$(a)\;\sin ^{-1} \sqrt 3 \\ (b)\;\sin ^{-1} \sqrt 2 \\ (c)\;\sin ^{-1} \frac{\sqrt 3}{2} \\ (d)\;\sin ^{-1} \frac{1}{\sqrt 3} $

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For the ray of graze the surface it should be incident at the critical angle.

$\angle OBA= \theta_c$

$\therefore \angle r= (90- \theta_c)$

So at 0

$1 \times \sin \theta =\sqrt 3 \sin (90 -\theta_c)$

$\sin \theta = \sqrt 3 \cos \theta_c$

$\qquad= \sqrt 3 \sqrt {1- \sin ^2 \theta_c}$

Since $\mu =\sqrt 3$

$\sin \theta_c=\large\frac{1}{\sqrt 3}$

$\therefore \sin \theta=\sqrt 3 \sqrt {1- (\Large\frac{1}{\sqrt 3})^2}$

$\qquad= \sqrt 3 \sqrt {\Large\frac{3-1}{3}}$

$\qquad= \sqrt 3 \large\frac{\sqrt 2}{\sqrt 3}$

$\qquad= \sqrt 2$

$\theta =\sin ^{-1} \sqrt 2$

Hence b is the correct answer.

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