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

Using Snell's Law at point $P$ we get:-

$\frac{sin60^o}{sin{r}_1}=\sqrt{3}$

$sin{r}_1=\frac{1}{2}⟹r_1=30^o$

As we know that $r_1=r_2⟹r_2=30^o$

Using Snell's Law at point $Q$ we get:-

$\frac{sin30^o}{sin{i}_2}=\frac{1}{\sqrt{3}}⟹i_2=60^o$

Since reflection at point $Q$ occurs

$⟹r_2^{\prime }=r_2=60^o$

Let angle between the refracted ray and reflected ray be $\alpha$

$\alpha =180^o-(r_2^{\prime }+r_2)=90^o$

Hence option $(B)$ is correct.