Question

A ray of monochromatic light is incident on one refracting face of a prism of angle 75 degree.it passes through the prism and is incident on the other face at the critical angle .if the refractive index of the material of the prism is root 2 the angle of incidence on the first face of the prism is

Answer 1

Angle of Incidence = 45°

Answer 2

The incident angle on the first face is 45°

Explanation:

According to Snell’s law,

$\frac{\sin i}{\sin r}=\mu$

Here, i and r are the incident and refracted angle respectively and μ is their corresponding refractive medium index.

At critical incident angle, i = 90°

So,

$\sin r=\frac{1}{\mu}$

$\text{Critical angle} =r=\sin ^{-1}\left(\frac{1}{\mu}\right)$

$\text{Critical angle of prism} = c=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

As the refractive index of the prism is given as $\sqrt{2}$

Thus c = 45°

The angle of the prism is 75°

We know that Angle of prism (A) = angle of refraction of first face (r) + angle of refraction or critical angle of second face (c)

$A= r+c$

$75=r+45$

$75-45=r$

Angle of refraction of first face, $r=30^{\circ}$

Thus by applying the snell’s law again,

Where,

r = 30°

$\text{Refractive index} = \sqrt{2}$

$\frac{\sin i}{\sin r}=\mu$

$\frac{\sin i}{\sin 30}=\sqrt{2}$

$\sin i=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}$

$\therefore i=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Thus, the angle of incidence is 45°