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

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.

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