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, $A=90^{circ}$
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, $A=90^{circ}$