The angle of minimum deviation of a glass prism is π – 2 A , where A is angle of prism. The angle of prism is _ _ _ _ _ _ _ _ _ _ _ _ .

    a )      2 tan μ

    b )      2 tan^-1 μ   

    c )       2 cot^-1 μ

    d )      2 cot μ

Solution :

Given :

δ _m = π – 2 A

The refractive index of prism can be given as

μ = { sin [ ( δ _m + A ) / 2 ] } / { sin ( A / 2 ) }

putting the value of δ _m in above equation we get,

μ  = { sin [ ( π – 2 A ) + A ) / 2 ] } /  { sin ( A / 2 ) }

μ  = { sin [ ( π + A ) / 2 ] } / { sin ( A / 2 ) }

μ  = { sin [ ( π / 2 ) + ( A / 2 ) ] } / { sin ( A / 2 ) }

but, We know that

sin [ ( π / 2 ) + θ ] = cos θ

Therefore above equation becomes,

μ  = { cos ( A / 2 ) } /  { sin ( A / 2 ) }

μ  = { cot ( A / 2 ) }

( A / 2 ) = cot ^{- 1} μ

A = 2 cot ^{– 1}  μ

The angle of prism = A = 2 cot ^{– 1}  μ.