Question:
In a Young's double slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is :
In a Young's double slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is :
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Compare the width of the central diffraction maximum (\(2\lambda D/a\)) with the fringe width (\(\lambda D/d\)).
Updated On: Nov 11, 2026
- 3
- 6
- 12
- 24
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The Correct Option is C
Solution and Explanation
Let slit width = a and slit separation d = 6.1a.
Central diffraction maximum extends over \(\sin\theta = \pm\lambda/a\), i.e. width \(2\lambda D/a\) on the screen.
Interference fringe width \(\beta_f = \frac{\lambda D}{d}\).
Number of fringes in the central maximum \(= \frac{2\lambda D/a}{\lambda D/d} = \frac{2d}{a} = 2\times6.1 = 12.2 \approx 12\).
Answer: (3) 12
Central diffraction maximum extends over \(\sin\theta = \pm\lambda/a\), i.e. width \(2\lambda D/a\) on the screen.
Interference fringe width \(\beta_f = \frac{\lambda D}{d}\).
Number of fringes in the central maximum \(= \frac{2\lambda D/a}{\lambda D/d} = \frac{2d}{a} = 2\times6.1 = 12.2 \approx 12\).
Answer: (3) 12
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