#### 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.

#### Solution

Young’s double slit experiment demonstrated the phenomenon of interference of light. Consider two fine slits S_{1} and S_{2} 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 S_{1} and S_{2} 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 = S_{2}P − S_{1}P

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

S_{2}P − S_{1}P = 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:

(S_{2}P)^{2}-(S_{1}P)^{2}=2xd

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

As d<<D, then S_{2}P + S_{2}P = 2D (∵ S_{1}P = S_{2}P ≡ 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`