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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.
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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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