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