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The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit).

1. 25 μm
2. 50 μm
3. 75 μm
4. 100 μm

Correct answer: 25 μm

Solution

Central-max angular width 60 deg means half-angle 30 deg, so lambda = a*sin30 = 1 um * 0.5 = 0.5 um. Fringe width beta = lambda*D/d, so d = lambda*D/beta = (0.5e-6 m * 0.5 m)/(0.01 m) = 25e-6 m = 25 um.

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