Question

The angle if minimum deviation measured with a prism is ${30}^{\circ }$ and the angle of prism is ${60}^{\circ }$. The refractive index of the prism material is.

• A. $\sqrt{2}$
• B. $2$
• C. $\frac{3}{2}$
• D. $\frac{4}{3}$

Answer 1

Answer: (A)

$\sqrt{2}$

The refractive index of prism material is given by ,

$n=\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$

Given , ${\delta }_m={30}^o,A={60}^o$

Hence , $n=\frac{\sin \left(\frac{60+30}{2}\right)}{\sin \frac{60}{2}}$

or $n=\frac{\sin \left(\frac{90}{2}\right)}{\sin 30}$

or $n=\frac{\sin 45}{\sin 30}$

or $n=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}$

or $n=\sqrt{2}$