Physics · Wave Optics

JEE Main 2026 — 6 April, Morning Shift — Question 17

An unpolarized light of intensity $I_0$ passes through polarizer and then through a certain optically active solution 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 solution with respect to analyser is ______ degrees.

Answer: 30

Numerical answer — enter this value.

Step-by-step solution

After polarizer, intensity = $I_0/2$ . Optically active solution rotates plane by $\theta$ . Then analyser at 0°, so angle between light polarization and analyser axis is $\theta$ . Intensity out = $(I_0/2)\cos^2\theta = 3I_0/8 \Rightarrow \cos^2\theta = 3/4 \Rightarrow \cos\theta = \sqrt{3}/2 \Rightarrow \theta = 30^\circ$ .

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Physics
Chapter: Wave Optics
Topic: Polarization of Light Waves