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For a prism kept in air, it is observed that for an angle of incidence $60^{o}$ , the angle of prism A, The angle of emergence e and the angle of deviation $δ$ become equal. If refractive index of the prism is $\sqrt{3}$ , angle of prism A is

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$45^{o}$

$120^{o}$

$90^{o}$

$60^{o}$

answer is D.

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Detailed Solution

$\begin{array}{l} δ = i + e - A \Rightarrow δ = i + A - A = i \\ Since δ = e, we con clude i = e \\ Since δ = e, we can write r_{1} = r_{2} = \frac{A}{2} \\ ∴ \frac{\sin 60^{o}}{sinA / 2} = \sqrt{3} \Rightarrow A = 60^{o} \end{array}$

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