In a Young's double-slit experiment, the two slits are 2 mm apart from the screen is positioned 140 cm away from the plane of the slits. The slits are illuminated with light of wavelength 600 nm. Find the distance of the third fringe from the central maximum in the interference pattern obtained on the screen. If the wavelength  of the incident  light were changed to 480 nm, find the shift in the position of the third bright fringe from the central maximum.
[Ans. 1.26 mm, 0.25 mm] 

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Distance between two slits :-2 x 10^-3m Distance between slit and screen :-1.4m Wavelength:-6 x 10^-7m Distance of third bright fringe:- 3 x 6 x 10^-7 x1.4/2*10^-3=12.6 x 10^-4m
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