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Learn7408Paper 1 · Paper 1WavesElectromagnetic spectrum and polarisation
Waves · Subtopic 4 of 6

Electromagnetic Spectrum and Polarisation

All electromagnetic waves are transverse and travel at 3.00×1083.00\times10^8 m s⁻¹ in free space. Wavelengths run from kilometres (radio) to less than 10−1110^{-11} m (gamma); visible light is 400400–700700 nm. Only transverse waves can be polarised. A polariser halves unpolarised light; after that, Malus's law gives I=I0cos⁡2θI=I_0\cos^2\theta.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH12AQA 7408 Paper 1≈ 30 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Recall the order and approximate wavelengths of the regions of the EM spectrum.
  • Use c=fλc=f\lambda with c=3.00×108c=3.00\times10^8 m s⁻¹.
  • Explain polarisation and why it shows waves are transverse.
  • Use Malus's law I=I0cos⁡2θI=I_0\cos^2\theta.
Pause and think

Look at a phone screen through polarising sunglasses and slowly tilt your head. The screen goes dark at one angle. Why?

Section 2 of 14

Core idea

Start hereSunglasses that cut glare

Light reflected off a lake is mostly vibrating horizontally. Polarising sunglasses only let vertical vibrations through, so the glare vanishes. Polarisation is also the proof that light is a transverse wave.

The big idea

Polarised: oscillation in one plane. Only transverse waves polarise. Malus's law: I=I0cos⁡2θI=I_0\cos^2\theta.

1. Polarising filters

Malus's law: intensity through an analyser at angle θ
501001500.20.40.60.810angle θ (degrees)I / I₀I = I₀cos²θ
Aligned: all gets through. At 90∘90^\circ (crossed): nothing. At 45∘45^\circ: half.
Malus's law
  1. A=A0cos⁡θA=A_0\cos\theta
    Only the component along the analyser's axis passes.
  2. I∝A2I\propto A^2
    Intensity ∝ amplitude².
  3. I=I0cos⁡2θI=I_0\cos^2\theta
    Combine.
Unpolarised light through one filter: half the intensity (average of cos⁡2\cos^2 over all angles is 12\frac12).
Unpolarised light through two polarising filters
all planesone plane, I₀/2I₀cos²θ
The first filter halves unpolarised light. The second passes only the component along its own axis.
Worked exampleTwo filters at 60°

Unpolarised 40 W m−2^{-2}.

Goal
Final intensity.
Strategy
Half, then Malus.
  1. 20 W m−220\text{ W m}^{-2}
    First filter halves.
  2. 20cos⁡260∘=20×14=5 W m−220\cos^260^\circ=20\times\tfrac14=5\text{ W m}^{-2}
    Second.
Answer55 W m−2^{-2}.
✓ CheckAt 90∘90^\circ it would be 0 ✓.
Worked exampleAngle for 75% transmission

Polarised light through an analyser.

Goal
θ\theta.
Strategy
Solve cos⁡2θ=0.75\cos^2\theta=0.75.
  1. cos⁡θ=0.75=0.866\cos\theta=\sqrt{0.75}=0.866
    Square root.
  2. θ=30∘\theta=30^\circ
    Inverse cos.
Answer30∘30^\circ.
✓ Checkcos⁡230∘=34\cos^230^\circ=\frac34 ✓.
Common misconceptioncos⁡θ\cos\theta not squared Amplitude uses cos⁡θ\cos\theta; intensity uses cos⁡2θ\cos^2\theta.
Pause and think

Why can't sound be polarised?

Summary

  • Transverse only.
  • I=I0cos⁡2θI=I_0\cos^2\theta.
Key words
Plane polarised
Oscillating in one plane.
HistoryThe Iceland spar mystery. In 16691669 Rasmus Bartholin noticed that a crystal of Iceland spar produced a double image. In 18081808 Étienne-Louis Malus, looking through such a crystal at sunlight reflected off a palace window, saw one image vanish as he turned it: reflected light was polarised. He gave his name to the law.
Fun factVikings may have used sunstones. Norse sagas mention a "sunstone" for navigation. Some researchers think it was Iceland spar: by finding the direction of polarisation of skylight, sailors could locate the Sun even through cloud or after sunset.
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