For an isosceles prism of angle A and refractive index $μ$ , it is found that the angle of minimum deviation $δ_{m} = A$ . Which of the following options is/are correct?

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At minimum deviation, the incident angle $i_{1}$ and the refracting angle $r_{1}$ at the first refracting surface are related by $r_{1} = (i_{1} / 2)$

For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_{1} = \sin^{- 1} [\sin A \sqrt{4 \cos^{2} \frac{A}{2} - 1} - \cos A]$

For this prism the refractive index $μ$ and the angle of prism A are related as $A = \frac{1}{2} \cos^{- 1} (\frac{μ}{2})$

For the angle of incidence $i_{1} = A$ , the ray inside the prism is parallel to the base of the prism

answer is A, B, C.

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Detailed Solution

$\begin{array}{l} δ_{m} = A & μ = \frac{\sin (\frac{A + {\bar{δ}}_{m}}{2})}{\sin A / 2} \\ \Rightarrow μ = \frac{\sin A}{\sin A / 2} \\ \Rightarrow μ = \frac{2 \sin A / 2 \cos A / 2}{\sin A / 2} = 2 \cos A / 2 \\ \Rightarrow \cos A / 2 = (μ / 2) \\ \Rightarrow (A / 2) = \cos^{- 1} (μ / 2) \\ \Rightarrow A = 2 \cos^{- 1} (μ / 2) (D) is not correct. \\ δ_{m} = (i_{1} + i_{2}) - A \\ \Rightarrow (i_{1} + i_{2}) = 2 A and i_{1} = i_{2} = A \end{array}$

So option A is correct.

$\begin{array}{l} r_{1} = r_{2} and μ = \frac{\sin i_{1}}{\sin r_{1}} \\ 2 \cos A / 2 = \frac{\sin A}{\sin r_{1}} \\ \Rightarrow \sin r_{1} = \frac{2 \sin A / 2 \cos A / 2}{2 \cos A / 2} \\ \Rightarrow r_{1} = A / 2 = i_{1} / 2 option (B) is correct \end{array}$

Emergent ray tangential to the surface means

$\begin{array}{l} i_{2} = 90^{\circ} \Rightarrow r_{2} = C (critical angle) \\ r_{1} = (A - C) \\ μ = \frac{\sin i_{1}}{\sin r_{1}} \\ \sin i_{1} = μ \sin r_{1} = 2 \cos A / 2 [\sin (A - C)] \\ = 2 \cos A / 2 (\sin A \cos C - \cos A \sin C) \\ = 2 \cos A / 2 [\sin A \sqrt{1 - \sin^{2} C} - \cos A \frac{1}{μ}] \\ = 2 \cos A / 2 [\sin A \sqrt{1 - \frac{1}{μ^{2}}} - \cos A \frac{1}{μ}] \\ = 2 \cos A / 2 [\sin A \sqrt{1 - \frac{1}{4 \cos^{2} A / 2}} - \cos A \frac{1}{2 \cos A / 2}] \\ = 2 \cos A / 2 [\frac{\sin A \sqrt{4 \cos^{2} A / 2 - 1}}{2 \cos A / 2} - \cos A \frac{1}{2 \cos A / 2}] \\ = [\sin A \sqrt{4 \cos^{2} A / 2 - 1} - \cos A] \\ \Rightarrow i_{1} = \sin^{- 1} [\sin^{- 1} A \sqrt{4 \cos^{2} A / 2 - 1} - \cos A] \end{array}$