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Q.

The angle of minimum deviation by prism is (180°–2A). Its critical angle will be

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a

sin-1tanA2

b

sin-1cotA2

c

cos-1cotA2

d

cos-1tanA2

answer is A.

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Detailed Solution

μ=sinA+180-2A2sin A/2 ; μ=cotA2 μ=1sin c ; 1sin c=cot A2⇒c=sin-1tanA2
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