Question

A right-angled prism of apex angle $4^{\circ }$ and refractive index $1.5$ is located in front of a vertical plane mirror as shown in figure. A horizontal ray of light is falling on the prism. Find the total deviation produced in the light ray as it emerges 2nd time from the prism.

• A. $8^{\circ }CW$
• B. $6^{\circ }CW$
• C. ${180}^{\circ }CW$
• D. ${176}^{\circ }CW$

Answer 1

The correct option is D ${176}^{\circ }CW$

Consider the path of the ray shown in the diagram


Let deviation produced after first refraction through the prism be $+\delta$ (clockwise)

Then the emergent is incident on the plane mirror at an angle $\delta$ to the normal.

Total deviation produced by the plane mirror $=\pi -2\delta$ (clockwise)

After second refraction through the prism, the deviation produced by the prism will be $-\delta$ (anticlockwise)

So , total deviation $=\delta +\pi -2\delta -\delta =\pi -2\delta$ (clockwise)

Angle of deviation of thin lens $\delta =(\mu -1)A$ where \mu is the refrative index of the prism and $A$ is the apex angle of the prism.

Considering the prism to be thin,

$\delta =(1.5-1)\times 4^0=2^0$
Therefore, total deviation $=\pi -2\ast 2^0={176}^0$ (clockwie)