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In a Young’s double slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference
- 11 λ/2
- 5 λ/2
- λ/2
- 9 λ/2
Correct answer: 9 λ/2
Solution
In a Young's double-slit experiment, the condition for a minimum is that the path difference is an odd multiple of λ/2. For the fifth minimum, the path difference is (2n+1)λ/2, where n=4 (since n starts from 0). Substituting n=4 gives (2×4+1)λ/2 = 9λ/2.
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