$(a)\;\frac{4}{3} \sin \theta \\ (b)\;\frac{1}{\sin \theta} \\ (c)\;\frac{4}{3} \\ (d)\;1 $

Since the ray finally emerges parallel to the water surface.

r must be the critical angle.

$\therefore \sin r =\large\frac{1}{\mu _{\Large water}}$

$\sin r =\large\frac{3}{4}$

At the water glass inerface

$\mu _g \sin \theta=\mu \sin r$

$\mu _g =\large\frac{\mu _w \times \sin r}{\sin \theta}$

$\qquad= \large\frac{\Large\frac{4}{3} \times \frac{3}{4}}{\sin \theta}$

$\qquad= \large\frac{1}{\sin \theta}$

Hence b is the correct answer.

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