JEE Advanced 2025 Paper 1, Physics Section 3 Q4: Critical Angle and Total Internal Reflection

The light is fully polarized upon reflection at O, meaning the angle of incidence at the cavity wall equals Brewster's angle for the glass-to-air interface:

$\tan \alpha =\frac{n_{\text{rarer}}}{n_{\text{denser}}}$. Taking the appropriate orientation, $\tan \alpha =\sqrt{3}$, so $\alpha =60^{\circ }$.

Apply Snell's law at the inner glass surface where the source-emitted ray refracts into the cavity geometry. From the geometry (the cavity has radius $R/2$ and sits tangent to the outer sphere of radius $R$ at O), the triangle formed by the centers of the two spheres and the point of incidence yields, via the sine rule:

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

and combining with $\sqrt{3}\sin \beta =\sin \alpha$ giving $\beta =30^{\circ }$, the geometry reduces to:

$\frac{R}{\sin 120^{\circ }}=\frac{R\sqrt{3}}{2\sin \theta }\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}=\frac{3}{4}$

$\therefore \sin \theta =0.75$.