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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?

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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_m)//2])/(sin (A//2))i.e., sqrt(2) = ( sin A)/( sin A//2) ` (or)
`sqrt(2) sin (A )/( 2 ) = 2 sin (A )/(2) cos (A )/(2) i.e., cos ""(A)/(2) = (1)/( sqrt(2)) `(or )
`A/2 = cos^(-1) [(1)/(sqrt2)]=45^@ ,i.e., ,A=90^@`
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