Question

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

Answer 1

Given,
The angle of the prism (A) = 60˚
The angle of deviation (δ_m) = 30˚
$$\begin{gathered} \mathrm{Refractive}\mathrm{index}, \\ \mathrm{\mu }\le \frac{\sin \left({\displaystyle \frac{A+{\delta }_{\mathrm{m}}}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)} \\ \mathrm{\mu }\le \frac{\sin \left({\displaystyle \frac{60°+{\delta }_{\mathrm{m}}}{2}}\right)}{\sin \left({\displaystyle \frac{60°}{2}}\right)} \\ \mathrm{\mu }\le 2\sin \left(\frac{60°+{\delta }_{\mathrm{m}}}{2}\right) \end{gathered}$$
As there is one ray that has been found which has deviated by 30˚, the angle of minimum deviation should be either equal to or less than 30˚ but it can not be more than 30˚.

Therefore,
$$\begin{gathered} \mathrm{\mu }\le 2\sin \left(\frac{60°+{\delta }_{\mathrm{m}}}{2}\right) \\ \mathrm{\mu }\le 2\sin \left(\frac{60°+30°}{2}\right)=2\sin \left(45°\right) \end{gathered}$$

Refractive index (μ) will be more if angle of deviation (δ_m) is more.
$\mathrm{\mu }\le 2\times \frac{1}{\sqrt{2}}$
$\Rightarrow \mathrm{\mu }\le \sqrt{2}$

Hence, the required limit of refractive index is $\sqrt{2}$