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MHT-CET Physics · Wave Optics

Polarisation: Malus' Law and Brewster's Law

A polaroid passes only the component of light along its axis: unpolarised light is halved, and polarised light is cut to I cos²θ; light reflected at Brewster's angle, tan θ = μ, is fully polarised.

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Why this matters

10 PYQs, one HARD. Two shapes: intensity through a chain of polaroids — halve once, then multiply a cos² for every angle between neighbours — and Brewster's angle, tied to the refractive index and to the speed of light in the medium.

Concept 1 of 2: Malus' Law and Chains of Polaroids

The first polaroid throws away half of unpolarised light whatever its angle. After that the light is polarised, and each polaroid passes the fraction cos²θ, θ being the angle between its axis and the previous one's. Crossed polaroids pass nothing, but slip a third one between them and light gets through.

Definition

  • Unpolarised through one polaroid: I02\dfrac{I_0}{2}. Polarised through a polaroid at θ\theta: Icos⁡2θI\cos^2\theta.
  • Chains: use the angle between NEIGHBOURS. Axes at 0°, 60°, 90°: I02⋅14⋅34=3I032\dfrac{I_0}{2}\cdot\dfrac{1}{4}\cdot\dfrac{3}{4} = \dfrac{3I_0}{32}.
  • Four polaroids each 30° from the last: I02(34)3=27I0128\dfrac{I_0}{2}\left(\dfrac{3}{4}\right)^3 = \dfrac{27I_0}{128}.
  • Crossed pair with a third at θ\theta between: I08sin⁡22θ\dfrac{I_0}{8}\sin^2 2\theta.

Malus' law

I=Iincos⁡2θI = I_{\text{in}}\cos^2\theta

Worked example

Unpolarised light of 40 W/m² passes two polaroids whose axes are 60° apart. Intensity out?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q47Moderate

Example 1 · Wave Optics · Polarisation — Malus and Brewster

Two polaroid are oriented with their planes perpendicular to incident light and transmission axis making an angle 30∘30^{\circ} with each other. What fraction of incident unpolarised light is transmitted?
(cos⁡30∘=3/2)\left( \cos30^{\circ}=\sqrt{3}/2 \right)

Measuring every angle from the first polaroid

Malus' law uses the angle between a polaroid and the ONE BEFORE it. At 60° then 90° from the first, the second step is only 30°: the factor is cos⁡230∘\cos^2 30^\circ, not cos⁡290∘=0\cos^2 90^\circ = 0.
Drill 6 more on malus' law and chains of polaroids

Concept 2 of 2: Brewster's Angle

At one special angle of incidence the reflected and refracted rays are at right angles, and the reflected light has no component in the plane of incidence — it is completely polarised. That angle obeys tan θ = μ, and since μ = c/v it also ties the angle to the speed of light in the medium.

Definition

  • tan⁡ip=μ=cv\tan i_p = \mu = \dfrac{c}{v}; reflected light fully polarised; ip+r=90∘i_p + r = 90^\circ.
  • So vsin⁡ip=ccos⁡ipv\sin i_p = c\cos i_p, and ip=cot⁡−1vci_p = \cot^{-1}\dfrac{v}{c}.
  • Reflected ray polarised at 60∘60^\circ ⇒ μ=3\mu = \sqrt{3}; at another incidence use Snell's law with that μ\mu.

Brewster's law

tan⁡ip=μ=cv\tan i_p = \mu = \frac{c}{v}

Worked example

Glass has μ=1.732\mu = 1.732. Brewster's angle, and the angle of refraction at it?
Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 1 · Q45Moderate

Example 2 · Wave Optics · Polarisation — Malus and Brewster

A beam of light is incident on a glass plate at an angle of 60∘60^{\circ}. The reflected ray is polarized. If angle of incidence is 45∘45^{\circ} then angle of refraction is

Using sin instead of tan

Brewster's law is tan⁡ip=μ\tan i_p = \mu; sin⁡ic=1μ\sin i_c = \frac{1}{\mu} is the critical angle. The two get swapped in the options.
Drill 2 more on brewster's angle

Summary — formulas & gotchas at a glance

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