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In single-slit Fraunhofer diffraction, let I0 be the intensity at the centre of the central maximum. Find the ratio I/I0 at a point on the screen for which the path difference between waves from the two edges of the slit is (3/4) of a wavelength.

1. 0.09
2. 0.045
3. 0.5
4. 0.81

Correct answer: 0.09

Solution

Define alpha = pi*(path difference)/lambda = pi*(3/4) = 3pi/4. The single-slit intensity ratio is (sin alpha / alpha)². With sin(3pi/4) = 0.7071 and alpha = 2.356, the ratio = (0.7071/2.356)² = (0.300)² = 0.090.

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