Animated Solution for Physics - Optics: A ray of light is incident on a prism $ABC$ of refractive index $\sqrt{3}$ as shown in figure. (a) Find the angle of incidence for which the deviation of light ray by the prism $ABC$ is minimum. (b) By what angle the second identical prism must be rotated, so that the final ray suffers net minimum deviation.

A ray of light is incident on it such that it undergoes minimum deviation. Our goal is to find this specific angle of incidence, and then determine how to orient a second identical prism to minimize the net deviation of the entire system.

This is a beautiful symmetric property of prisms. Because the ray is parallel to the base, the angle of refraction at the first surface, $r_1$, is exactly half of the prism angle $A$.

Since it's an equilateral prism, $A=60^{\circ }$, making $r_1$ equal to $30^{\circ }$:

Now, let's apply Snell's law at the first interface. The refractive index $\mu$ is the ratio of the sine of the angle of incidence to the sine of the angle of refraction:

We substitute the given refractive index, $\sqrt{3}$, and $r_1$ as $30^{\circ }$:

Solving for the angle of incidence, we multiply $\sqrt{3}$ by $\sin 30^{\circ }$, which is $\frac{1}{2}$. This gives:

Therefore, the angle of incidence $i_1$ must be $60^{\circ }$. This answers the first part of our question.

We need to find how much to rotate it so that the net deviation of the final ray is minimum.

Think about it... what is the absolute minimum deviation a light ray can suffer? Zero! This happens when the two prisms are oriented oppositely, effectively forming a rectangular glass slab. In a glass slab, the emergent ray is perfectly parallel to the incident ray.

To achieve this opposite orientation, we must rotate the second prism. If we rotate it by $60^{\circ }$ anti-clockwise, its faces will align perfectly with the first prism, creating a parallelogram shape. The ray now travels straight through the interface and emerges parallel to the original incident ray. The net deviation is zero. Thus, the required rotation is $60^{\circ }$ anti-clockwise.