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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. - Physics

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प्रश्न

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:

  1. The distance of the second bright fringe from the central maximum pertaining to the light of 600 nm.
  2. The least distance from the central maximum at which bright fringes due to both wavelengths coincide.
योग
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उत्तर

(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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2021-2022 (March) Term 2 - Delhi Set 1

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