Question

a prism of refractive index √2 has a refracting angle of 60° at what angle must a ray incident on it so that it suffers minimum deviation​

Answer 1

Answer:

First, use prism formula:

$n = \frac{sin(\frac{Sm + A}{2}) }{sin(\frac{A}{2}) }$

$\sqrt{2} = \frac{sin(\frac{Sm + 60}{2} )}{sin 30}$

$\frac{\sqrt{2} }{2} = sin(\frac{Sm + 60}{2} )$

$sin(\frac{1}{\sqrt{2} } ) = \frac{Sm + 60}{2}$

45 x 2 = Sm + 60

90 - 60 = Sm

Sm = 30 degree

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now,

Incident angle =  $\frac{Sm + A}{2}$

∠I = $\frac{30 + 60}{2}$

∠I = 45°