Skip to content
dronacharya.ai
← JEE Advanced 2025 · Paper 1
PhysicsNumerical+4Question 27 of 48

JEE Advanced 2025 · Paper 1, Physics question 11

A solid glass sphere of refractive index n=3n = \sqrt3 and radius RR contains a spherical air cavity of radius R2\dfrac{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 _____.

A glass sphere (grey, n = √3) of radius R in air, with its radius drawn from the centre to the upper right. Inside it, an air cavity of radius R/2 touches the bottom of the sphere at point O, so the cavity's lowest point and the sphere's lowest point coincide at O. A source S in the upper-left part of the glass sends a ray up towards the sphere's surface, making angle θ with the dotted normal there.
Answer and worked solution

Official answer: 0.5

Same angle along the path. Any chord of a sphere meets the surface at the same angle to the radius at both ends. So the ray hits the sphere's surface at angle θ\theta, and after reflection it reaches O also at angle θ\theta to the radius.

At O. O lies on both the outer sphere and the cavity. The two spheres touch there and share the normal. So at O the light meets a glass→air interface at incidence θ\theta.

Brewster condition. For the reflected light to be fully polarized, θ\theta is the Brewster angle from glass to air:
tan⁡θ=13  ⟹  θ=30∘,sin⁡θ=0.5.\tan\theta=\frac{1}{\sqrt3}\implies\theta=30^\circ,\qquad\sin\theta=0.5.

The official key also accepts 0.75.

Topics:Polarisation and Brewster's lawTotal internal reflection
← Previous questionNext question →