Obtain the expression for fringe width in the case of interference of light waves.

Explanation

Let A and B be two slits separated by a distance .d.. Let .`lambda`. be the wavelength of light . Let .D. be the distance between the screen and the double slit.
Let .C. be a point on BP such that `AP~~CP`. Path difference between the two waves reaching .P. is given by BP-AP=BC=d
From the D BFP , `BP^2 =BF^2 +FP^2` and from the `D AEP , AP^2 = AE^2 +EP^2`
`BP^2 - AP^2 =(D^2 + FP^2) -(D^2 - EP^2) `
i.e., `BP^2 -AP^2 =FP^2 -EP^2 `
`=(x+d/2)^2 - (x-d/2)^2 `
(BP-AP)(BP+AP)=`2(2x.d/2)`
For a point .P. close to .O.,
BP > > AP=D
(BP-AP)(2D)=2x.d
But (BP-AP)=BC=d
i.e., `d=(xd)/D`
or `x=(deltaD)/d`
For a constructive interference `d=nlambda`
Distance of `n^(th)` bright fringe from the central bright fringe `x_n=n(lambdaD)/d`
Distance of `(n+1)^(th)` bright fringe from the central bright fringe `x_(n+1)=(n+1)(lambdaD)/d`
By definition, Fringe width is the distance between two consecutive bright or dark fringes .
`x_(n+1)-x_(n)=b=lambda/d(n+1-n)`
i.e, `beta=(lambdaD)/d`
Fringe width `beta prop D , beta prop lambda` and `beta prop 1/d`.
We can show that the fringe width between any two dark fringes is also `beta=(lambdaD)/d` .