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The angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of the given prism. Then the angle of the prism is ______. (sin 48° 36' = 0.75)

Asked in GUJCET 2015 · Deviation and minimum deviation

Answer: (4) 82° 48'

Step-by-step solution

Given: n = 1.5 and the angle of minimum deviation δₘ = A.

Idea: at minimum deviation n = (sin ((A+δₘ)/2))/(sin (A/2)).

Put δₘ = A: n = (sin A)/(sin (A/2)) = (2 sin (A/2)cos (A/2))/(sin (A/2)) = 2 cos (A/2).

So cos (A/2) = n/2 = (1.5)/2 = 0.75.

The datum gives sin 48° 36' = 0.75, and cos 41° 24' = sin 48° 36', so A/2 = 41° 24'.

Hence A = 2× 41° 24' = 82° 48'.

Why the other options are wrong

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