## In This post you will get answer of Q.

A prism is made up of material of refractive index $sqrt{2}$. The angle of the prism is A. If the

angle of minimum deviation is equal to the angle of the prism, the value of A is

Solution:

## $because mu=frac{sin left[frac{left(A+delta_{m}right)}{2}right]}{sin left(frac{A}{2}right)}$

Here, $mu$ is refractive index $delta_{m}$ is minimum deviation angle, $A$ is angle of prism.

$therefore sqrt{2}=frac{sin left[frac{A+A}{2}right]}{sin frac{A}{2}}$

$Rightarrow sqrt{2} sin frac{A}{2}=sin, A$

Hence, $A=90^{circ}$