White light is used in Young's double slit experiment, as shown in the figure. At a point on the screen directly in front of slit S₂,certain Wavelengths are producing destructive interference (i.e. they are missing in the diffraction pattern). Find these wavelengths, corresponding to first and second order diffraction. Here, d << D. [Ans: (i) d²/D, n=1, (ii) d²/3D, n=2]
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path difference , ∆x =
∆x =
∆x = D√{1 + d²/D^2} - D
∆x = D(1 + d²/D²)^½ - D
use binomial expansion ,
here, 1 >> d²/D²
so, (1 + d²/D²)^½ = 1 + d²/2D²
now, ∆x = D(1 + d²/2D²) - D
∆x = D + d²/2D - D = d²/2D
we know, path difference of distractive interference is given by
where n is positive integers
now, (2n - 1)λ/2 = d²/2D
(2n - 1)λ = d²/D
λ = d²/(2n - 1)D
hence, wavelength corresponding to first order is λ = d²/(2 - 1)D = d²/D
wavelength corresponding to 2nd order is
λ = d²/(2×2-1)D = d²/3D
∆x =
∆x = D√{1 + d²/D^2} - D
∆x = D(1 + d²/D²)^½ - D
use binomial expansion ,
here, 1 >> d²/D²
so, (1 + d²/D²)^½ = 1 + d²/2D²
now, ∆x = D(1 + d²/2D²) - D
∆x = D + d²/2D - D = d²/2D
we know, path difference of distractive interference is given by
where n is positive integers
now, (2n - 1)λ/2 = d²/2D
(2n - 1)λ = d²/D
λ = d²/(2n - 1)D
hence, wavelength corresponding to first order is λ = d²/(2 - 1)D = d²/D
wavelength corresponding to 2nd order is
λ = d²/(2×2-1)D = d²/3D
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