JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Wave Optics

A solid glass sphere of refractive index

n = 3

and radius

R

contains a spherical air cavity of radius

\frac{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 = 1

) remains inside the glass sphere. An unpolarized, unidirectional and monochromatic light source

S

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

. The value of

sin θ

is ____.

Answer: 0.75

Solution

For the reflected light to be fully polarized, it must be reflected at Brewster's angle. At the glass-air interface:

tan α = n_{ai r} / n_{g l a ss} = \frac{1}{3}

α = 60°

Applying Snell's law at the glass-air interface:

3 sin β = 1 \times sin 60°

sin β = \frac{3 /2}{3} = \frac{1}{2}

β = 30°

Using the sine rule in the triangle formed:

\frac{R /2}{sin 30°} = \frac{x}{sin 120°}

where x is the distance from the source to point O.

For the ray from S to reach O:

\frac{R}{sin 120°} = \frac{R 3}{2 \times sin θ}

sin θ = \frac{3}{2} \times \frac{3}{2} = \frac{3}{4} = 0.75

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Numerical answer

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JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Wave Optics

Solution

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