Question

If the refractive index of a material of equilateral prism is $\sqrt{3}$, then angle of minimum deviation of the prism is

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

$A={60}^{\circ },n=\sqrt{3}n=\frac{sin\left(\frac{A+\delta }{2}\right)}{sin\frac{A}{2}}$
where n is the refractive index, δm is the angle of minimum deviation, A is the prism angle
$\Rightarrow sin\left(\frac{A+\delta }{2}\right)=\frac{\sqrt{3}}{2}\frac{A+\delta }{2}={60}^{\circ }\Rightarrow \delta ={60}^{\circ }$