A ray of light passing through a prism having refractive index `sqrt2` suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. What is the angle of prism?

Explanation

As the prism is in ‘the position of minimum deviation
`delta_(m) = (2i- A)` with r=A/2
According to given problem
i = 2r = A[as r = A/2]
`delta_(m)=2A-A=A` and hence from
`mu=(sin(A+ delta)//2)/(sin(A//2))` i.e `sqrt(2)= (sinA)/(sin(A//2))` (or)
`sqrt(2)= sin(A/2)=2sin(A/2)cos(A/2)` i.e, `cos(A/2)=1/sqrt(2)` or
`A/2=cos^(-1)[1/sqrt(2)]=45^@` i.e `A=90^@`