When a glass prism of refracting angle 60° is immersed in a liquid its angle of minimum deviation is 30°. The critical angle of glass with respect to the liquid medium is

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52°

50°

45°

42°

answer is B.

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Detailed Solution

$A = 60^{\circ}, δ_{m} = 30^{\circ} so μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})} μ = \frac{\sin (\frac{60^{\circ} + 30^{\circ}}{2})}{\sin (\frac{60^{\circ}}{2})} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \sqrt{2} Also μ = \frac{1}{\sin C} \Rightarrow C = \sin^{- 1} (\frac{1}{μ}) \Rightarrow C = 45^{\circ}$

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