JEE Physics 2023 Question with Solution

Given: The pass-axes of polaroids A and B are perpendicular. Polaroid C is placed between them so that it bisects the angle, hence the angle between A and C is $45^\circ$ and the angle between C and B is also $45^\circ$. The incident light is unpolarised with intensity $I_0$.

Find: The intensity after passing through polaroid B.

For unpolarised light passing through the first polaroid,

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

Using Malus' law for polaroid C,

$$I_C = I_A \cos^2 45^\circ = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$$

Again applying Malus' law for polaroid B,

$$I_B = I_C \cos^2 45^\circ = \frac{I_0}{4} \times \frac{1}{2} = \frac{I_0}{8}$$

Therefore, the transmitted intensity is $\frac{I_0}{8}$. The solution states the correct option is A, but this value corresponds to option C in the given options. Hence the defensible correct option is C.

Given: Three polaroids are arranged in sequence. A and B are crossed, and C is inserted midway between their axes.

Find: Final transmitted intensity.

1. After the first polaroid A, unpolarised light becomes plane polarised and its intensity becomes

$$\frac{I_0}{2}$$

2. Since C bisects the angle between A and B, the angle between A and C is $45^\circ$.
3. By Malus' law,

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

Since

$$\cos^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$$

we get

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

4. The angle between C and B is also $45^\circ$, so

$$I_B = I_C \cos^2 45^\circ = \frac{I_0}{4} \times \frac{1}{2} = \frac{I_0}{8}$$

Therefore, the final transmitted light intensity is $\frac{I_0}{8}$.