Question

For a prism of refractive index $$1.732$$, the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is

Answer 1

Correct option is C. $$60^{o}$$
$$\mu=\dfrac{\sin \dfrac{A+\delta_{m}}{2}}{\sin \dfrac{A}{2}}=\dfrac{\sin \dfrac{A+A}{2}}{\sin \dfrac{A}{2}}=\dfrac{\sin A}{\sin \dfrac{A}{2}}$$
$$=\dfrac{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}{\sin \dfrac{A}{2}}=2\cos \dfrac{A}{2}$$
So, $$\sqrt{3}=2\cos \dfrac{A}{2}\Rightarrow \dfrac{\sqrt{3}}{2}=\cos \dfrac{A}{2}\Rightarrow A=60^{o}$$