Question

The angle of minimum deviation for prism of angle $\frac{\pi }{3}$ is$\frac{\pi }{6}$. Calculate the velocity in the material of the prism, if the velocity of light in vacuum is $3\times {10}^{8}m{s}^{-1}$.

Answer 1

We know that,

Refractive index is

$\mu =\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\mu =\frac{\sin \left(\frac{60+30}{2}\right)}{\sin \left(\frac{60}{2}\right)}$

$\mu =\frac{\sin 45}{\sin 30}$

$\mu =\frac{2}{\sqrt{2}}$

$\mu =\sqrt{2}$

Now, speed in material of prism

$v_m=\frac{v_v}{\mu }$

$v_m=\frac{3\times {10}^8}{\sqrt{2}}$

$v_m=2.1\times {10}^{8}m/s$

Hence, the velocity in material of prism is $2.1\times {10}^{8}m/s$