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Question
In Young's double slit experiment using light of wavelength 600 nm, the slit separation is 0.8 mm and the screen is kept 1.6 m from the plane of the slits. Calculate
- the fringe width
- the distance of (a) third minimum and (b) fifth maximum, from the central maximum.
Solution
(i) Fringe width = β = `(λ"D")/"d"`
β = `(600 xx 10^-9 xx 1.6)/(0.8 xx 10^-3)`
∴ β = 12 × 10−4 m
(ii)
- Distance of 3rd minimum from the central maximum = `5/2`β = 30 × 10−4 m
- Distance of 5th maximum from central fringe = 5β = 60 × 10−4 m
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ASSERTION (A): In an interference pattern observed in Young's double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.
REASON (R): Fringe width is proportional to (d/D).
In Young's double slit experiment the two slits are 0.6 mm distance apart. Interference pattern is observed on a screen at a distance 80 cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be ______ nm.
