$(a)\;2\;cm \\ (b)\;6\;cm \\ (c)\;5.2\;cm \\ (d)\;3.26\;cm $

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Let d be diameter of refracted beam. Then,

$d=PQ \cos 60^{\circ}$

and $d'= PQ \cos r$

ie $\large\frac{d'}{d} =\frac{\cos r}{\cos 60^{\circ}}$

$\qquad= 2 \cos r$

or $d'= 2 d \cos r$

$\sin r =\large\frac{\sin i}{\mu}$

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

$\qquad= \large\frac{1}{\sqrt 3}$

$\therefore \cos r =\sqrt {1- \sin ^2 r}= \sqrt {\large\frac{2}{3}}$

$\therefore d' =(2)(2) \sqrt {\large\frac{2}{3}}=4 \sqrt {\large\frac{2}{3}}$$cm$

$\qquad = 3.26\;cm$

Hence d is the correct answer.

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