AQA A-Level Physics: Waves (7408)

This study guide teaches section 3.3 Waves of the AQA AS and A-level Physics specification (7407/7408, version 1.4, AS and A-level exams June 2016 onwards). It covers 3.3.1 Progressive and stationary waves and 3.3.2 Refraction, diffraction and interference. Waves is AS and A-level content, not A-level only: at A-level it is assessed in Paper 1 (sections 1 to 5 and 6.1) and assumed knowledge in Paper 2. It is not one of the optional topics (sections 3.9 to 3.13) that you choose between for Paper 3 Section B.

Follow it with the revision notes and practice questions. The course hub lists every topic, and the printable checklist tracks each outcome.

What this topic covers

Specification	What you must be able to do
3.3.1.1 Progressive waves	Wave terms; c = fλ, f = 1/T; phase difference in radians, degrees or cycles
3.3.1.2 Longitudinal and transverse waves	Both types with examples; EM speed in a vacuum; polarisation, Polaroid, aerials
3.3.1.3 Superposition and stationary waves	Graphical formation; nodes, antinodes, harmonics; f = (1/2l)√(T/μ); Required practical 1
3.3.2.1 Interference	Path difference, coherence, lasers; w = λD/s; white light; sound and EM; Required practical 2
3.3.2.2 Diffraction	Single-slit patterns; derive and use d sin θ = nλ; uses of gratings
3.3.2.3 Refraction at a plane surface	n = c/c_s; Snell’s law; critical angle; step-index fibres, dispersion, absorption

Not expected: Malus’s law, the single-slit intensity graph, how a laser works, the spectrometer, and the terms “fundamental” and “overtone”.

3.3.1.1 Progressive waves

A progressive wave transfers energy without transferring matter. Particles of the medium oscillate about fixed equilibrium positions.

• Amplitude — maximum displacement from equilibrium.
• Wavelength λ — distance between two adjacent points oscillating in phase (for example crest to crest).
• Frequency f — number of complete oscillations per second (Hz). Period T — time for one oscillation, so f = 1/T.
• Wave speed c — distance the wave travels per second: c = fλ.
• Phase — the stage a point has reached in its cycle. Phase difference between two points a distance x apart along a wave is x/λ of a cycle, which is 2πx/λ radians or 360x/λ degrees.

Worked example 1

A sound wave has frequency 250 Hz and wavelength 1.36 m. Find its speed, its period, and the phase difference between two points 0.34 m apart along the wave.

c = fλ = 250 × 1.36 = 340 m s⁻¹
T = 1/f = 1/250 = 4.0 × 10⁻³ s (4.0 ms)
x/λ = 0.34/1.36 = 0.25 of a cycle
phase difference = 0.25 × 2π = π/2 rad = 90°

3.3.1.2 Longitudinal and transverse waves

In a transverse wave the particles (or fields) oscillate at right angles to the direction of energy propagation: waves on a string, and EM waves (electric and magnetic fields).

In a longitudinal wave the particles oscillate parallel to the direction of energy propagation, forming compressions and rarefactions. Example: sound.

All electromagnetic waves travel at the same speed in a vacuum, about 3.00 × 10⁸ m s⁻¹.

Polarisation

A plane polarised wave oscillates in one plane only. Only transverse waves can be polarised; a longitudinal wave can only oscillate along its direction of travel. So polarisation of light is evidence that light is transverse. Sound cannot be polarised.

• Polaroid material transmits oscillations in one plane only. Two filters with axes at 90° block the light. Polaroid sunglasses cut glare from partly polarised reflected light.
• Aerials: transmitters send polarised radio waves, so a receiving aerial must be aligned with the plane of polarisation for the strongest signal.

3.3.1.3 Superposition and stationary waves

Principle of superposition: where two waves meet, the resultant displacement at any point is the vector sum of the individual displacements.

A stationary wave forms when two progressive waves of the same frequency travel in opposite directions, usually a wave and its reflection. For the graphical explanation, draw both waves at instants a quarter of a period apart and add them: some points always cancel, others oscillate with maximum amplitude.

• Node — point of zero amplitude (permanent destructive superposition).
• Antinode — point of maximum amplitude.
• Adjacent nodes (or adjacent antinodes) are λ/2 apart; a node and the next antinode are λ/4 apart.
• Between adjacent nodes all points oscillate in phase; points in adjacent loops are in antiphase (π rad).
• A stationary wave stores energy; it does not transfer energy along the medium.

Harmonics on a string

A string fixed at both ends must have a node at each end. The first harmonic has one loop, so λ = 2l. The nth harmonic has n loops, λ = 2l/n and frequency nf₁. For a string of tension T and mass per unit length μ, the wave speed is √(T/μ), giving for the first harmonic:

f = (1/2l) √(T/μ)

Worked example 2

A string of length 0.80 m has tension 40 N and mass per unit length 2.5 × 10⁻³ kg m⁻¹. Find the first-harmonic frequency, the third-harmonic frequency, and the new first harmonic if the tension is increased to 160 N.

√(T/μ) = √(40 / 2.5 × 10⁻³) = √16 000 = 126.5 m s⁻¹
f₁ = 126.5 / (2 × 0.80) = 79.1 Hz
f₃ = 3 × 79.06 = 237 Hz
T × 4 → √(T/μ) × 2 → f₁ = 2 × 79.06 = 158 Hz

Microwaves and sound

A microwave transmitter facing a metal plate sets up a stationary wave; a probe finds nodes λ/2 apart. A loudspeaker facing a hard wall does the same with sound.

Worked example 3

A detector moves from the 1st node to the 5th node of a microwave stationary wave, a distance of 7.0 cm. Find λ and f.

1st to 5th node = 4 node spacings = 4 × λ/2 = 2λ
2λ = 7.0 cm → λ = 3.5 cm = 0.035 m
f = c/λ = 3.00 × 10⁸ / 0.035 = 8.6 × 10⁹ Hz

Required practical 1

Vary length, tension and mass per unit length one at a time. A string runs over a pulley to hanging masses (T = mg) and is driven by a vibration generator. Adjust the frequency until the first harmonic is clear. Graphs of f against 1/l, or f against √T, should be straight lines through the origin. Find μ by weighing a measured length of string.

3.3.2.1 Interference

Path difference is the difference in distance from two sources to a point. For two sources in phase:

• constructive interference (maximum) when path difference = nλ
• destructive interference (minimum) when path difference = (n + ½)λ

Coherent sources have the same frequency and a constant phase difference. A stable interference pattern needs coherent sources.

A laser gives monochromatic, coherent light. Safety: never look into the beam or its reflections, remove reflective surfaces, and keep the beam below eye level.

Young’s double-slit experiment

A single source lights two narrow slits a distance s apart, which act as coherent sources. Fringes appear on a screen a distance D away, with spacing:

w = λD/s

With white light, the central fringe is white (zero path difference for every wavelength). Other fringes are spectra with blue nearest the centre, since shorter λ gives smaller w.

Worked example 4

Laser light of wavelength 633 nm passes through slits 0.50 mm apart onto a screen 1.80 m away. Find w.

w = λD/s = (633 × 10⁻⁹ × 1.80) / (0.50 × 10⁻³) = 2.28 × 10⁻³ m = 2.3 mm

Measuring across 10 spacings (22.8 mm) and dividing by 10 cuts the percentage uncertainty. The measurements revision notes cover how to combine uncertainties.

Interference with sound and microwaves

Two loudspeakers on one signal generator are coherent: walking past them you hear loud and quiet regions. Microwaves through two gaps in a metal sheet give maxima and minima at a probe. Explain both using path difference.

How ideas about EM radiation changed

Newton’s corpuscular (particle) theory was long preferred to Huygens’ wave theory. Young’s fringes are an interference effect, supporting the wave model. The photoelectric effect (section 3.2.2) later showed a particulate nature too, giving wave–particle duality.

Required practical 2

Investigate interference effects, including Young’s slits and a diffraction grating: measure w, s and D to find λ, then use a grating to find λ from d sin θ = nλ.

3.3.2.2 Diffraction

Single slit, monochromatic light: a wide, bright central maximum with narrower, much dimmer maxima either side, separated by dark fringes. The central maximum is twice as wide as the others.

Single slit, white light: a white central maximum; the side maxima show spectra, with blue nearer the centre.

Qualitatively, the central maximum gets wider when:

• the wavelength increases (red spreads more than blue)
• the slit width decreases

The diffraction grating

A plane transmission grating has many slits a distance d apart. At normal incidence, light from adjacent slits arriving at angle θ has path difference d sin θ. A bright maximum needs this to equal a whole number of wavelengths:

d sin θ = nλ

where n is the order. Because sin θ ≤ 1, the highest order is the largest whole number not exceeding d/λ.

Worked example 5

A grating has 300 lines per mm. Light of wavelength 589 nm is incident normally. Find the first-order angle and the highest order seen.

d = 1 / (300 × 10³) = 3.33 × 10⁻⁶ m
sin θ₁ = λ/d = 589 × 10⁻⁹ / 3.33 × 10⁻⁶ = 0.177 → θ₁ = 10.2°
d/λ = 5.66 → highest order n = 5

Applications: analysing spectra from stars and discharge tubes to identify elements, and measuring wavelengths. Grating maxima are sharper and brighter than double-slit fringes.

3.3.2.3 Refraction at a plane surface

Refractive index: n = c/c_s, where c_s is the speed of light in the substance. Recall that n for air is approximately 1.

Snell’s law: n₁ sin θ₁ = n₂ sin θ₂ (angles measured from the normal).

Total internal reflection happens when light in the denser medium (n₁ > n₂) meets the boundary at more than the critical angle, where sin θc = n₂/n₁.

Worked example 6

Light passes from air into glass of n = 1.52 at an angle of incidence of 35°. Find the angle of refraction and the speed in the glass.

1.00 × sin 35° = 1.52 × sin θ₂ → sin θ₂ = 0.377 → θ₂ = 22.2°
c_s = c/n = 3.00 × 10⁸ / 1.52 = 1.97 × 10⁸ m s⁻¹

Optical fibres (step index only)

A step-index fibre has a core surrounded by cladding of lower refractive index. Light stays in the core by total internal reflection. For a core of n = 1.48 and cladding of n = 1.46, sin θc = 1.46/1.48, so θc = 80.6°.

The cladding:

• keeps n lower than the core so total internal reflection occurs
• protects the core from scratches, which would let light escape
• stops signals crossing between touching fibres

Absorption reduces signal amplitude along the fibre, so signals need amplifying at intervals.

Pulse broadening: a pulse spreads out in time. Overlapping pulses cannot be distinguished, which limits the pulse rate.

• Material dispersion: different wavelengths travel at different speeds in the core. Reduce it with monochromatic light.
• Modal dispersion: rays taking different paths travel different distances. Reduce it with a very narrow core (single-mode fibre).

Common errors

• Using the distance between adjacent nodes as λ instead of λ/2.
• Counting nodes, not spacings: 1st to 5th node is 4 spacings.
• Forgetting to convert mm and nm in w = λD/s.
• Using “lines per mm” as d without taking the reciprocal and converting to metres.
• Rounding the highest grating order up instead of down.
• Writing sin θc = n₁/n₂ (gives a value above 1).

Next steps

Use the revision notes, then the practice questions. See the exam preparation guide and the free diagnostics.

Official syllabus

AQA AS and A-level Physics specification (7407/7408), version 1.4, July 2026, for AS and A-level exams June 2016 onwards, published by AQA. Section 3.3 Waves.