Calculate the angle of minimum deviation of a 60o prism of a refractive index [sin-10.75 = 49o]

Calculate the angle of minimum deviation of a 60o prism of a refractive index [sin-10.75 = 49o]

Answer Details

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