Question

Light from a denser medium $ I $ goes to a rarer medium $ II $ . When angle of incidence is $ \theta $ , the reflected and refracted rays are perpendicular to each other. The critical angle is :

• A. $ \sin ^{-1}(\cos \theta ) $
• B. $ \sin ^{-1}(\cot \theta ) $
• C. $ \sin ^{-1}(\tan \theta ) $
• D. $ \sin ^{-1}(1) $

Answer 1

Answer: (C)

From laws of reflection, we know that angle of incidence is equal to angle of reflection.

Let $r$ be the angle of refraction, along line $A O B$, we have
$\theta+90^{\circ}+r=180^{\circ}$
$\Rightarrow r=90-\theta^{\circ}$
Also from Snell's law
${ }_{1} \mu_{2}=\frac{\sin i}{\sin r}$
In this case $i=\theta, r=90^{\circ}-\theta$
${ }_{1} \mu_{2}=\frac{\sin \theta}{\sin \left(90^{\circ}-\theta\right)}=\frac{\sin \theta}{\cos \theta}=\tan \theta$
Also from definition of critical angle, we have
$\sin C =\frac{1}{{ }_{2} \mu_{1}} $
$\sin C =\tan \theta$
$ \Rightarrow C =\sin ^{-1}(\tan \theta)$