$(a)\;\tan ^{-1} 2/3 \\ (b)\;\sin ^{-1} 2/3 \\ (c)\;90^{\circ} \\ (d)\;\cos^{-1} 1/3 $

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$OA=OB$ radius of the circle, $AO$ and $BO$ normal to the sphere at A and B respectively.

Since the incident ray emerge out at B, the emergent ray BQ is tangential we must have $\angle ABO $ = critical angle $\theta_c$ and $\therefore \angle OAB=\theta_c$

$\sin \theta_c=\large\frac{1}{\mu}$

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

At A $\mu =\large\frac{\sin i}{\sin \theta_c}$

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

$\therefore \sin i =\large\frac{3}{2}$$ \sin \theta_c$

$\qquad= \large\frac{3}{2} \times \frac{2}{3} =$$1$

$\therefore i= 90^{\circ}$

Hence c is the correct answer.

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