Question

A light ray, going through a prism with the angle of prism ${60}^o$, is found to deviate by ${30}^o$. What limit on the refractive index can be put from these data ?

Answer 1

Given $A={60}^{\circ }$ and $\delta ={30}^{\circ }$

we know that,

$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin A/2}=\frac{\sin \frac{{60}^{\circ }+{\delta }_m}{2}}{\sin {30}^{\circ }}=2\sin \frac{{60}^{\circ }+{\delta }_m}{2}$

Since, one ray has been found out which has deviated by ${30}^{\circ }$, the angle of minimum deviation should be either equal or less than ${30}^{\circ }$. (It can not be more than ${30}^{\circ }$).

So, $\mu \le 2\sin \frac{{60}^{\circ }+{\delta }_m}{2}$ (because $\mu$ will be more if ${\delta }_m$ will be more)

or, $\mu \le 2\times 1/\sqrt{2}$ or, $\mu \le \sqrt{2}$.