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 ${60}^{\circ }$

Given , $\mu =\sqrt{3},A={60}^o$ (as prism is equilateral)

The refractive index of a prism is given by ,

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

$\sqrt{3}=\frac{\sin \left(\frac{60+{\delta }_m}{2}\right)}{\sin \frac{60}{2}}$

or $\sqrt{3}=\frac{\sin \left(\frac{60+{\delta }_m}{2}\right)}{\sin 30}$

or $\sqrt{3}=\frac{\sin \left(\frac{60+{\delta }_m}{2}\right)}{\frac{1}{2}}$

or

$\frac{\sqrt{3}}{2}=\sin \left(\frac{60+{\delta }_m}{2}\right)$

or $\sin 60=\sin \left(\frac{60+{\delta }_m}{2}\right)$

or $60=\left(\frac{60+{\delta }_m}{2}\right)$

or ${\delta }_m=120-60={60}^o$