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A circular beam of diameter $d=2 \;cm$ falls on a plane surface of glass . The angle of incidence is $60^{\circ}$ and refractive index of glass is $ \mu=\large\frac{3}{2}$. The diameter of the refracted beam is ,

$(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.
answered Feb 19, 2014 by meena.p
edited Jul 28, 2014 by meena.p

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