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Question:

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

100µm

50µm

75µm

25µm

Solution:

Angular width of the central maxima θc=60°
Half angle θ=60°/2=30°
Given : b=1µm
Using sinθ=λ/b
Or sin30°=λ/1µm ⇒ λ=0.5×1µm
Given : D=50 cm=0.5 m
Fringe width β=1 cm=0.01 m
Using β=λD/d
∴0.01=0.5×10⁻⁶×0.5/d
0.01=25µm