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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Physics
Chapter
Ray Optics

JEE Advanced 2025 · Paper 1 · Physics · Ray Optics

A solid glass sphere of refractive index n= 3 and radius R contains...

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 ______ [given : θ > 30°]
A solid glass sphere of refractive index n= 3 and radius R contains a spherical air cavity of radius R 2 , as shown in

Correct answer

0

Step-by-step solution

We added a condition θ > 30° to make this question clear for students, otherwise 0.5 and 0.75 both are correct answers A solid glass sphere of refractive index n= 3 and radius R contains a spherical air cavity of radius R 2 , as shown in
tan⁡α=3α=60∘\begin{aligned} & \tan \alpha=\sqrt{3} \\ & \alpha=60^{\circ}\end{aligned}3sin⁡β=1×sin⁡α⇒β=30∘R2sin⁡30∘=xsin⁡120∘Rsin⁡120∘=R32×sin⁡θ⇒sin⁡θ=32×32sin⁡θ=34\begin{aligned} & \sqrt{3} \sin \beta=1 \times \sin \alpha \Rightarrow \beta=30^{\circ} \\ & \frac{\mathrm{R}}{2 \sin 30^{\circ}}=\frac{\mathrm{x}}{\sin 120^{\circ}} \\ & \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 \theta=\frac{3}{4}\end{aligned}$
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