In Young's double slit experiment, what is the path difference Delta x for constructive interference (bright fringe)?
Answer and explanation
Correct answer: A. n lambda
- Step 1: Constructive interference occurs when waves arrive in phase. Step 2. The path difference must be an integer multiple of the wavelength: Delta x = n lambda (where n = 0, 1, 2, ...). Answer: n lambda. Why not the others: (2n + 1) lambda / 2 is the condition for destructive interference (dark fringe); (n + 1/2) lambda is dark fringe; (2n - 1) lambda is odd wavelengths.
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