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For a flint glass prism of refractive index $\sqrt{3}$ , the angle of minimum deviation is found equal to the angle of the prism. The angle of incidence on the first refracting surface of the prism in the minimum deviation position is

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$60^{\circ}$

$30^{\circ}$

$45^{\circ}$

$90^{\circ}$

answer is B.

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Detailed Solution

A = D = 2r

$μ = \frac{\sin (\frac{A + D}{2})}{S i n \frac{A}{2}}$

$\sqrt{3} = \frac{S i n A}{S i n \frac{A}{2}}$

$S i n (\frac{A}{2}) \times \sqrt{3} = 2 \sin \frac{A}{2} . C o s \frac{A}{2} \Rightarrow \frac{A}{2} = \frac{\sqrt{3}}{2}$

$\Rightarrow \frac{A}{7} = 30 °$

$\Rightarrow A = 60 °$

$\Rightarrow r = 30 °$

$\Rightarrow D = 60 °$

$D = 2 i - A$

$\begin{aligned} 60 = 2 i - 60 \\ 2 i = 120^{\circ} \\ i = 60^{\circ} \end{aligned}$

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