Question:
In a Young's double slit experiment with light of wavelength \(\lambda\) the separation of slits is \(d\) and distance of screen is \(D\) such that \(D \gt\gt d \gt\gt \lambda\). If the Fringe width is \(\beta\), the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is :
In a Young's double slit experiment with light of wavelength \(\lambda\) the separation of slits is \(d\) and distance of screen is \(D\) such that \(D \gt\gt d \gt\gt \lambda\). If the Fringe width is \(\beta\), the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is :
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Use \(I = I_{max}\cos^2(\phi/2)\); half intensity means \(\phi = \pi/2\), i.e. path difference \(\lambda/4\), and the fringe width is \(\beta = \lambda D/d\).
Updated On: Nov 11, 2026
- \(\frac{\beta}{2}\)
- \(\frac{\beta}{4}\)
- \(\frac{\beta}{3}\)
- \(\frac{\beta}{6}\)
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The Correct Option is B
Solution and Explanation
Intensity at a point with phase difference \(\phi\): \(I = I_{max}\cos^2\left(\frac{\phi}{2}\right)\).
For \(I = \frac{I_{max}}{2}\): \(\cos^2(\phi/2) = \frac12\), so \(\phi/2 = \frac{\pi}{4}\) and \(\phi = \frac{\pi}{2}\).
Path difference \(\Delta = \frac{\lambda}{2\pi}\phi = \frac{\lambda}{4}\).
For a point at height y, \(\Delta = \frac{yd}{D}\), so \(y = \frac{\lambda D}{4d}\).
Fringe width \(\beta = \frac{\lambda D}{d}\), hence \(y = \frac{\beta}{4}\).
Answer: (2) \(\frac{\beta}{4}\)
For \(I = \frac{I_{max}}{2}\): \(\cos^2(\phi/2) = \frac12\), so \(\phi/2 = \frac{\pi}{4}\) and \(\phi = \frac{\pi}{2}\).
Path difference \(\Delta = \frac{\lambda}{2\pi}\phi = \frac{\lambda}{4}\).
For a point at height y, \(\Delta = \frac{yd}{D}\), so \(y = \frac{\lambda D}{4d}\).
Fringe width \(\beta = \frac{\lambda D}{d}\), hence \(y = \frac{\beta}{4}\).
Answer: (2) \(\frac{\beta}{4}\)
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