The refractive index of a prism with apex angle A is cot A/2. The angle of minimum deviation is :

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$δ_{m} = 180^{0} - 2 A$

$δ_{m} = 180^{0} - A$

$δ_{m} = 180^{0} - 3 A$

$δ_{m} = 180^{0} - 4 A$

answer is D.

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Detailed Solution

$μ = \frac{\sin (\frac{A + δ m}{2})}{\sin \frac{A}{2}} \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}} = \frac{\sin (\frac{A + δ m}{2})}{\sin \frac{A}{2}} \sin (\frac{π}{2} - \frac{A}{2}) = \sin (\frac{A + δ_{m}}{2}) δ_{m} = π - 2 A$

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