Question:

In the Young's double slit experiment a monochromatic source of wavelength $\lambda$ is used. The intensity of light passing through each slit is $I _{0}$. The intensity of light reaching the screen $S _{ C }$ at a point $P$, a distance $X$ from $O$ is given by (Take $d < < D$ )

Updated On: Nov 14, 2025

$I _{ o } \cos ^{2}\left(\frac{\pi D }{\lambda d } x \right)$
$4 I _{ o } \cos ^{2}\left(\frac{\pi d }{\lambda D } x \right)$
$I_{o} \sin ^{2}\left(\frac{\pi d}{2 \lambda D} x\right)$
$4 I _{ o } \cos \left(\frac{\pi d }{2 \lambda D } x \right)$

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The Correct Option is B

Solution and Explanation

To solve the given problem on the Young's double slit experiment, we need to derive the expression for the intensity of light at a point on the screen based on the intensity through each slit and the interference pattern created. Let's break down the steps involved:

Conceptual Background:

In Young's double slit experiment, two coherent light sources interfere, producing an interference pattern of bright and dark fringes. The intensity at any point on the screen depends on the path difference between the light waves originating from the two slits and reaching that point.

Interference Formula:

The intensity of light at a point on the screen can be given as:

$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos(\phi)$

Where:

$I_1$ and $I_2$ are the intensities from each slit (here both are $I_0$).
$\phi$ is the phase difference between the two waves.

Phase Difference:

The phase difference $\phi$ is given by:

$\phi = \frac{2\pi}{\lambda}\Delta x$

Where:

$\lambda$ is the wavelength of light.
$\Delta x$ is the path difference between the waves from the two slits.

For a point located at a distance $X$ from the central maximum $O$ on the screen:

$\Delta x = \frac{d X}{D}$

where $d$ is the distance between the slits and $D$ is the distance from the slits to the screen.

Intensity Calculation:

Substitute the values in the intensity formula:

$I = 2I_0 + 2I_0\cos\left(\frac{2\pi d X}{\lambda D}\right)$

Simplify to:

$I = 4I_0 \cos^2\left(\frac{\pi d X}{\lambda D}\right)$

Conclusion:

Thus, the correct expression for the intensity at point $P$ is $4 I _{ o } \cos ^{2}\left(\frac{\pi d }{\lambda D } x \right)$, which matches the option:

$4 I _{ o } \cos ^{2}\left(\frac{\pi d }{\lambda D } x \right)$

This derivation is in line with the principle of superposition and the conditions given (i.e., $d << D$).

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Concepts Used:

Young’s Double Slit Experiment

Considering two waves interfering at point P, having different distances. Consider a monochromatic light source ‘S’ kept at a relevant distance from two slits namely S₁ and S₂. S is at equal distance from S₁ and S₂. SO, we can assume that S₁ and S₂ are two coherent sources derived from S.
The light passes through these slits and falls on the screen that is kept at the distance D from both the slits S₁ and S₂. It is considered that d is the separation between both the slits. The S₁ is opened, S₂ is closed and the screen opposite to the S₁ is closed, but the screen opposite to S₂ is illuminating.
Thus, an interference pattern takes place when both the slits S₁ and S₂ are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged, to reach point P the light waves from slits S₁ and S₂ must travel at different distances. It implies that there is a path difference in the Young double-slit experiment between the two slits S₁ and S₂.