This mini-lesson covers AQA 3.3 — Waves: progressive and stationary waves, superposition and interference, Young’s double slit, diffraction gratings (nλ = d sinθ), refraction, total internal reflection in optical fibres, and polarisation.
Section 3.3 is compulsory and appears on Paper 1.
Where this sits: AQA sections 3.1–3.8 are compulsory for every A-level Physics student. (Sections 3.9–3.13 are the five options — you study just one of those.)
Work through each screen, answer the questions as you go (many are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
3.3.1 · progressive waves
Progressive waves and the wave equation
A progressive wave transfers energy from one place to another without transferring matter.
c = fλ · T = 1 ÷ famplitude A · wavelength λ · frequency f · period T · phase difference in radians
Transverse — oscillations are perpendicular to the direction of energy transfer (all EM waves, waves on a string).
Longitudinal — oscillations are parallel to the direction of travel; they have compressions and rarefactions (sound).
Phase difference: two points one whole wavelength apart are in phase (2π rad); half a wavelength apart they are in antiphase (π rad).
Key distinction: only transverse waves can be polarised, because only they have oscillations in more than one plane perpendicular to travel. Sound (longitudinal) can never be polarised — a favourite exam question.
3.3.1 · stationary waves
Stationary (standing) waves
A stationary wave forms when two waves of the same frequency and amplitude travelling in opposite directions superpose — usually a wave and its own reflection.
Nodes — points of permanently zero amplitude (destructive superposition every instant).
Antinodes — points of maximum amplitude, half a wavelength apart from the next antinode.
Node-to-node distance = λ/2. No net energy is transferred along a stationary wave.
Between adjacent nodes all points oscillate in phase; across a node they are in antiphase.
f = (1 ÷ 2L) × √(T ÷ µ)first harmonic of a stretched string: L = length, T = tension, µ = mass per unit length
Worked example
L = 0.60 m, T = 40 N, µ = 1.0×10−3 kg m−1.
√(40 ÷ 1.0×10−3) = √40000 = 200 m s−1; f = 200 ÷ (2 × 0.60) = 167 Hz.
Sort it
Progressive, stationary, or true of both?
Tap a statement, then tap the box it belongs in.
➡️ Progressive only
🎻 Stationary only
🔁 True of both
Quick check
Think it through
?A stationary wave on a string has adjacent nodes 0.25 m apart. What is the wavelength of the waves forming it?
3.3.2 · superposition
Superposition, coherence and path difference
Principle of superposition: where waves meet, the resultant displacement is the vector sum of the individual displacements.
For a stable interference pattern the sources must be coherent:
Same frequency (and, for light, essentially the same wavelength).
A constant phase difference between them.
Similar amplitude, so cancellation is complete at the minima.
Why a laser or a single slit first? Two separate lamps are incoherent — their phase difference changes randomly billions of times a second, so the pattern washes out. Young used a single slit to make two coherent sources from one wavefront.
3.3.2 · Young double slit
Young’s double-slit experiment
Monochromatic, coherent light through two narrow slits produces evenly spaced bright and dark fringes.
w = λD ÷ sw = fringe spacing · D = slit-to-screen distance · s = slit separation
Increase λ (red rather than blue) → fringes get wider apart.
Increase D → fringes get wider apart.
Increase s → fringes get closer together.
Worked example
λ = 600 nm, D = 2.0 m, s = 0.50 mm.
w = (600×10−9 × 2.0) ÷ (0.50×10−3) = 1.2×10−6 ÷ 5.0×10−4 = 2.4×10−3 m = 2.4 mm.
Safety & significance: never look along a laser beam. Historically, Young’s fringes were the decisive evidence that light is a wave — particles cannot cancel each other out.
Calculate
Your turn — calculation 1
1In a double-slit experiment, λ = 600 nm, the slit separation is 0.50 mm and the screen is 2.0 m away. Calculate the fringe spacing in mm.
mm
Hint: w = λD ÷ s = (600×10−9 × 2.0) ÷ (0.50×10−3), then convert m → mm.
Quick check
Think it through
?In a double-slit experiment, the slit separation s is doubled and nothing else is changed. What happens to the fringe spacing?
3.3.2 · diffraction gratings
Diffraction and the diffraction grating
Waves spread out when they pass through a gap. Spreading is greatest when the gap is about the same size as the wavelength. A single slit gives a wide, bright central maximum (twice the width of the others) with dimmer side maxima.
d sinθ = nλd = grating spacing = 1 ÷ (lines per metre) · n = order · θ = angle from the straight-through beam
A grating gives sharper, brighter maxima than two slits, because thousands of slits interfere.
Maximum order: sinθ ≤ 1, so n ≤ d/λ (round down to a whole number).
With white light, each order is a spectrum: red (longest λ) is deviated most; the zero order stays white.
Worked example
Grating with 300 lines mm−1 → d = 1 ÷ (300×103 m−1) = 3.33×10−6 m.
2A diffraction grating has 300 lines per mm. Light of wavelength 500 nm is shone normally at it. Calculate the angle θ of the first-order maximum, in degrees.
°
Hint: d = 1 ÷ (300 000 lines per m) = 3.33×10−6 m. sinθ = λ/d = 0.150; θ = sin−1(0.150).
Quick check
Think it through
?White light passes through a diffraction grating. In the first-order spectrum, which colour is deviated through the largest angle?
3.3.2 · refraction & TIR
Refraction, total internal reflection and optical fibres
Light changes speed when it enters a new medium, so it changes direction. Refractive index:
n = c ÷ cs · n1 sinθ1 = n2 sinθ2critical angle: sinθc = n2 ÷ n1 (going from dense n1 to less dense n2)
Total internal reflection happens only when light travels from a more to a less optically dense medium and the angle of incidence exceeds θc.
An optical fibre has a high-index core and a lower-index cladding. The cladding protects the core, keeps the fibre strong, and (crucially) prevents light crossing between touching fibres.
Modal dispersion (rays taking different path lengths) and material dispersion (different wavelengths travelling at different speeds) both blur pulses — fixed by a narrow core and monochromatic light.
Worked example — critical angle
Glass n = 1.50, air n = 1.00: sinθc = 1.00 ÷ 1.50 = 0.667 → θc = 41.8°.
Calculate
Your turn — calculation 3
3Light travels from glass of refractive index 1.50 into air (n = 1.00). Calculate the critical angle in degrees.
5A stretched string of length 0.60 m has tension 40 N and mass per unit length 1.0 × 10−3 kg m−1. Calculate the frequency of its first harmonic, in Hz.
An unpolarised transverse wave oscillates in all planes perpendicular to its direction of travel. A polarised wave oscillates in one plane only.
Pass light through a polarising filter: rotating a second filter makes the transmitted intensity rise and fall, reaching zero when the filters are crossed (90° apart).
Light reflected from a shiny surface (water, road) is partially polarised — which is why Polaroid sunglasses cut glare.
TV and radio aerials must be aligned with the plane of polarisation of the transmitted wave, or the signal is weak.
The decisive point: the fact that light can be polarised proves it is a transverse wave. Sound cannot be polarised, because its oscillations are already confined to the direction of travel.
Match it
Match each observation to the physics
Tap a card on the left, then its partner on the right.
Statement
Answer
Quick check
Think it through
?Which statement about sound waves is correct?
Recap
The big ideas to know
Wave basics: c = fλ · transverse vs longitudinal · phase difference in radians
Stationary waves: two counter-propagating waves · nodes and antinodes · node spacing = λ/2 · f = (1/2L)√(T/µ)
Superposition: coherent = same frequency + constant phase difference · constructive nλ, destructive (n+½)λ
Young: w = λD/s
Gratings: d sinθ = nλ · d = 1/(lines per metre) · red deviates most