Q: A ray of light passing through a prism having μ = √2 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.

Sol: As the prism is in ‘the position of minimum deviation δ_{m} = (2i-A) with r = A/2

According to given problem,

i = 2r = A [as r = A/2 ]

δ_{m} = 2A – A = A

Hence from , $\large \mu = \frac{sin(A + \delta_m)/2}{sin(A/2)} $

$\large \sqrt{2} = \frac{sinA}{sin(A/2)} $

$\large \sqrt{2}sin(A/2) = 2sin(A/2) cos(A/2) $

$\large cos(A/2) = \frac{1}{\sqrt{2}} $

A/2 = 45° ⇒ A = 90°