Question:
In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is \(\frac{1}{8}\) th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to :
In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is \(\frac{1}{8}\) th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to :
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Phase difference \(\phi = \frac{2\pi}{\lambda}\Delta x\); intensity \(I = I_0\cos^2(\phi/2)\) where \(I_0\) is the maximum intensity.
Updated On: Nov 5, 2026
- 0.94
- 0.85
- 0.74
- 0.80
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The Correct Option is B
Solution and Explanation
Path difference \(\Delta x = \lambda/8\), so phase difference \(\phi = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{8} = \frac{\pi}{4}\).
For two coherent sources of equal intensity, \(I = I_{max}\cos^2\left(\frac{\phi}{2}\right)\).
\(\frac{I}{I_{max}} = \cos^2\left(\frac{\pi}{8}\right) = \cos^2(22.5^\circ) = (0.9239)^2 = 0.854\).
This is close to 0.85.
Answer: (2) 0.85
For two coherent sources of equal intensity, \(I = I_{max}\cos^2\left(\frac{\phi}{2}\right)\).
\(\frac{I}{I_{max}} = \cos^2\left(\frac{\pi}{8}\right) = \cos^2(22.5^\circ) = (0.9239)^2 = 0.854\).
This is close to 0.85.
Answer: (2) 0.85
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