The maximum intensity in a… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Optics

2026mcqmedium

The maximum intensity in a Young's double slit experiment is $I_0$ . Distance between the slits ( $d$ ) is $5\lambda$ , where $\lambda$ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at $D = 10d$ is _______.

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{I_0}{4}$
B
$\dfrac{I_0}{2}$
C
$I_0$
D
$\dfrac{3I_0}{4}$

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Solution

The position of the point on the screen exactly opposite to one of the slits is at a distance $y = \dfrac{d}{2}$ from the central maximum.

The path difference $\Delta x$ at this point is given by:

$\Delta x = \dfrac{yd}{D}$

Substituting $y = \dfrac{d}{2}$ and $D = 10d$ :

$\Delta x = \dfrac{\left(\dfrac{d}{2}\right)d}{10d} = \dfrac{d^2}{20d} = \dfrac{d}{20}$

Given that the distance between the slits is $d = 5\lambda$ , we have:

$\Delta x = \dfrac{5\lambda}{20} = \dfrac{\lambda}{4}$

The corresponding phase difference $\Delta \phi$ is:

$\Delta \phi = \dfrac{2\pi}{\lambda} \Delta x = \dfrac{2\pi}{\lambda} \left(\dfrac{\lambda}{4}\right) = \dfrac{\pi}{2}$

The intensity at this point is given by:

$I = I_0 \cos^2\left(\dfrac{\Delta \phi}{2}\right)$

Substituting $\Delta \phi = \dfrac{\pi}{2}$ :

$I = I_0 \cos^2\left(\dfrac{\pi}{4}\right) = I_0 \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_0}{2}$

Answer: $\dfrac{I_0}{2}$

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Solution

The position of the point on the screen exactly opposite to one of the slits is at a distance $y = \dfrac{d}{2}$ from the central maximum.

The path difference $\Delta x$ at this point is given by:

$\Delta x = \dfrac{yd}{D}$

Substituting $y = \dfrac{d}{2}$ and $D = 10d$ :

$\Delta x = \dfrac{\left(\dfrac{d}{2}\right)d}{10d} = \dfrac{d^2}{20d} = \dfrac{d}{20}$

Given that the distance between the slits is $d = 5\lambda$ , we have:

$\Delta x = \dfrac{5\lambda}{20} = \dfrac{\lambda}{4}$

The corresponding phase difference $\Delta \phi$ is:

$\Delta \phi = \dfrac{2\pi}{\lambda} \Delta x = \dfrac{2\pi}{\lambda} \left(\dfrac{\lambda}{4}\right) = \dfrac{\pi}{2}$

The intensity at this point is given by:

$I = I_0 \cos^2\left(\dfrac{\Delta \phi}{2}\right)$

Substituting $\Delta \phi = \dfrac{\pi}{2}$ :

$I = I_0 \cos^2\left(\dfrac{\pi}{4}\right) = I_0 \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_0}{2}$

Answer: $\dfrac{I_0}{2}$

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Options

Did you get this right?

Solution

Your note

JEE Main Optics in other years

More JEE Main Optics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

Keep practicing Optics