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Calculate the angle of minimum deviation of a 6... - JAMB Physics 2017 Question

Calculate the angle of minimum deviation of a 60o prism of a refractive index [sin-10.75 = 49o]
A
38.00o
B
19.47o
C
16.25o
D
49o
correct option: 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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