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

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) = 49^o

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

60 + D = 2 * 49 = 98^o

D = 98^o - 60^o

D = 38^o