Three identical polaroids P1, P2, and P3 are placed one after ano… — Solution | JEE Physics

NVAEasyJEE 2023Diffraction & Polarisation

Three identical polaroids $P_1$ , $P_2$ , and $P_3$ are placed one after another. The pass axis of $P_2$ and $P_3$ are inclined at angles of $60^\circ$ and $90^\circ$ with respect to the axis of $P_1$ . The source $S$ has an intensity of $256 \, \text{W/m}^2$ . The intensity of light at point $O$ is —— $\text{W/m}^2$ .

Answer

Correct answer:24

Step-by-step solution

Standard Method

Given: The source intensity is $I_0 = 256 \, \text{W/m}^2$ . Three identical polaroids are placed successively with pass-axis angles $0^\circ$ , $60^\circ$ , and $90^\circ$ relative to $P_1$ .

Find: The intensity of light at point $O$ after passing through all three polaroids.

For unpolarized light passing through the first polaroid,

I_1 = \frac{I_0}{2} = \frac{256}{2} = 128 \, \text{W/m}^2

Using Malus's law for the second polaroid,

I_2 = I_1 \cos^2 60^\circ

Since

\cos 60^\circ = \frac{1}{2}

we get

I_2 = 128 \left(\frac{1}{2}\right)^2 = 128 \cdot \frac{1}{4} = 32 \, \text{W/m}^2

The angle between $P_2$ and $P_3$ is

90^\circ - 60^\circ = 30^\circ

Again applying Malus's law,

I_3 = I_2 \cos^2 30^\circ

Since

\cos 30^\circ = \frac{\sqrt{3}}{2}

we get

I_3 = 32 \left(\frac{\sqrt{3}}{2}\right)^2 = 32 \cdot \frac{3}{4} = 24 \, \text{W/m}^2

Therefore, the intensity of light at point $O$ is $24 \, \text{W/m}^2$ .

Common mistakes

Using $90^\circ$ as the angle between $P_2$ and $P_3$ is incorrect because both angles are given with respect to $P_1$ . The correct relative angle is $30^\circ$ . Always subtract the two given orientations to find the angle between successive polaroids.
Forgetting that unpolarized light intensity becomes half after the first polaroid is incorrect. The first transmission is not given by Malus's law from the source intensity. First use $I_1 = I_0/2$ , then apply Malus's law for subsequent polaroids.

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