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For a prism of refractive index 1.732. The angle of minimum deviation is equal to the angle of prism. The angle of prism is $(\sqrt{3} = 1.732)$

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80 °

60 °

70 °

50 °

answer is C.

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Detailed Solution

μ = \frac{\sin \frac{(A + A)}{2}}{\sin \frac{A}{2}}

$= \frac{\sin 2 \frac{A}{2}}{\sin \frac{A}{2}} = \frac{sinA}{sin \frac{A}{2}} = \frac{2 \sin \frac{A}{2} . \cos \frac{A}{2}}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2}$

$1.732 = 2 \cos \frac{A}{2}$

$\frac{\sqrt{3}}{2} = \cos \frac{A}{2}$

$\cos 60 = \cos \frac{A}{2}$

$60 ° = A$

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