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 S1 overlaps that from S2.


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 ΔPo

(where ΔPo is optical path difference, ΔPo=ΔPg; ΔPg  being the geometrical path difference.)


= 2π/λ  [ S2P - S1P ] (here λ = 1 in air)

As     As, D >> d,

S2P - S1P ≈  λ d sinθ

sin 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 W 

It is the separation of two consecutive maxima or two consecutive minima.

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

W = yn+1 – yn [yn gives the position of nth maxima on screen]

 = λD/d

Intensity Variation on Screen

If A and Io 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 = Io­ + Io + 2√Io2 cosθ,

When θ = 2π(dsinθ)/ λ = 4Io cos2 Φ/2

Illustration 1:

A beam of light consisting of two wavelengths 6500 oA and 5200 oA 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 oA.

(b)    What is the least distance from the central maxima where the bright fringes due to both the wavelengths coincide?


(i) y3 = n. Dλ/d = 3 x 1.2m x 6500 x 10-10m / 2 x 10-3m  = 0.12cm

Let nth maxima of light with wavelength 6500 Å coincides with that of mth maxima of 5200Å.

(ii) m x 6500Ao x D/d = n x 5200Ao x D/d ⇒ m/n = 5200/6500 = 4/5

Least distance = y4 = 4.D (6500Ao)/d = 4 x 6500 x 10-10 x 1.2/ 2 x 10-3m = 0.16cm


Contact askiitians experts to get answers to your queries by filling up the form given below:

We promise that your information will be our little secret. To know more please see our Privacy Policy
We promise that your information will be our little secret. To know more please see our Privacy Policy


Sign Up with Facebook

Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”

Related Resources
Diffraction and Polarization

DIFFRACTION a) Diffraction is the bending or...

Wave Nature of Light

WAVE NATURE OF LIGHT Huygens’ Wave Theory...

Polarisation and Brewsters Law

POLARISATION Polarisation of two interfering wave...

Displacement of Fringes in Youngs Double Slit Experiment

Displacement of Fringes When a film of...

Questions and Solutions part 2

Questions and Solutions part 2 11. Angular width...

Questions and Solutions part 3

Questions and Solutions part 3 1. When light is...

Interference by Thin Film

INTERFERENCE BY THIN FILM A ray of light incident...

Questions and Solutions part 1

Questions and Solutions Part 1 1. In Young's...