589\times10^{-9})/(2.50\times10^{-6})=0.2356$ . Therefore $\theta=13.6^{\circ}$ .	3
02.1	The central maximum is white, with coloured edges in which red is farther from the centre than violet because the longer red wavelength diffracts more.	At the centre, the maxima for all visible wavelengths overlap, producing white light. The angular width increases with wavelength, so the red component spreads farther from the centre than the violet component and colours appear at the edges.	3
03.1	The path difference between rays from adjacent slits is $d\sin\theta$ ; a principal maximum requires this difference to be a whole number of wavelengths, $n\lambda$ ; therefore $d\sin\theta=n\lambda$ .	Draw rays from adjacent slits travelling in the direction at angle $\theta$ to the normal. The extra distance travelled by one ray is the projection of the slit separation onto the ray direction, equal to $d\sin\theta$ . Constructive interference occurs when this path difference is $n\lambda$ , giving $d\sin\theta=n\lambda$ .	3

Tier 3 · Hard

Mark scheme for 3.3.2.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Third order at $59.8^{\circ}$	The spacing is $d=1/(600\times10^3)=1.667\times10^{-6}\,\text{m}$ . Since $n\lambda\le d$ , $n\le d/\lambda=(1.667\times10^{-6})/(480\times10^{-9})=3.47$ , so the largest integer order is $n=3$ . Then $\sin\theta=3(480\times10^{-9})/(1.667\times10^{-6})=0.864$ , giving $\theta=59.8^{\circ}$ . For $n=4$ , $\sin\theta=1.152>1$ , so no real diffraction angle exists.	5
02.1	$675\,\text{nm}$ and $6.72^{\circ}$	Coincident maxima have the same $d\sin\theta$ , so $2\lambda_{\rm unknown}=3(450\,\text{nm})$ and $\lambda_{\rm unknown}=675\,\text{nm}$ . The grating spacing is $d=1/(500\times10^3)=2.00\times10^{-6}\,\text{m}$ . For first order, $\theta_{450}=\sin^{-1}[(450\times10^{-9})/d]=13.0^{\circ}$ and $\theta_{675}=\sin^{-1}[(675\times10^{-9})/d]=19.7^{\circ}$ . Their angular separation is $19.7246^{\circ}-13.0029^{\circ}=6.72^{\circ}$ to three significant figures.	5
03.1	$37.4^{\circ}$ to $38.6^{\circ}$ ; first order $760$ – $780\,\text{nm}$ and second order $380$ – $390\,\text{nm}$	The grating spacing is $d=1/(800\times10^3)=1.25\times10^{-6}\,\text{m}$ . At a shared angle, $\lambda_1=2\lambda_2$ . Both orders are therefore present only when the first-order wavelength is $760$ – $780\,\text{nm}$ and the second-order wavelength is $380$ – $390\,\text{nm}$ . The limiting values of $d\sin\theta$ are $760\,\text{nm}$ and $780\,\text{nm}$ , giving $\theta_{\min}=\sin^{-1}[(760\times10^{-9})/d]=37.4^{\circ}$ and $\theta_{\max}=\sin^{-1}[(780\times10^{-9})/d]=38.6^{\circ}$ .	5
04.1	$(574\pm8)\,\text{nm}$	The spacing is $d=1/(725\times10^3)=1.37931\times10^{-6}\,\text{m}$ . Thus $\lambda=d\sin24.6^{\circ}=5.74180\times10^{-7}\,\text{m}=574.180\,\text{nm}$ . Recalculate at the extremes. With $N=720$ and $\theta=24.8^{\circ}$ , $\lambda_{\max}=582.6\,\text{nm}$ ; with $N=730$ and $\theta=24.4^{\circ}$ , $\lambda_{\min}=565.9\,\text{nm}$ . Half the range is $8.3\,\text{nm}$ , i.e. $1.45\%$ .	5
05.1	$0.251^{\circ}$ (accept $0.25^{\circ}$ to $0.26^{\circ}$ ); the wavelengths are resolved in third order	The grating spacing is $d=1/(640\times10^3)=1.5625\times10^{-6}\,\text{m}$ . For third order, $\theta_1=\sin^{-1}[3(487.3\times10^{-9})/d]=69.3279^{\circ}$ and $\theta_2=\sin^{-1}[3(488.1\times10^{-9})/d]=69.5786^{\circ}$ . Their angular separation is $69.5786^{\circ}-69.3279^{\circ}=0.250751^{\circ}=0.251^{\circ}$ . Since this exceeds $0.240^{\circ}$ , the spectrometer resolves the two maxima.	5

3.3.2.3 · Refraction at a plane surface

Tier 1 · Easy

Mark scheme for 3.3.2.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.50$	Use $n=c/v=(3.00\times10^8)/(2.00\times10^8)=1.50$ . Refractive index has no unit.	2
02.1	$1.85\times10^8\,\text{m s}^{-1}$	Use $n=c/v$ , so $v=c/n=(3.00\times10^8)/1.62=1.8519\times10^8\,\text{m s}^{-1}=1.85\times10^8\,\text{m s}^{-1}$ to three significant figures.	2

Tier 2 · Standard

Mark scheme for 3.3.2.3 Tier 2 · Standard
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01.1	Yes; the critical angle is $41.1^{\circ}$ .	For glass to air, $\sin c=n_{\rm air}/n_{\rm glass}=1/1.52$ . Hence $c=\sin^{-1}(1/1.52)=41.1^{\circ}$ . The ray travels from higher to lower refractive index and $44.0^{\circ}>41.1^{\circ}$ , so total internal reflection occurs.	3
02.1	$22.6^{\circ}$ ; a change of $12.4^{\circ}$ towards the normal	Snell's law gives $n_{\rm air}\sin i=n_{\rm acrylic}\sin r$ . Hence $\sin r=\sin35.0^{\circ}/1.49=0.385$ , so $r=22.6^{\circ}$ to three significant figures. The ray changes direction by $35.0^{\circ}-22.6^{\circ}=12.4^{\circ}$ towards the normal.	3
03.1	$0.500\,\mu\text{s}$ ; modal dispersion	The speed in the fibre is $v=c/n=(3.00\times10^8)/1.50=2.00\times10^8\,\text{m s}^{-1}$ . The path difference is $0.100\,\text{km}=100\,\text{m}$ , so the arrival-time difference is $\Delta t=100/(2.00\times10^8)=5.00\times10^{-7}\,\text{s}=0.500\,\mu\text{s}$ . Broadening caused by different ray path lengths is modal dispersion.	3

Tier 3 · Hard

Mark scheme for 3.3.2.3 Tier 3 · Hard
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01.1	The critical angle is $78.5^{\circ}$ , so the ray undergoes total internal reflection; wavelength-dependent speeds and different ray path lengths produce pulse broadening.	At the core-cladding boundary, $\sin c=n_{\rm clad}/n_{\rm core}=1.47/1.50=0.980$ , so $c=78.5^{\circ}$ . The ray is travelling from higher to lower refractive index and its incidence angle is $80.0^{\circ}>c$ , so total internal reflection guides it. Material dispersion occurs because different wavelengths have different refractive indices and therefore different speeds. Modal dispersion occurs because rays at different angles travel different path lengths. Their arrival times spread out, broadening the received pulse and limiting the data rate.	6
02.1	$1.58$ ; $39.4^{\circ}$ (accept $39.3^{\circ}$ to $39.5^{\circ}$ ); $1.90\times10^8\,\text{m s}^{-1}$	From Snell's law, $n_{\rm oil}=\sin52.0^{\circ}/\sin30.0^{\circ}=1.57602=1.58$ . At an oil-air boundary, $\sin c=1/n_{\rm oil}$ , so using the unrounded index gives $c=39.3836^{\circ}=39.4^{\circ}$ . Finally, $v=c/n=(3.00\times10^8)/1.57602=1.9035\times10^8\,\text{m s}^{-1}=1.90\times10^8\,\text{m s}^{-1}$ .	5
03.1	$74.4\,\text{ns}$ and $13.4\,\text{MHz}$	The axial route has length $L$ , while the angled route has length $L/\cos10.0^{\circ}$ . Their delay is $\Delta t=(nL/c)(1/\cos10.0^{\circ}-1)=[1.50(900)/(3.00\times10^8)](1/\cos10.0^{\circ}-1)=6.942\times10^{-8}\,\text{s}=69.4\,\text{ns}$ . Adding the input duration gives a shortest output duration of $69.4+5.00=74.4\,\text{ns}$ . Non-overlapping pulses must be separated by at least this time, so $f_{\max}=1/(74.4\times10^{-9})=1.34\times10^7\,\text{Hz}=13.4\,\text{MHz}$ .	5
04.1	$30.4^{\circ}$ ; $48.0^{\circ}$ ; $2.04\times10^8\,\text{m s}^{-1}$	At entry, $1.00\sin48.0^{\circ}=1.47\sin r$ , giving $r=30.3673^{\circ}=30.4^{\circ}$ . At the parallel second face, reversing Snell's law gives an emergent angle of $48.0^{\circ}$ . The speed in acrylic is $v=c/n=(3.00\times10^8)/1.47=2.04082\times10^8\,\text{m s}^{-1}=2.04\times10^8\,\text{m s}^{-1}$ .	4
05.1	$10.8\,\text{km}$ ; a fibre with a core narrow enough to support one path removes modal dispersion but not wavelength-dependent material dispersion	The pulse period is $1/(4.00\times10^6)=250\,\text{ns}$ . Since each input pulse already lasts $35.0\,\text{ns}$ , the greatest permitted material-dispersion delay is $250-35.0=215\,\text{ns}$ . Using $\Delta t=L\Delta n/c$ gives $L=(215\times10^{-9})(3.00\times10^8)/(1.478-1.472)=1.075\times10^4\,\text{m}=10.8\,\text{km}$ . Supporting one path removes different geometrical routes, but the two wavelengths still travel at different speeds in the material.	6