A ray of monochromatic light is incident on one refracting face of a prism of angle $75^o$. 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 $\sqrt{2}$, the angle of incidence on the first face of the prism is :

The correct option is B

Hint: Consider snells law and Critical angle

Critical angle is given as,

$c={\sin }^{-1}\left(\frac{1}{\mu }\right)$ 

Where, $\mu$ is the refractive index.

In order to describe the relation between the angle of incidence and reflection snells law is used.

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

Where $i$ is the angle of incidence and $r$ is the angle of reflection.

Explanation for correct answer:

Step 1: Information required

Critical angle is given as,

$c={\sin }^{-1}\left(\frac{1}{\mu }\right)$ 

Where, $\mu$ is the refractive index.

In order to describe the relation between the angle of incidence and reflection snells law is used.

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

Here, $i$ is angle of incidence 

$r$ is the angle of reflection.

Angle of prism $A=75^o$ 

Refractive index $\mu =\sqrt{2}$ 

Step 2: Calculate the angle of reflection,

The angle of the prism is the sum of reflection and critical angle.

Then,

$A=r+c$ 

Substitute the value of critical angle and angle of prism in the above equation 

$A=r+{\sin }^{-1}\left(\frac{1}{\mu }\right)$ 

$75=r+{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$75=r+45$

$r=75-45$ 

$r=45^o$ 

So, the angle of reflection is $r=45^o$.

Step 3: Calculate angle of incidence,

Consider snells law,

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

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

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

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

$i={\sin }^{-1}\frac{1}{\sqrt{2}}$

$i=45^o$

Hence the angle of incidence is $i=45^o$.