Question

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]

Answer 1

path difference , ∆x = $S_1P-S_2P$
∆x = $\sqrt{D^2+d^2}-D$
∆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
$\Delta x = \frac{(2n-1)\lambda}{2}$
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