3.3 Waves — AQA 7408 revision notes

3.3.1.1

Year 1

Progressive waves

Notes

Evidence from your answers: none yet

Your confidence:

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

Definition

A progressive wave transfers energy through a medium whose particles oscillate about their equilibrium positions without travelling with the wave. Amplitude is the maximum displacement from equilibrium; wavelength $\lambda$ is the shortest distance between two points in phase; frequency $f$ is the number of complete cycles per second and period $T=1/f$ ; wave speed $c=f\lambda$ . Phase difference between two points may be given in radians, degrees or fractions of a cycle: a separation of $x$ along the wave is a phase difference of $\dfrac{2\pi x}{\lambda}$ radians.

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

Two loudspeakers P and Q are connected to the same signal generator, which produces a frequency of

1650\,\text{Hz}

. A microphone at point X is

3.20\,\text{m}

from P and

3.35\,\text{m}

from Q. The speed of sound in air is

330\,\text{m s}^{-1}

. Calculate the wavelength of the sound, and the phase difference between the two waves arriving at X.

[4 marks]

Total for this question: 4

1.Wavelength: $\lambda=\dfrac{c}{f}=\dfrac{330}{1650}=0.200\,\text{m}$ .
2.Path difference: $3.35-3.20=0.15\,\text{m}$ .
3.As a fraction of a wavelength: $\dfrac{0.15}{0.200}=0.75$ of a cycle.
4.Phase difference $=0.75\times2\pi=1.5\pi\,\text{rad}$ (or $270^{\circ}$ , or three-quarters of a cycle).

Answer: $\lambda=0.200\,\text{m}$ ; phase difference $=1.5\pi\,\text{rad}$ ( $270^{\circ}$ )

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.

Mark-scheme wording to use

“same frequency/wavelength and constant phase difference” — AQA 7408, November 2020 Paper 1, Q03.1
the oscillations of particles in the wave are at right angles … at right angles to the direction of energy transfer — paraphrase of AQA 7408, June 2025 Paper 1, Q02.1
in phase, zero phase difference or 2pi phase difference — paraphrase of AQA 7408, June 2024 Paper 1, Q07.3
“180 degrees OR π radians … reject ‘half a cycle’” — AQA 7408, June 2017 Paper 3, Q01.1
use v=f lambda with path difference as one wavelength — paraphrase of AQA 7408, November 2020 Paper 1, Q03.4

What the examiner wants

The November 2020 Paper 1 report highlighted candidates confusing a phase difference of $\pi$ with $2\pi$ : points a whole wavelength apart are in phase ( $0$ or $2\pi$ ), half a wavelength apart are in antiphase ( $\pi$ radians, $180^{\circ}$ ) — and 'half a cycle' is rejected as a phase difference. Define a transverse wave by the DIRECTION of the oscillations relative to the direction of energy transfer, and 'coherent' means same frequency AND constant phase difference.

Exam tip

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

Tier 1 · Easy

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.3.1.2

Year 1

Longitudinal and transverse waves

Notes

Evidence from your answers: none yet

Your confidence:

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

Explanation

In a transverse wave, 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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.3.1.3

Year 1

Principle of superposition of waves and formation of stationary waves

Notes

Evidence from your answers: none yet

Your confidence:

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

Definition

A stationary wave is formed when two progressive waves of the same frequency (and similar amplitude) travelling in opposite directions superpose — on a string, the wave from the vibration generator and its reflection from the fixed end. Nodes are points of zero displacement at all times, antinodes points of maximum amplitude; adjacent nodes are half a wavelength apart, so a string of length $l$ vibrating in its first harmonic has $\lambda=2l$ and $f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$ , where $T$ is the tension and $\mu$ the mass per unit length. Higher harmonics have $n$ loops and frequency $nf_1$ .

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 wire is stretched between two bridges

0.600\,\text{m}

apart. The wire has a mass per unit length of

6.60\times10^{-4}\,\text{kg m}^{-1}

and the tension in it is

24.0\,\text{N}

. Calculate the frequency of the first harmonic of the wire, and the frequency of its third harmonic.

[4 marks]

Total for this question: 4

1.First harmonic: $f_1=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}=\dfrac{1}{2\times0.600}\sqrt{\dfrac{24.0}{6.60\times10^{-4}}}$ .
2. $\sqrt{3.64\times10^{4}}=190.7\,\text{m s}^{-1}$ (the wave speed), so $f_1=\dfrac{190.7}{1.20}=159\,\text{Hz}$ .
3.The third harmonic has three loops, so $\lambda=\dfrac{2l}{3}$ and $f_3=3f_1$ .
4. $f_3=3\times159=477\,\text{Hz}$ (accept $480\,\text{Hz}$ to 2 significant figures).

Answer: $f_1=159\,\text{Hz}$ ; $f_3=477\,\text{Hz}$

Required practical 1

Investigation into the variation of the frequency of stationary waves on a string with length, tension and mass per unit length of the string

Apparatus

Vibration generator driven by a signal generator with a frequency readout
String (or wire) passing over a pulley to a mass hanger and slotted masses; a movable bridge
Metre rule, balance and a long length of the same string to find $\mu$

Method

Find $\mu$ by weighing a measured length of the string; set a known tension with the hanging masses ( $T=mg$ ).
Increase the generator frequency slowly until the string vibrates in one loop with a stable, maximum-amplitude pattern; record $f_1$ and the length $l$ between the generator and the bridge.
Vary the length by moving the bridge (constant $T$ and $\mu$ ) and record $f_1$ each time; plot $f_1$ against $1/l$ — a straight line through the origin.
Restore the length and vary the tension by changing the hanging mass; plot $f_1$ against $\sqrt{T}$ .
Repeat with strings of different mass per unit length; plot $f_1$ against $1/\sqrt{\mu}$ and compare each gradient with the prediction from $f=\dfrac{1}{2l}\sqrt{T/\mu}$ .

Variables

Independent:: length $l$ , then tension $T$ , then mass per unit length $\mu$ — one at a time
Dependent:: frequency of the first harmonic
Control:: the two variables not being tested; amplitude setting of the generator; judging the same maximum-amplitude pattern each time

Hazards and precautions

Hanging masses can fall on feet and the string can snap under tension: keep the total mass moderate, stand clear of the load and wear eye protection.
The vibration generator and signal generator get warm and the mains leads are near the pulley: keep leads clear of moving parts and switch off between readings.

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.

Mark-scheme wording to use

waves from the vibration generator and the reflected wave travel in opposite directions along the wire and superpose — paraphrase of AQA 7408, June 2025 Paper 1, Q02.2
Do not accept 'interfere' in place of 'superpose' — paraphrase of AQA 7408, June 2025 Paper 1, Q02.2
mass per unit length of string … length of string between the vibration generator and pulley — paraphrase of AQA 7408, June 2025 Paper 3, Q01.1
wavelength 0.70 m from node spacing — paraphrase of AQA 7408, June 2024 Paper 1, Q07.2
“appreciation of reducing diameter when string is stretched … lower mass per unit length so (constant of proportionality and hence) frequency is higher (than would be predicted)” — AQA 7408, June 2017 Paper 1, Q02.3

What the examiner wants

Say 'superpose', not 'interfere', and name both waves — the incident wave from the generator and the wave reflected from the fixed end. The November 2021 Paper 1 report found candidates treating a stationary-wave question as a progressive wave: on a stationary wave all particles between two adjacent nodes are in phase, and those in the next loop are in antiphase. Get the wavelength from the node spacing (node to node is $\lambda/2$ ) before using $c=f\lambda$ , and evaluate the square root before dividing by $2l$ .

Exam tip

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

Tier 1 · Easy

ORIGINAL

State the difference between a node and an antinode in a stationary wave.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.3.2.1

Year 1

Interference

Notes

Evidence from your answers: none yet

Your confidence:

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

Explanation

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

ORIGINAL

State the two conditions that two sources must satisfy to be coherent.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.3.2.2

Year 1

Diffraction

Notes

Evidence from your answers: none yet

Your confidence:

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

Explanation

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.3.2.3

Year 1

Refraction at a plane surface

Notes

Evidence from your answers: none yet

Your confidence:

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

Definition

The refractive index of a substance is $n=\dfrac{c}{c_s}$ , the ratio of the speed of light in a vacuum to its speed in the substance; for air $n\approx1$ . At a plane boundary Snell's law gives $n_1\sin\theta_1=n_2\sin\theta_2$ with angles measured from the normal. Light travelling from a higher to a lower refractive index is totally internally reflected when the angle of incidence exceeds the critical angle, $\sin\theta_c=\dfrac{n_2}{n_1}$ . An optical fibre's core is surrounded by cladding of lower refractive index so that total internal reflection occurs at the boundary.

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

A step-index optical fibre has a core of refractive index

1.52

and cladding of refractive index

1.44

. A ray of light in air enters the flat end of the fibre at an angle

A

to the axis of the fibre. Calculate the critical angle at the core–cladding boundary, and hence the largest value of

A

for which the ray is totally internally reflected inside the core.

[4 marks]

Total for this question: 4

1.Critical angle: $\sin\theta_c=\dfrac{n_2}{n_1}=\dfrac{1.44}{1.52}=0.947$ , so $\theta_c=71.3^{\circ}$ (keep the unrounded $71.33^{\circ}$ in the calculator).
2.For the ray to just reach the boundary at $\theta_c$ , the angle of refraction at the flat end face is $90^{\circ}-\theta_c=18.67^{\circ}$ to the axis.
3.Snell's law at the end face (air to core): $1.00\times\sin A=1.52\times\sin18.67^{\circ}=0.4866$ .
4. $A=\sin^{-1}(0.4866)=29.1^{\circ}$ — any larger entry angle strikes the cladding below the critical angle and some light escapes. (Rounding $\theta_c$ to $71.3^{\circ}$ first gives $29.2^{\circ}$ ; AQA credits either, but carry the unrounded value.)

Answer: $\theta_c=71.3^{\circ}$ ; largest entry angle $A=29.1^{\circ}$ ( $29.1$ – $29.2^{\circ}$ depending on rounding)

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.

Mark-scheme wording to use

critical angle 41.1 degrees … 45>critical so TIR — paraphrase of AQA 7408, June 2019 Paper 1, Q02.3
incidence angle decreases … it falls below critical angle so light escapes — paraphrase of AQA 7408, November 2021 Paper 1, Q03.3
Cladding has lower refractive index so total internal reflection occurs and protects the boundary from scratching — paraphrase of AQA 7408, Specimen Paper 1, Q05.1
Blue travels slower than red because it has greater refractive index. Red reaches the end before blue, causing material pulse broadening — paraphrase of AQA 7408, Specimen Paper 1, Q05.2
same material/index/optical density … no gap/clean contact/index-matched bond — paraphrase of AQA 7408, June 2019 Paper 1, Q02.2

What the examiner wants

A total-internal-reflection answer is two linked steps: calculate the critical angle, then compare the actual angle of incidence with it — the comparison mark is contingent on the calculation. Measure every angle from the normal, and when a fibre is bent say the angle of incidence at the boundary decreases below the critical angle so SOME light escapes (AQA rejects 'all the light escapes'). Distinguish material dispersion (speed depends on wavelength) from modal dispersion (different path lengths).

Exam tip

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

Tier 1 · Easy

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

Progressive waves

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Longitudinal and transverse waves

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Principle of superposition of waves and formation of stationary waves

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Interference

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Diffraction

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Refraction at a plane surface

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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