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  4. If the refractive index of a material of equilateral prism is √ 3 , then angle of minimum deviation of the prism is

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Q. If the refractive index of a material of equilateral prism is $\sqrt 3 $ , then angle of minimum deviation of the prism is

AIPMTAIPMT 1999

Solution:

$A= 60^\circ , \mu = \sqrt 3 , {\delta}_m =?$
$ \mu = \frac{sin\bigg( \frac{A + {\delta}_m }{2}\bigg)}{ sin \frac{A}{2}} \, \, \therefore \, \, {\delta}_m = 60 ^ \circ $