Unpolarized light with intensity I₀ passes through three polaroids such that the transmission axis of the last polaroid is crossed with the first. If the intensity of emerging light is I₁(I₁ = 3/32I₀). The angle between the first two polaroids and the angle for which the transmitted intensity is maximum are respectively given, cos (90) = 0 (a) 1 ⁻¹ √(3) (b) ⁻¹ √(3) /2sin /2 and 45° sin /2 and 90° (c) 1 ⁻¹ √(3) (d) ⁻¹ √(3) /2cos /2 and 45° cos /2 and 90°

Answer: (a). With θ between the first two and the third crossed with the first: I₁ = I₀/2cos ²θsin ²θ = I₀/8sin ² 2θ I₀/8sin ² 2θ = (3I₀)/(32) ⇒ sin 2θ = √(3)/2 ⇒ θ = 1/2sin ⁻¹(√(3)/2) = 30° Intensity is maximum when sin 2θ = 1, i.e. θ = 45°. So the answer is (a) 1/2sin ⁻¹(√(3)/2) and 45°.

Unpolarized light with intensity I₀ passes through three polaroids… | MHT CET 2026 (12 May, Shift 1)

Q4MHT CET 2026 · 12 MAY · SHIFT 1Medium

Polarisation

Unpolarized light with intensity $I_0$ passes through three polaroids such that the transmission axis of the last polaroid is crossed with the first. If the intensity of emerging light is $I_1\left(I_1 = \frac{3}{32}I_0\right)$ . The angle between the first two polaroids and the angle for which the transmitted intensity is maximum are respectively $\{$ given, $\cos(90) = 0\}$

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Answer: (a)

With $\theta$ between the first two and the third crossed with the first: $I_1 = \frac{I_0}{2}\cos^2\theta\sin^2\theta = \frac{I_0}{8}\sin^2 2\theta$
$\frac{I_0}{8}\sin^2 2\theta = \frac{3I_0}{32} \Rightarrow \sin 2\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{1}{2}\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ$
Intensity is maximum when $\sin 2\theta = 1$ , i.e. $\theta = 45^\circ$ .
So the answer is (a) $\frac{1}{2}\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$ and $45^\circ$ .

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Polarisation question from MHT CET 2026 (12 May, Shift 1)

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