In the Young's double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is

(2n + 1) λ / 2
(2n + 1) λ / 4
(2n + 1) λ / 8
(2n + 1) λ / 16

Correct answer: (2n + 1) λ / 4

Solution

Half the peak intensity corresponds to a phase difference of π/2, which translates to a path difference of (2n + 1)λ/4. This is the condition for the given intensity.

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