Edexcel A-level Physics (9PH0) · Topic 5: Waves and Particle Nature of Light
Mini-Lesson
Waves & Particle Nature of Light
Topic 5 is the biggest topic on the paper. It runs from progressive and stationary waves, through superposition, interference and diffraction, into refraction and polarisation — and then detonates the whole wave model with the photoelectric effect and de Broglie.
v = fλ · λ = wx/D · nλ = d sin θ · n₁ sin θ₁ = n₂ sin θ₂ · hf = φ + KEmax · λ = h/pthe six equations this topic is built on
The punchline: light diffracts and interferes like a wave, yet ejects electrons like a particle. Electrons behave like particles, yet diffract like waves. Neither model alone is complete — that is wave–particle duality.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
Topic 5 · wave basics
Progressive waves and polarisation
v = fλ · f = 1/Tspeed = frequency × wavelength; period is the reciprocal of frequency
Transverse: oscillations are perpendicular to the direction of energy transfer (all EM waves, waves on a string, S-waves).
Longitudinal: oscillations are parallel to the energy transfer, with compressions and rarefactions (sound, P-waves).
Phase difference in radians: 2π × (path difference ÷ λ). In phase = 0 or 2π; antiphase = π.
Polarisation restricts the oscillations to a single plane. Only transverse waves can be polarised — that is the decisive experimental evidence that light is transverse. Sound, being longitudinal, cannot be polarised at all.
Applications: polarising sunglasses cut horizontally polarised glare reflected from water. Rotate a polarising filter in front of an LCD screen and the picture goes black — the screen is already emitting polarised light.
Calculate
Your turn — the wave equation
1A sound wave of frequency 850 Hz travels at 340 m s⁻¹. Calculate its wavelength in m.
m
Hint: v = fλ, so λ = v ÷ f = 340 ÷ 850.
Topic 5 · superposition
Superposition, coherence and stationary waves
Principle of superposition: when two waves meet, the resultant displacement is the vector sum of the individual displacements.
Constructive: path difference = nλ (in phase) → large amplitude.
Coherent sources have the same frequency and a constant phase difference. Without coherence the interference pattern washes out.
A stationary wave forms when two identical progressive waves travel in opposite directions and superpose — usually an incident wave and its reflection.
fundamental: λ = 2Lon a string fixed at both ends: nodes at the ends, one antinode in the middle
Nodes: permanently zero amplitude. Antinodes: maximum amplitude. Adjacent nodes are λ/2 apart.
A stationary wave transfers no net energy along its length; a progressive wave does.
All points between two adjacent nodes are in phase with each other; points either side of a node are in antiphase.
Quick check
Only one kind can be polarised
?Sound waves cannot be polarised, but light waves can. What does this tell you?
Quick check
Stationary waves
?Which statement about a stationary wave on a string is correct?
Topic 5 · double slit
Young's double-slit experiment
λ = w x / Dw = slit separation · x = fringe spacing · D = slit-to-screen distance
Two coherent sources produce a pattern of equally spaced bright and dark fringes. Young's result was the decisive nineteenth-century evidence that light is a wave.
Precision tip: measure across ten fringe spacings and divide by ten. That divides your percentage uncertainty by ten too.
Calculate
Your turn — Young's slits
2In a double-slit experiment the slits are 0.25 mm apart and the screen is 2.4 m away. The fringes are 6.0 mm apart. Calculate the wavelength of the light. Give your answer in nm.
nm
Hint: λ = wx/D = (0.25 × 10⁻³ × 6.0 × 10⁻³) ÷ 2.4 = 6.25 × 10⁻⁷ m. Multiply by 10⁹ for nm.
Topic 5 · gratings
The diffraction grating
n λ = d sin θd = grating spacing = 1 ÷ (lines per metre) · n = order number (0, 1, 2 …)
A grating has thousands of slits, so the maxima are far sharper and brighter than double-slit fringes — which makes it the instrument of choice for measuring wavelength precisely (spectroscopy).
A grating quoted as 600 lines per mm has d = 1 ÷ (600 × 10³ m⁻¹) = 1.67 × 10⁻⁶ m.
The maximum order is set by sin θ ≤ 1, so nmax = d/λ (rounded down).
White light gives a white zero order and a spectrum in each higher order, with red deviated most (longest λ, largest θ) — the opposite way round from a prism.
3A grating with 600 lines per mm is illuminated with light of wavelength 500 nm. Calculate the angle of the second-order maximum. Give your answer in degrees to 3 significant figures.
°
Hint: d = 1 ÷ (600 × 10³) = 1.667 × 10⁻⁶ m. sin θ = nλ/d = (2 × 500 × 10⁻⁹) ÷ 1.667 × 10⁻⁶ = 0.600. Now take the inverse sine.
Topic 5 · refraction & TIR
Refraction, critical angle and TIR
n = c / v · n₁ sin θ₁ = n₂ sin θ₂ · sin C = n₂ / n₁angles always measured from the NORMAL, never from the surface
Light slows on entering a denser medium; frequency stays the same but wavelength shortens, so it bends towards the normal.
Total internal reflection needs two conditions: the light must be going from a more dense to a less dense medium, and the angle of incidence must exceed the critical angle C.
Going from a medium of index n into air: sin C = 1/n.
Optical fibres: a high-index core is surrounded by a lower-index cladding so that TIR keeps the light in. The cladding also prevents light leaking between touching fibres (crosstalk) and protects the core surface.
Calculate
Your turn — Snell's law
4Light travels from air (n = 1.00) into glass of refractive index 1.50, striking the surface at 40° to the normal. Calculate the angle of refraction in the glass. Give your answer in degrees to 3 significant figures.
°
Hint: 1.00 × sin 40° = 1.50 × sin θ₂. sin 40° = 0.6428, so sin θ₂ = 0.6428 ÷ 1.50 = 0.4285.
Calculate
Your turn — critical angle
5Calculate the critical angle for the boundary between that same glass (n = 1.50) and air. Give your answer in degrees to 3 significant figures.
°
Hint: sin C = 1 ÷ n = 1 ÷ 1.50 = 0.6667. Take the inverse sine.
Topic 5 · photoelectric effect
The photoelectric effect
hf = φ + KEmaxphoton energy = work function + maximum kinetic energy of the emitted electron
Shine light on a clean metal surface and electrons may be emitted. Three observations the wave model cannot explain:
Below a threshold frequency f₀, no electrons are emitted, however intense the light or however long you wait.
Emission is instantaneous above f₀, even at very low intensity.
KEmax depends only on frequency, never on intensity. Greater intensity gives more electrons, not faster ones.
Einstein's fix: light arrives as photons of energy E = hf. One photon transfers all its energy to one electron. If hf < φ the electron cannot escape — and stacking up more photons does not help, because they are absorbed one at a time.
Worked example
Metal with work function φ = 2.30 eV, illuminated with light of λ = 400 nm.
6A metal has work function 2.30 eV. It is illuminated with light of wavelength 400 nm. Calculate the maximum kinetic energy of the emitted electrons. Give your answer in eV to 2 significant figures. (h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, 1 eV = 1.60 × 10⁻¹⁹ J)
eV
Hint: E = hc/λ = 4.97 × 10⁻¹⁹ J = 3.11 eV. Then KE_max = 3.11 − 2.30.
Quick check
Turn up the brightness
?Light above the threshold frequency causes photoemission from a metal. The intensity is doubled at the same frequency. What happens?
Topic 5 · duality
Energy levels, spectra and de Broglie
Electrons in an atom occupy discrete energy levels (measured in eV, and always negative because the electron is bound). A photon is emitted when an electron drops between levels:
hf = E₁ − E₂this is why emission spectra are LINE spectra — only certain photon energies are possible
Emission spectrum: bright lines on black — from electrons falling down.
Absorption spectrum: dark lines on a continuous background — photons of exactly the right energy are absorbed and re-emitted in all directions.
De Broglie then turned duality around: if a wave can behave as a particle, a particle can behave as a wave.
λ = h / p = h / mvthe de Broglie wavelength — significant only when λ is comparable to the gap it passes through
Evidence: fire electrons at a thin graphite film and you get diffraction rings — the atomic spacing acts as the grating. Speed the electrons up and p rises, so λ falls, and the rings shrink. That is exactly what the wave model predicts, and no particle model can explain.
Calculate
Your turn — de Broglie wavelength
7An electron travels at 3.0 × 10⁶ m s⁻¹. Calculate its de Broglie wavelength. Give your answer in nm to 2 significant figures. (me = 9.11 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s)
nm
Hint: p = mv = 9.11 × 10⁻³¹ × 3.0 × 10⁶ = 2.73 × 10⁻²⁴ kg m s⁻¹. λ = h/p = 6.63 × 10⁻³⁴ ÷ 2.73 × 10⁻²⁴ = 2.4 × 10⁻¹⁰ m. Convert to nm.
Sort it
Which model does the evidence support?
Tap a piece of evidence, then tap the model it supports.
🌊 Wave nature of light
⚛️ Particle nature of light
🔬 Wave nature of matter
Match it
Match the term to its meaning
Tap an item on the left, then its partner on the right.
Term
Meaning
Quick check
Constructive interference
?Two coherent sources emit waves of wavelength λ. At a point P the path difference is 1.5λ. What is observed at P?
Recap
The big ideas to know
Waves: v = fλ; only transverse waves polarise — so light is transverse
Superposition: constructive at nλ, destructive at (n + ½)λ; coherence = same f, constant phase difference
Stationary waves: two waves in opposite directions; nodes λ/2 apart; no net energy transfer
Young's slits: λ = wx/D · Grating: nλ = d sin θ, sharper maxima, red deviated most
Refraction: n = c/v · n₁ sin θ₁ = n₂ sin θ₂ · TIR above the critical angle, sin C = 1/n into air
Photoelectric effect: hf = φ + KE_max; threshold frequency; one photon → one electron; intensity changes number, not KE
Duality: hf = E₁ − E₂ gives line spectra · λ = h/p gives electron diffraction
That is the whole of Edexcel Topic 5 — the largest topic on the specification, done. Press Finish to see your score.
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