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The angle of incidence for a ray of light at a refracting surface of a prism is 45° . The angle of prism is 60°. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are:

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$45^{\circ}, \frac{1}{\sqrt{2}}$

$30^{\circ}, \sqrt{2}$

$45^{\circ}, \sqrt{2}$

$30^{\circ}, \frac{1}{\sqrt{2}}$

answer is B.

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Detailed Solution

Given: Angle of incidence and angle of prism:

i=45°; A= 60°;

Angle of minimum deviation : $δ_{m} = 2 i - A = 30^{\circ}$

Refractive index of material of prism:

$\begin{array}{l} μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})} \\ = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1}{\sqrt{2}} \cdot \frac{2}{1} = \sqrt{2} \end{array}$

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