A light ray enters through a right angled prism at point $P$ with the angle of incidence $30^{\circ}

Question

A light ray enters through a right angled prism at point $P$ with the angle of incidence $30^{\circ}$ as shown in figure. It travels through the prism parallel to its base $B C$ and emerges along the face $A C$. The refractive index of the prism is

Professor PC Thomas Classes · Answer

(b) : Let the refractive index of the prism be $\mu$. Applying Snell's law at point $P$, $$ \begin{align*} & 1 \sin 30^{\circ}=\mu \sin r_1 \ & \sin r_1=\frac{1}{2 \mu}  	ag{i}\ & r_1+r_2=90^{\circ} \Rightarrow r_2=90^{\circ}-r_1 \end{align*} $$ Applying Snell's law at point $Q$, $\mu \sin r_2=1 	imes \sin 90^{\circ}$ $$ \begin{aligned} & \mu 	imes \sin \left(90^{\circ}-r_1ight)=1 \ & \mu \cos r_1=1 \end{aligned} $$ $\mu 	imes \sqrt{1-\sin ^2 r_1}=1 ; \mu \sqrt{1-\frac{1}{4 \mu^2}}=1$ (Using equation (i)) $$ 1-\frac{1}{4 \mu^2}=\frac{1}{\mu^2} ; \frac{5}{4 \mu^2}=1 ; \mu=\frac{\sqrt{5}}{2} $$