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LearnYPH11WPH12 · Unit 2: Waves and ElectricityWaves and the particle nature of lightThe diffraction grating
Waves and the particle nature of light · Subtopic 6 of 9

The Diffraction Grating

A diffraction grating has thousands of equally spaced slits. Maxima appear only at angles where light from every slit arrives in phase: dsin⁡θ=nλd\sin\theta=n\lambda, with dd the slit spacing. The maxima are much sharper and brighter than with two slits, so gratings measure wavelengths precisely and split light into spectra.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH12AQA 7408 Paper 1≈ 35 minRevision 6 · updated 2026-11-01
Section 1 of 14

Learning objectives

  • Find the slit spacing from the number of lines per mm.
  • Use dsin⁡θ=nλd\sin\theta=n\lambda to find angles and wavelengths.
  • Find the highest order visible.
  • Explain why gratings give sharp maxima and how they split white light.
Pause and think

Tilt a CD or DVD under a lamp and you see rainbows. There's no prism. Where do the colours come from?

Section 2 of 14

Core idea

Start hereRainbows from a CD

Tilt a CD in sunlight and rainbows appear. Its thousands of evenly spaced tracks act like thousands of slits. With so many sources, maxima become razor-sharp, perfect for measuring wavelengths precisely.

The big idea

dsin⁡θ=nλd\sin\theta=n\lambda; highest order n≤dλn\le\frac d\lambda.

1. The equation

Orders of diffraction from a grating
θ₁gratingn = 0n = 1n = 1n = 2n = 2
With 500 lines per mm, d=2×10−6d=2\times10^{-6} m. For λ=600\lambda=600 nm: sin⁡θ1=0.3\sin\theta_1=0.3 (17.5∘17.5^\circ) and sin⁡θ2=0.6\sin\theta_2=0.6 (36.9∘36.9^\circ), as drawn.
dsin⁡θ=nλd\sin\theta=n\lambda
  1. Δ=dsin⁡θ\Delta=d\sin\theta
    Path difference between adjacent slits.
  2. dsin⁡θ=nλd\sin\theta=n\lambda
    All slits reinforce when this is a whole number of wavelengths.
  3. sin⁡θ≤1⇒n≤dλ\sin\theta\le1\Rightarrow n\le\frac d\lambda
    Maximum order.
Worked example500 lines per mm, 600 nm

Angles and number of maxima.

Goal
θn\theta_n; total maxima.
Strategy
dd first.
  1. d=10−3500=2.0×10−6 md=\frac{10^{-3}}{500}=2.0\times10^{-6}\text{ m}
    Spacing.
  2. sin⁡θn=0.3n: 17.5∘, 36.9∘, 64.2∘\sin\theta_n=0.3n:\ 17.5^\circ,\ 36.9^\circ,\ 64.2^\circ
    n=1,2,3n=1,2,3.
  3. dλ=3.33⇒nmax⁡=3\frac d\lambda=3.33\Rightarrow n_{\max}=3
    Round down.
  4. 2(3)+1=72(3)+1=7
    Both sides plus the centre.
Answer7 maxima.
✓ Checkn=4n=4 would need sin⁡θ=1.2\sin\theta=1.2: impossible ✓.
sin θ against order n: gradient λ/d
0.511.522.533.540.20.40.60.810order nsin θgradient = λ/d
Plotting all measured orders and taking the gradient gives a more reliable wavelength than a single order.
Common misconceptionRounding n up Always round d/λd/\lambda down.
Pause and think

Why is a grating better than double slits for λ?

Summary

  • dsin⁡θ=nλd\sin\theta=n\lambda.
  • nmax⁡=⌊d/λ⌋n_{\max}=\lfloor d/\lambda\rfloor.
Key words
Order
The integer nn of a maximum.
HistoryFraunhofer's dark lines. In 18211821 Joseph von Fraunhofer made the first fine diffraction gratings from wires and measured the wavelengths of dark lines in the Sun's spectrum. Those lines later revealed the elements in the Sun, including one unknown on Earth: helium, named after the Sun.
Fun factGratings find exoplanets. Astronomers use high-precision grating spectrographs to detect the tiny Doppler shifts in a star's spectrum as an orbiting planet tugs it back and forth, sometimes at less than 11 m s⁻¹, slower than walking pace.
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