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˚

\[Refractive  index, \] 

\[  \mu \leq \frac{\sin\left( \frac{A + \delta_m}{2} \right)}{\sin\left( \frac{A}{2} \right)}\] 

\[\mu \leq \frac{\sin\left( \frac{60^\circ + \delta_m}{2} \right)}{\sin\left( \frac{60^\circ}{2} \right)}\] 

\[\mu \leq 2\sin\left( \frac{60^\circ\ + \delta_m}{2} \right)\]
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,
\[\mu \leq 2\sin \left( \frac{60^\circ  + \delta_m}{2} \right)\]
\[\mu \leq 2\sin \left( \frac{60^\circ \ + 30^\circ}{2} \right) = 2\sin \left( 45^\circ\ \right)\]
Refractive index (μ) will be more if angle of deviation (δ_m) is more.
\[\mu \leq 2 \times \frac{1}{\sqrt{2}}\]
\[\Rightarrow \mu \leq \sqrt{2}\]
Hence, the required limit of refractive index is \[\sqrt{2}\]