Question

For 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

• A. 80${}^o$
• B. 70${}^o$
• C. 60${}^o$
• D. 50${}^o$

Answer 1

The correct option is C 60${}^o$

We know that the refractive index ($\mu$) of a prism is related to the prism angle ($A$) and the minimum deviation angle ($D_m$) by the relation $\mu =\frac{\sin \frac{A+D_m}{2}}{\sin \frac{A}{2}}$

Given that the minimum deviation angle and the prism angle are equal. i.e., $A=D_m$ and

Refractive index of the prism is $\mu =1.732$

Thus, $1.732=\frac{\sin A}{\sin \left(A/2\right)}\Rightarrow 1.732=\frac{2\sin \left(A/2\right)\cos \left(A/2\right)}{\sin \left(A/2\right)}\Rightarrow \cos \left(A/2\right)=0.866\Rightarrow A={60}^{\circ }$