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Question
A slit of width 0.6 mm is illuminated by a beam of light consisting of two wavelengths 600 nm and 480 nm. The diffraction pattern is observed on a screen 1.0 m from the slit. Find:
- The distance of the second bright fringe from the central maximum pertaining to the light of 600 nm.
- The least distance from the central maximum at which bright fringes due to both wavelengths coincide.
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Solution
(i) Distance of 2nd bright fringe from central maximum = `(2λ"D")/"d"`
= `(2 xx 600 xx 10^-19 xx 1)/(0.6 xx 10^-3)`
= 20 × 10−4 m
(ii) If nth bright fringe due to 600 nm coincides with (n + 1)th bright fringe due to 480 nm, then
`("n"λ_1"D")/"d" = (("n" - 1)λ_2"D")/"d"`
Or, nλ1D = (n + 1)λ2D
Or, `"n"/(("n" - 1)) = λ_2/λ_1`
Or, `"n"/(("n" - 1)) = 600/480`
`"n" xx 480 = 600 ("n" - 1)`
`480 "n" = 600 "n" - 600`
600 = `600 "n" - 480 "n"`
600 = `600 "n" - 480 "n"`
600 = `120 "n"`
∴ n = `600/120`
n = 5
So, the least distance from the central maximum = `(5 xx 480 xx 10^-9 xx 1)/(0.6 xx 10^-3)`
= 4 × 10−3 m
= 4 mm
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