In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ are orthogonal to each other. The polarizer $\mathrm{P}_{3}$ covers both the slits with its transmission axis at $45^{\circ}$ to those of $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$. An unpolarized light of wavelength $\lambda$ and intensity $\mathrm{I}_{0}$ is incident on $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$. The intensity at a point after $P_{3}$ where the path difference between the light waves from $\mathrm{s}_{1}$ and $\mathrm{s}_{2}$ is $\frac{\lambda}{3}$, is

Wave Optics — JEE Main · 2025 · 24 Jan 2025 · Shift 2 Question with Solution

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

4 5^{\circ}

to those of

P_{1}

and

P_{2}

. An unpolarized light of wavelength

λ

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}

\frac{λ}{3}

, is

Answer: (B)

Unpolarized light of intensity

I_{0}

falls on

P_{1}

(covering slit

S_{1}

) and on

P_{2}

(covering slit

S_{2}

), with

P_{1} ⊥ P_{2}

. Passing unpolarized light through a polarizer always halves its intensity, so the light reaching each slit has intensity

I_{1} = I_{2} = \frac{I _{0}}{2},

and the two beams are polarized along mutually perpendicular directions (along

P_{1}

and

P_{2}

respectively).

Both beams then pass through

P_{3}

, whose transmission axis is at

4 5^{\circ}

to both

P_{1}

and

P_{2}

. By Malus's law, each beam is attenuated by

cos^{2} (4 5^{\circ}) = \frac{1}{2}

I_{1}^{'} = I_{2}^{'} = \frac{I _{0}}{2} \times \frac{1}{2} = \frac{I _{0}}{4} .

After

P_{3}

both beams share the same polarization direction, so they can now interfere at the screen. At the point where the path difference is

λ /3

, the phase difference is

Δ ϕ = \frac{2 π}{λ} \cdot \frac{λ}{3} = \frac{2 π}{3} .

With equal component intensities

I_{1}^{'} = I_{2}^{'} = I_{0} /4

, the resultant intensity is

I_{r es} = I_{1}^{'} + I_{2}^{'} + 2 I_{1}^{'} I_{2}^{'} cos Δ ϕ = \frac{I _{0}}{4} + \frac{I _{0}}{4} + 2 (\frac{I _{0}}{4}) cos \frac{2 π}{3} .

Since

cos (2 π /3) = - \frac{1}{2}

I_{r es} = \frac{I _{0}}{2} + \frac{I _{0}}{2} (- \frac{1}{2}) = \frac{I _{0}}{2} - \frac{I _{0}}{4} = \frac{I _{0}}{4} .

The intensity at that point is

\frac{I _{0}}{4}

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option