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 δm = (2i-A) with r = A/2
According to given problem,
i = 2r = A [as r = A/2 ]
δm = 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°