Calculate the angle of minimum deviation of a 60º prism of a refractive index 1.5[ $sin^{-1}$ (0.75) = 49º]

38^o

19.47^o

16.25^o

49^o

Correct answer is A

For a refractive index () = $\frac{\sin\frac{1}{2} (A + D)}{\sin\frac{1}{2}A}$

D = angle of minimum deviation

A = refractive angle of the prism

1.5 = $\frac{\sin\frac{1}{2} (60 + D)}{\sin\frac{1}{2} \times 60}$

1.5 = $\frac{\sin\frac{1}{2} (60 + D)}{\sin 30}$

Sin $\frac{1}{2}$ (60 + D) = 1.5 * Sin 30

Sin $\frac{1}{2}$ (60 + D) = 0.75

$\frac{1}{2}$ (60 + D) = Sin-1 (0.75)

but Sin-1 (0.75) = 49o

$\frac{1}{2}$ (60 + D) = 49

60 + D = 2 * 49 = 98o

D = 98o - 60o

D = 38o

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