A ray incident at a point at an angle of incidence of 60° enters a glass sphere of R.I. n = $\sqrt{3}$ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is

90°

Refraction at P.

$\frac{\sin 60°}{\sin r_1}=\sqrt{3}$

$⇒\sin r_1 = \frac{1}{2}$

$⇒r_1 = 30°$

Since $r_2 = r_1$

$∴ r_2 = 30°$

Refraction at Q, $\frac{\sin r_2}{\sin i_2}=\frac{1}{\sqrt{3}}$

Putting $r_2= 30°$ we obtain $i_2 = 60°$

Reflection at Q

$r'_2=r_2=30°$

$∴α=180°-(r'_2+i_2)$

= 180° - (30° + 60°)

= 90°