In a Young double slit… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Optics

2026mcqmedium

In a Young double slit experiment, the wavelength of incident light is $6000$ Å, the separation between slits $S_1$ and $S_2$ is $5$ cm and the distance between slits plane and screen is $50$ cm, as shown in the figure below. If the resultant intensity at $P$ is equal to the intensity due to individual slits, the path difference between interfering waves is __________ Å.

JEE Main 2026 Physics, Wave Optics — question figure

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$4000$
B
$3000$
C
$2000$
D
$1000$

Did you get this right?

Solution

Let the intensity due to each individual slit be $I_0$ .

The resultant intensity $I_R$ at a point where the phase difference is $\phi$ is given by:

$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$

Given that $I_1 = I_2 = I_0$ and the resultant intensity $I_R = I_0$ , we can substitute these values into the equation:

$I_0 = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos \phi$

$I_0 = 2I_0 + 2I_0 \cos \phi$

$I_0 = 2I_0 (1 + \cos \phi)$

$1 + \cos \phi = \dfrac{1}{2}$

$\cos \phi = -\dfrac{1}{2}$

The minimum phase difference satisfying this condition is:

$\phi = \dfrac{2\pi}{3}$

The relationship between phase difference $\phi$ and path difference $\Delta x$ is:

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

Substituting the value of $\phi$ :

$\dfrac{2\pi}{3} = \dfrac{2\pi}{\lambda} \Delta x$

$\Delta x = \dfrac{\lambda}{3}$

Given the wavelength $\lambda = 6000 \text{ \AA}$ :

$\Delta x = \dfrac{6000}{3} = 2000 \text{ \AA}$

Answer: $2000$

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Solution

Let the intensity due to each individual slit be $I_0$ .

The resultant intensity $I_R$ at a point where the phase difference is $\phi$ is given by:

$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$

Given that $I_1 = I_2 = I_0$ and the resultant intensity $I_R = I_0$ , we can substitute these values into the equation:

$I_0 = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos \phi$

$I_0 = 2I_0 + 2I_0 \cos \phi$

$I_0 = 2I_0 (1 + \cos \phi)$

$1 + \cos \phi = \dfrac{1}{2}$

$\cos \phi = -\dfrac{1}{2}$

The minimum phase difference satisfying this condition is:

$\phi = \dfrac{2\pi}{3}$

The relationship between phase difference $\phi$ and path difference $\Delta x$ is:

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

Substituting the value of $\phi$ :

$\dfrac{2\pi}{3} = \dfrac{2\pi}{\lambda} \Delta x$

$\Delta x = \dfrac{\lambda}{3}$

Given the wavelength $\lambda = 6000 \text{ \AA}$ :

$\Delta x = \dfrac{6000}{3} = 2000 \text{ \AA}$

Answer: $2000$

Options

Did you get this right?

Solution

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JEE Main Optics in other years

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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