JEE Physics 2024 Question with Solution

Given: Unpolarized light of intensity $I_0$ passes first through polaroid A and then through polaroid B. The angle between their principal planes is $45^\circ$.

Find: The intensity of the emergent light after both polaroids.

When unpolarized light passes through a polaroid, the transmitted intensity becomes

$$I_A = \frac{I_0}{2}$$

Now apply Malus's law for the second polaroid:

$$I_B = I_A \cos^2 \theta$$

With $\theta = 45^\circ$,

$$I_B = \left( \frac{I_0}{2} \right) \cos^2 45^\circ$$

Since

$$\cos 45^\circ = \frac{1}{\sqrt{2}}$$

we get

$$I_B = \left( \frac{I_0}{2} \right) \left( \frac{1}{\sqrt{2}} \right)^2 = \left( \frac{I_0}{2} \right) \left( \frac{1}{2} \right) = \frac{I_0}{4}$$

Therefore, the intensity of emergent light is $I_0/4$. The correct option is A.

Given: Initial light is unpolarized with intensity $I_0$. Two polaroids are used, and the second is inclined at $45^\circ$ to the first.

Find: Final transmitted intensity.

For the first polaroid, unpolarized light always emerges with half the original intensity:

$$I_1 = \frac{I_0}{2}$$

This light is now plane-polarized along the transmission axis of polaroid A.

For the second polaroid, use Malus's law:

$$I_2 = I_1 \cos^2 \theta$$

Substitute $I_1 = \frac{I_0}{2}$ and $\theta = 45^\circ$:

$$I_2 = \frac{I_0}{2} \cos^2 45^\circ$$

Since

$$\cos^2 45^\circ = \frac{1}{2}$$

we obtain

$$I_2 = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$$

Hence, the final intensity is $I_0/4$.

The solution states one place as option C, but the working clearly gives $I_0/4$, which matches option A. Therefore, the defensible correct option is A.