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Question
In Young's double slit experiment, the distance of the 4th bright fringe from the centre of the interference pattern is 1.5 mm. The distance between the slits and the screen is 1.5 m, and the wavelength of light used is 500 nm. Calculate the distance between the two slits.
Solution
Given that D = 1.5 m, λ = 500 nm = 500 × 10-9 m, n = 4, d = ?
In, YDSE, the formula for bright fringes (constructive interference) is
`"Y"_"n" = ("n"lambda"D")/"d"`
`Delta"Y" = (4 xx 500 xx 10^-9 xx 1.5)/"d" = 1.5 xx 10^-3`
⇒ `"d" = (4 xx 50 xx 10^-9 xx 1.5)/(1.5 xx 10^-3)`
⇒ d = 200 × 10-6 m
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