Question

A ray of light is incident on a prism made of a material of refractive index $\sqrt{2}$ at an angle $\left(i\right)$. Angle of the prism is ${30}^{\circ }$ and if the second refracting surface is silvered, a ray of light entering the prism will retrace its path $($after reflection from the silvered surface$)$ then find the angle of incidence.

[Assume, surrounding medium to be air]

• A. ${60}^{\circ }$
• B. ${45}^{\circ }$
• C. ${30}^{\circ }$
• D. Zero

Answer 1

The correct option is B ${45}^{\circ }$

Given,
Refractive index of material of the prism, ${\mu }_{\text{prism}}=\sqrt{2}$

Angle of prism, $A={30}^{\circ }$

Applying snell's law at first face, we get

${\mu }_{\text{air}}\sin i={\mu }_{\text{prism}}\sin r$

Substituting the given data,

$1\sin i=\sqrt{2}\sin r......\left(1\right)$


From the $\Delta \text{ABC}$,

$\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow {30}^{\circ }+\angle B+{90}^{\circ }={180}^{\circ }$

$\therefore \angle B={60}^{\circ }$

From figure we can write,

$r={90}^{\circ }-{60}^{\circ }={30}^{\circ }$

Now, substituting the value of $r$ in equation $\left(1\right)$, we have

$1\sin i=\sqrt{2}\sin {30}^{\circ }$

$\Rightarrow \sin i=\sqrt{2}\left(\frac{1}{2}\right)$

$\therefore i={45}^{\circ }$

So, angle of incidence will be ${45}^{\circ }$.

Thus, option (b) is the correct answer.

Why is Question : This question tests the conceptual understanding of refraction and reflection when one of the face of the prism is made reflecting surface instead of refracting surface.