A ray of light falls on a glass sphere of refractive index $\sqrt{3}$ such that the directions of the incident ray and emergent ray when produced meet the surface at the same point on the surface. Find the value of angle of incidence, in degree.

The ray diagram for the situation is shown in figure.

Since, $\mathrm{\angle }\mathrm{PCQ}=\mathrm{\pi }-2\mathrm{r},\mathrm{\angle }\mathrm{PRQ}=\mathrm{\pi }-2(\mathrm{i}-\mathrm{r})$

From the property of a circle, we get,

 $2\left(\mathrm{\angle }\mathrm{PRQ}\right)+\mathrm{\angle }\mathrm{PCQ}=2\mathrm{\pi }$

$\Rightarrow 2\mathrm{\pi }-4(\mathrm{i}-\mathrm{r})+\mathrm{\pi }-2\mathrm{r}=2\mathrm{\pi }$

$\Rightarrow \mathrm{r}=2\mathrm{i}-\frac{\mathrm{\pi }}{2}$

According to Snell’s Law, we have

$\mathrm{\mu }=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\frac{\sin \mathrm{i}}{\sin \left(2\mathrm{i}-\frac{\mathrm{\pi }}{2}\right)}$

$\Rightarrow \sqrt{3}=\frac{\sin \mathrm{i}}{-\cos 2\mathrm{i}}=\frac{\sin \mathrm{i}}{2{\sin }^2\mathrm{i}-1}$

$\Rightarrow 2\sqrt{3}{\sin }^2\mathrm{i}-\sin \mathrm{i}-\sqrt{3}=0$

$\Rightarrow \sin \mathrm{i}=\frac{1\pm \sqrt{1+24}}{4\sqrt{3}}=\frac{1\pm 5}{4\sqrt{3}}$

Rejecting the negative value, we get,

 $\mathrm{sini}=\frac{\sqrt{3}}{2}\Rightarrow \mathrm{i}={60}^0$