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In Young's double-slit experiment with monochromatic light of wavelength lambda, the path difference (expressed using an integer n) at a point where the intensity is half the peak intensity is:

1. (2n+1)*lambda/2
2. (2n+1)*lambda/4
3. (2n+1)*lambda/8
4. (2n+1)*lambda/16

Correct answer: (2n+1)*lambda/4

Solution

I = I_max*cos²(phi/2). Half intensity: cos²(phi/2) = 1/2 -> phi/2 = (2n+1)*pi/4 -> phi = (2n+1)*pi/2. Path difference dx = (lambda/(2*pi))*phi = (lambda/(2*pi))*(2n+1)*pi/2 = (2n+1)*lambda/4.

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