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A transparent solid rod in the shape of a cuboid has refractive index $\sqrt 3$ The cuboid is kept in air, light enters midpoint (0) of one end of rod as shown . What should be the incident angle $\theta$ Such that when the refracted ray grazes along the wall.

 

$(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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1 Answer

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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.
answered Jan 17, 2014 by meena.p
edited Jul 28, 2014 by meena.p
 

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