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A ray of light is incident on a prism of refractive index $\mu =\sqrt 2$ It is found that the deviation caused is $30^{\circ}$ The angle of prism is $60^{\circ}$ What will be the angle between the ray inside the prism and the base of the prism .

$(a)\;0^{\circ} \\ (b)\;45^{\circ} \\ (c)\;30^{\circ} \\ (d)\;60^{\circ} $

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We know that $\mu =\large\frac{\sin \bigg( \Large\frac{A + \delta _m}{2}\bigg)}{\sin \large \frac{A}{2}}$
$\sqrt 2 =\large\frac{\sin \bigg( \Large\frac{60 + \delta _m}{2}\bigg)}{\sin 30}$
From this condition we observe that if $ \delta=30 ^{\circ} $ the expression hold c.
Also given that the deviation caused is $30^{\circ}$
$\therefore $ At minimum deviation condition the ray travels parallel to the base inside the prism.
The angle between the ray and base inside the prism is zero.
Hence a is the correct answer.

 

answered Jan 24, 2014 by meena.p
edited Jul 28, 2014 by thagee.vedartham
 

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