A beam of light of intensity I₀ falls on a system of three polaroids which are arranged in succession such that the pass (transmission) axis is turned through 60° with respect to preceding one. The fraction of the incident light intensity that passes through the system is (cos 60° = 1/2) (a) 1/8 (b) 1/32 (c) 1/16 (d) 1/2

Answer: (b). Taking the incident beam as unpolarised, the first polaroid passes I₀/2. Each next polaroid is turned 60°: I = I₀/2cos ² 60°cos ² 60° = I₀/2·1/4·1/4 I/I₀ = 1/32 So the answer is (b) 1/32.

A beam of light of intensity I₀ falls on a system of three polaroids… | MHT CET 2025 (5 May, Shift 2)

Q17MHT CET 2025 · 5 MAY · SHIFT 2Medium

Polarisation

A beam of light of intensity $I_0$ falls on a system of three polaroids which are arranged in succession such that the pass (transmission) axis is turned through $60^\circ$ with respect to preceding one. The fraction of the incident light intensity that passes through the system is $\left(\cos 60^\circ = \frac{1}{2}\right)$

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Answer: (b)

Taking the incident beam as unpolarised, the first polaroid passes $I_0/2$ .
Each next polaroid is turned $60^\circ$ : $I = \frac{I_0}{2}\cos^2 60^\circ\cos^2 60^\circ = \frac{I_0}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}$
$\frac{I}{I_0} = \frac{1}{32}$
So the answer is (b) $\frac{1}{32}$ .

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Polarisation question from MHT CET 2025 (5 May, Shift 2)

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