JEE Advanced 2019 — Paper 2 — Question 13

When angle of incidence on first face of the prism is $60^{\circ}$ the angle of incidence on the other surface

of the prism will be slightly greater than critical angle.

For refraction at first surface of the prism

$\sin 60^{\circ}=\sqrt{3} \sin r_{1}$

$\Rightarrow r_{1}=30^{\circ}$

For second surface $r_{2}=75^{\circ}-30^{\circ}=45^{\circ}$

Since $r_{2} \approx \theta_{C}$

$\Rightarrow \sin 45^{\circ}=\frac{n}{\sqrt{3}}$

$\Rightarrow \mathrm{n}^{2}=1.50$