Question

A guarter cylinder having an index of refraction $\sqrt{3}$ and radius $R$ is kept in vacuum and is in the shape of quarter circle of radius $R$. A light ray parallel to the base of the material is incident from the left at a distance $\frac{\sqrt{3}R}{2}$ above the base and emerges out of the material at an angle $\theta$. The total deviation suffered by the emergent ray

Answer 1

$\sin i=\frac{\sqrt{3}R}{2R}=\frac{\sqrt{3}}{2}$

$\therefore i=60$

For the refraction at $A$,

$\sin i=\sqrt{3}\sin r$

$\frac{\sqrt{3}}{2}=\sqrt{3}\sin r$

$\therefore r^{\prime }={30}^{\underset{–}{\circ }}$

From the figure,

$r=r^{\prime }={30}^{\underset{–}{\circ }}$


$\therefore$ For the refraction at $B$,

$\sqrt{3}\sin 30{}^{\text{o}}=1\sin \theta$

$\frac{\sqrt{3}}{2}=\sin \theta$

$\therefore \theta ={60}^{\underset{–}{\circ }}$

The emergent ray $BS$ makes angle $\theta ={60}^{\underset{–}{\circ }}$ with the direction of incident ray

$\therefore$ deviation $\delta ={60}^{\underset{–}{\circ }}$