A solid glass sphere of refractive index and radius contains a spherical air cavity of radius , as shown in the figure. A very thin glass layer is present at the point O so that the air cavity (refractive index ) remains inside the glass sphere. [2025

Question

A solid glass sphere of refractive index $n = \sqrt{3}$ 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 _______. [2025]

Smart Achievers · Accepted Answer

(0.75)

Using Brewster's law, $\tan θ = μ$

$\tan α = \sqrt{3}$ $∴$ $α = 60 °$

From Snell's law,

$\sqrt{3} \sin β = 1 \times \sin α$ $∴$ $β = 30 °$

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

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

$∴$ $\sin θ = \frac{3}{4} = 0.75$

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (0.75) Using Brewster's law, tanθ=μ tanα=3 ∴ α=60° From Snell's law, 3sinβ=1×sinα ∴ β=30° R2sin30°=xsin120° Rsin120°=R32sinθ⇒sinθ=32×32 ∴ sinθ=34=0.75

Want Us to Email you About Special Offers & Updates?