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Physics · Wave Optics

JEE Advanced 2025 — Paper 1 — Question 9

A solid glass sphere of refractive index n=3n=\sqrt{3} and radius RR contains a spherical air cavity of radius R2\frac{\mathrm{R}}{2}, as shown in the figure. A very thin glass layer is present at the point O

so that the air cavity (refractive index n=1n=1 ) remains inside the glass sphere. An unpolarized, unidirectional and monochromatic light source SS emits a light ray from a point inside

the glass sphere towards the periphery of the glass sphere. If the light is reflected from the point O and is fully polarized, then the angle of incidence at the inner surface of the glass sphere is θ\theta. The value of sin⁡θ\sin \theta is \qquad

Question figure

Answer: 0.75

Numerical answer — enter this value.

Step-by-step solution

figure

tan⁡α=3\tan \alpha=\sqrt{3}

α=60∘\alpha=60^{\circ}

3sin⁡β=1×sin⁡α⇒β=30∘\sqrt{3} \sin \beta=1 \times \sin \alpha \Rightarrow \beta=30^{\circ}

R2sin⁡30∘=xsin⁡120∘\frac{\mathrm{R}}{2 \sin 30^{\circ}}=\frac{\mathrm{x}}{\sin 120^{\circ}}

Rsin⁡120∘=R32×sin⁡θ⇒sin⁡θ=32×32\frac{\mathrm{R}}{\sin 120^{\circ}}=\frac{\mathrm{R} \sqrt{3}}{2 \times \sin \theta} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}

sin⁡θ=34\sin \theta=\frac{3}{4}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2025
Paper
Paper 1
Subject
Physics
Chapter
Wave Optics
Topic
Polarization of Light Waves

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