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In the Fraunhofer single-slit diffraction pattern (normal incidence), what is the phase difference, in radians, between the wavelets coming from the two opposite edges of the slit at the angular position of the first minimum?

1. pi
2. 2*pi
3. pi/4
4. pi/2

Correct answer: 2*pi

Solution

The condition for the first single-slit diffraction minimum is a*sin(theta) = lambda, i.e. the path difference between waves from the two edges of the slit equals one wavelength. The corresponding phase difference is (2*pi/lambda) * lambda = 2*pi radians.

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