Question

The refractive index of a given glass is $\sqrt{2}$ find the angle of minimum deviation
produced by an equilateral prism made of this glass.

Answer 1

Concept: I would like to introduce this formula of an angle of minimum deviation for a prism as an angle of minimum deviation
$D=2{\sin }^{-1}\left(\sin \frac{A}{2}\right)-A$
Here, $n$ is the refractive index
$A$ is the angle of the prism
Solution:
Given: refractive index of glass,n=$\sqrt{2}$
A=${60}^0$

To find: the angle of minimum deviation
So,
$D=2{\sin }^{-1}\left(\sqrt{2}\sin 30°\right)-60°$
=$2{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)-60°$
=$2\times 45°-60°$
$D=30°$ or $\frac{\pi }{6}$