In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of P₁ and P₂ are orthogonal to each other. The polarizer P₃ covers both the slits with its transmission axis at 45° to those of P₁ and P₂. An unpolarized light of wavelength λ and intensity I₀ is incident on P₁ and P₂. The intensity at a point after P₃ where the path difference between the light waves from s₁ and s₂ is (λ)/(3), is (a) I₀ (b) I₀/3 (c) I₀/4 (d) I₀/2

Answer: (c). After P₁ and P₂ each slit gets intensity I₀/2, polarised in perpendicular directions (they cannot interfere yet). P₃ at 45° to both: each beam becomes I₀/2cos ² 45° = I₀/4, now with the same polarisation. Path difference λ/3 ⇒ φ = 2π/3. I = I₀/4 + I₀/4 + 2₀/4cos 2π/3 = I₀/2 - I₀/4 = I₀/4 So the answer is (c) I₀/4.

In a Young's double slit experiment, three polarizers are kept as… | JEE Main 2025 (24 Jan, Shift 2)

Q26JEE Main 2025 · 24 JAN · SHIFT 2Difficult

Polarisation

In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of $P_1$ and $P_2$ are orthogonal to each other. The polarizer $P_3$ covers both the slits with its transmission axis at $45^\circ$ to those of $P_1$ and $P_2$ . An unpolarized light of wavelength $\lambda$ and intensity $I_0$ is incident on $P_1$ and $P_2$ . The intensity at a point after $P_3$ where the path difference between the light waves from $s_1$ and $s_2$ is $\frac{\lambda}{3}$ , is

Show answer and solution

Answer: (c)

After $P_1$ and $P_2$ each slit gets intensity $I_0/2$ , polarised in perpendicular directions (they cannot interfere yet).
$P_3$ at $45^\circ$ to both: each beam becomes $\frac{I_0}{2}\cos^2 45^\circ = \frac{I_0}{4}$ , now with the same polarisation.
Path difference $\lambda/3 \Rightarrow \phi = \frac{2\pi}{3}$ .
$I = \frac{I_0}{4} + \frac{I_0}{4} + 2\cdot\frac{I_0}{4}\cos\frac{2\pi}{3} = \frac{I_0}{2} - \frac{I_0}{4} = \frac{I_0}{4}$
So the answer is (c) $\frac{I_0}{4}$ .

Report a mistake

Polarisation question from JEE Main 2025 (24 Jan, Shift 2)

More questions on Polarisation