# Light enters a triangular at an incident angle of $\theta$ at the point Q with an angle of refraction equal to $90^{\circ}$ , What is the refractive index of prism in terms of $\theta$ . The angle of the prism is $90^{\circ}$

$(a)\;(1+\sin ^2 \theta) \\ (b)\;(\sin ^2 \theta-1)^{1/2} \\ (c)\;(1+\sin ^2 \theta)^{1/2} \\ (d)\;(\sin ^2 \theta-1)$

## 1 Answer

Since angle of prison $A=90^{\circ}$
$\angle OPQ +\angle OQP=90^{\circ}$
Also since the emergent angle is $90^{\circ}$
$\angle OQP= \theta _c$ the critical angle
$\therefore \angle OPQ=90 - \theta _c$
Now $\mu = \large\frac{\sin \theta}{\sin (90 - \theta_l)}$
$\mu =\large\frac{\sin \theta}{\cos \theta_l}$
$\qquad= \large\frac{\sin \theta }{\sqrt {1- \sin^2 \theta_c}}$
We have $\mu =\large\frac{1}{\sin \theta_c}$
$\mu \sqrt {1- \large\frac{1}{\mu^2}}=\sin \theta$
$\mu^2 (1- \large\frac{1}{\mu^2})$$=\sin \theta$
$\mu^2-1=\sin ^2 \theta$
$\mu^2= 1+\sin ^2 \theta$
$\mu =(1+ \sin ^2 \theta )^{1/2}$
Hence c is the correct answer
answered Jan 20, 2014 by

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