Question
The minimum deviation produced by a hollow prism filled with a certain liquid is found to be $30^o $. The light ray is also found to be refracted at angle of $30^o $. The refractive index of the liquid is
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Answer
(a) For minimum angle of deviation for a prism
$A = 2r$, $\therefore $ $A = {60^o}$
Now $\mu = \frac{{\sin \frac{{60 + 30}}{2}}}{{\sin \frac{{60}}{2}}} = \frac{{\sin {{45}^o}}}{{\sin {{30}^o}}} = \frac{1}{{\sqrt 2 }} \times \frac{2}{1} = \sqrt 2 $
$A = 2r$, $\therefore $ $A = {60^o}$
Now $\mu = \frac{{\sin \frac{{60 + 30}}{2}}}{{\sin \frac{{60}}{2}}} = \frac{{\sin {{45}^o}}}{{\sin {{30}^o}}} = \frac{1}{{\sqrt 2 }} \times \frac{2}{1} = \sqrt 2 $
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