C.3 · IB Physics
Polarisation and Malus's law
Unpolarised light meets a stack of polarisers. Each one passes only the component along its axis, and a filter at 45° placed between crossed polarisers lets light through. The graph shows the final intensity as you turn polariser 2.
Predict first
Start from the default settings (Polariser 2 angle 30°, no polariser 3). Move Polariser 2 angle to 60°. How does the intensity after P2 change?
After P1 the intensity is 50 % of I₀. Then I = 50 cos²θ: cos²30° = 0.75 gives 37.5 %, cos²60° = 0.25 gives 12.5 %, which is one third as much.
Check it in the model: Move the Polariser 2 angle slider from 30° to 60° and compare the After P2 readout (37.5 % then 12.5 % of I₀).
Gold lines show each polariser's axis. Arrows show the direction of the electric field; the beam is brighter where more intensity is transmitted.
Malus's law: I = I₀ cos²θ
I₂ = I₁ cos²(θ₂ − 0°)
I₃ = I₂ cos²(θ₃ − θ₂)
Try this
- Press "Crossed: dark", then "Malus at 60°". Check that the After P2 reading matches ½ cos²θ for each angle.
- Press "45° smuggler". Why does inserting polariser 3 at 90° reduce the light reaching the screen compared with having only polariser 2, yet light still gets through?
- With P2 at 45° and polariser 3 inserted, sweep P3 from 45° to 135°. At which angles is the final intensity largest and smallest?
Exam tip: unpolarised light loses half its intensity at the first polariser whatever its axis; only after that apply I = I₀cos²θ, with θ measured between successive axes rather than from the first polariser.