An unpolarized light of intensity I₀ passes through polarizer and then through a certain optically active and finally it goes to analyser. If the angle between analyser and polariser is 0° and intensity of light emerged from analyser is 3/8I₀, the angle of rotation of the light by the with respect to analyser is ______ degrees. Numerical answer type. Work it out, then open the answer.

Answer: 30. After the polariser: I₁ = I₀/2. The optically active medium rotates the plane by θ; the analyser is parallel to the polariser, so I = I₀/2cos ²θ (Malus's law). I₀/2cos ²θ = 3/8I₀ ⇒ cos ²θ = 3/4 ⇒ θ = 30° So the angle of rotation is 30°.

An unpolarized light of intensity I₀ passes through polarizer and… | JEE Main 2026 (6 Apr, Shift 1)

Q13JEE Main 2026 · 6 APR · SHIFT 1Medium

Polarisation

An unpolarized light of intensity $I_0$ passes through polarizer and then through a certain optically active and finally it goes to analyser. If the angle between analyser and polariser is $0^\circ$ and intensity of light emerged from analyser is $\frac{3}{8}I_0$ , the angle of rotation of the light by the with respect to analyser is ______ degrees.

Numerical answer type. Work it out, then open the answer.

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Answer: 30

After the polariser: $I_1 = I_0/2$ .
The optically active medium rotates the plane by $\theta$ ; the analyser is parallel to the polariser, so $I = \frac{I_0}{2}\cos^2\theta$ (Malus's law).
$\frac{I_0}{2}\cos^2\theta = \frac{3}{8}I_0 \Rightarrow \cos^2\theta = \frac{3}{4} \Rightarrow \theta = 30^\circ$
So the angle of rotation is $30^\circ$ .

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