Question:medium

In a Young’s double slit experiment, the intensity at a point is \[ \left(\frac{1}{4}\right)^{\text{th}} \] of the maximum intensity, the minimum distance of the point from the central maximum is ____ $\mu$m.
(Given: $\lambda = 600 \, \text{nm}, \, d = 1.0 \, \text{mm}, \, D = 1.0 \, \text{m}$)

Updated On: Jan 13, 2026
Show Solution

Correct Answer: 200

Solution and Explanation

In Young's double-slit experiment, the intensity \( I \) at a point is described by \( I = I_0 \cos^2(\frac{\pi d \sin \theta}{\lambda}) \), where \( I_0 \) is the maximum intensity, \( d \) is the slit separation, \( \theta \) is the angle of the point from the center, and \( \lambda \) is the wavelength.

  • \( I_0 \) denotes maximum intensity.
  • \( d \) represents the slit separation distance.
  • \( \theta \) is the angular position of the point relative to the center.
  • \( \lambda \) is the wavelength of the light.

Given that \( I = \left(\frac{1}{4}\right) I_0 \), we find that:
\[ \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right) = \frac{1}{4} \] 
This simplifies to: 
\[ \cos\left(\frac{\pi d \sin \theta}{\lambda}\right) = \frac{1}{2} \] 
Consequently: 
\[ \frac{\pi d \sin \theta}{\lambda} = \frac{\pi}{3} \text{ or } \frac{2\pi}{3} \] 
Solving for \( \sin \theta \) yields: 
\[ \sin \theta = \frac{\lambda}{3d} \text{ (for } \frac{\pi}{3}) \text{ or } \sin \theta = -\frac{\lambda}{3d} \text{ (for } \frac{2\pi}{3}) \] 
Applying the small angle approximation (\( \sin \theta \approx \tan \theta \approx \frac{x}{D} \), where \( x \) is the distance from the central maximum), we get: 
\[ \frac{x}{D} = \frac{\lambda}{3d} \] 
Therefore, the distance \( x \) is given by: 
\[ x = \frac{\lambda D}{3d} \] 
Substituting the given values \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \), \( d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \), and \( D = 1.0 \, \text{m} \): 
\[ x = \frac{600 \times 10^{-9} \times 1.0}{3 \times 1.0 \times 10^{-3}} \] 
This results in: 
\[ x = 200 \times 10^{-6} \, \text{m} = 200 \, \mu\text{m} \] 
The minimum distance calculated is \( 200 \, \mu\text{m} \), which falls within the specified range (200,200).

Was this answer helpful?
0


Questions Asked in JEE Main exam