Question

If the refractive index of material of equilateral prism is $\sqrt{3}$ then angle of minimum deviation of the prism will be:

• A. $25{}^{\circ }$
• B. $60{}^{\circ }$
• C. $45{}^{\circ }$
• D. $30{}^{\circ }$

Answer 1

Answer: (B)

Refractive index of the material $\mu =\sqrt{3}$ The relation for refractive index is $\mu =\frac{\sin i}{\sin r}$ $\sqrt{3}=\frac{\sin i}{\sin r}$ $\sqrt{3}=\frac{\sin \frac{A+{\delta }_m}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{A+{\delta }_m}{2}}{\sin \frac{60{}^{\circ }}{2}}$ $\upsilon =\frac{\sin \frac{A+{\delta }_m}{2}}{\frac{1}{2}}$ Or $\sin \frac{A+{\delta }_m}{2}=\frac{\sqrt{3}}{2}$ $=\sin 60{}^{\circ }$ (where A is the angle of equilateral prism) Hence, $\frac{A+{\delta }_m}{2}=60{}^{\circ }$ Thus, $A+{\delta }_m=120$ ${\delta }_m=120{}^{\circ }-60{}^{\circ }$ ${\delta }_m=60{}^{\circ }$