YOUNG’S DOUBLE SLIT EXPERIMENT

A train of plane light waves is incident on a barrier containing two narrow slits separated by a distance’d’. The widths of the slits are small compared with wavelength of the light used, so that interference occurs in the region where the light from S

_{1}overlaps that from S_{2}.

A series of alternately bright and dark bands can be observed on a screen placed in this region of overlap.

The variation in light intensity along the screen near the centre O shown in the figure

Now consider a point P on the screen. The phase difference between the waves at P is θ, where

θ= 2π/λ ΔP

_{o}(where ΔP

_{o}is optical path difference, ΔP_{o}=ΔP_{g}; ΔP_{g }being the geometrical path difference.)

= 2π/λ [ S

_{2}P - S_{1}P ] (here λ = 1 in air)As As, D >> d,S

_{2}P - S_{1}P ≈ λ d sinθsin θλ ≈ tanθ( = y/D).

[for very small θ]

Thus, θ = 2π/λ (dy/D)

For constructive interference,θ = 2nλ (n = 0, 1, 2...)

⇒ 2π/λ (dy/D) = 2nπ ⇒ y = n λD/d

Similarly for destructive interference,y = (2n - 1) λD/2d (n = 1, 2 ...)

Fringe Width WIt is the separation of two consecutive maxima or two consecutive minima.

Near the centre O [where θ is very small],W = y

_{n+1}– y_{n}[y_{n}gives the position of nth maxima on screen]= λD/d

Intensity Variation on ScreenIf A and I

_{o}represent amplitude of each wave and the associated intensity on screen, then, the resultant intensity at a point on the screen corresponding to the angular position θ as in above figure, is given by

I = I

_{o} + I_{o}+ 2√I_{o}^{2}cosθ,When θ = 2π(dsinθ)/ λ = 4I

_{o}cos^{2}Φ/2

Illustration 1:A beam of light consisting of two wavelengths 6500

^{o}A and 5200^{o}A is used to obtain interference fringes in YDE. The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm.(a) Find the distance of the third bright fringe on the screen from the central maxima for the wavelength 6500

^{o}A.(b) What is the least distance from the central maxima where the bright fringes due to both the wavelengths coincide?

Solution:(i) y

_{3}= n. Dλ/d = 3 x 1.2m x 6500 x 10^{-10}m / 2 x 10^{-3}m = 0.12cm

Let nth maxima of light with wavelength 6500 Å coincides with that of m

^{th}maxima of 5200Å.(ii) m x 6500A

^{o}x D/d = n x 5200A^{o}x D/d ⇒ m/n = 5200/6500 = 4/5Least distance = y

_{4}= 4.D (6500A^{o})/d = 4 x 6500 x 10^{-10}x 1.2/ 2 x 10^{-3}m = 0.16cm