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 S1 overlaps that from S2.

1915_Deubleslit.JPG

 

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

226_fringe-width.JPG

 

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.)

 1679_YOUNG’S DOUBLE SLIT EXPERIMENT.JPG

 

= 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

 

 2204_Intensity.JPG

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?

 

Solution:

(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

 

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