Question

In young’s double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence, deduce the expression for the fringe width.

Answer 1

Young’s double slit experiment demonstrated the phenomenon of interference of light. Consider two fine slits S₁ and S₂ at a small distance d apart. Let the slits be illuminated by a monochromatic source of light of wavelength λ. Let GG′ be a screen kept at a distance D from the slits. The two waves emanating from slits S₁ and S₂ superimpose on each other resulting in the formation of an interference pattern on the screen placed parallel to the slits.

Let O be the centre of the distance between the slits. The intensity of light at a point on the screen will depend on the path difference between the two waves reaching that point. Consider an arbitrary point P at a distance x from O on the screen.

Path difference between two waves at P = S₂P − S₁P

The intensity at the point P is maximum or minimum as the path difference is an integral multiple of wavelength or an odd integral multiple of half wavelength

For the point P to correspond to maxima, we must have

S₂P − S₁P = n, n = 0, 1, 2, 3...

From the figure given above

`(S_2P)^2-(S_1P)^2=D^2+(x+d/2)^2-D^2+(x-d/2)^2`

On solving we get:

(S₂P)²-(S₁P)²=2xd

`S_2P-S_1P=(2xd)/(S_2P+S_1P)`

As d<<D, then S₂P + S₂P = 2D  (∵ S₁P = S₂P ≡ D when d<<D)

`:.S_2P-S_1P=(2xd)/(2D)=(xd)/D`

Path difference, `S_2P-S_1P=(xd)/D`

Hence, when constructive interfernce occur, bright region is formed.

For maxima or bright fringe, path difference = `xd/D=nlambda`

i.e `x=(nlambdaD)/d`

 where n=0,± 1, ±2,........

During destructive interference, dark fringes are formed:

Path difference, `(xd)/D=(n+1/2)lambda`

`x=(n+1/2)(lambdaD)/d`

The dark fringe and the bright fringe are equally spaced and the distance between consecutive bright and dark fringe is given by:

β = x_n+1-x_n

`beta=((n+1)lambdaD)/d-(nlambdaD)/d`

`beta=(lambdaD)/d`

Hence the fringe width is given by `beta = (lambdaD)/d`