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=delta.`
`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)/(x))`
For a point 'P' close to 'O' , `BP~~ AP=D`
`(BP-AP)(2D)2x.d.` But `(BP-AP)=BC=delta`
i.e. , `delta =(xd)/(D)`

or `x=(deltaD)/(d)`
For a constructive interference `delta=n lambda`.
Distance of `n^(th)` bright fringe from the central bright fringe `x_n=n(lambda D)/(d)`. By definition, Fringe width is the distance between two consecutive bright or dark fringes.
`x_(n+1)-x_(n)=beta=(lambda)/(d)(n+1-n)`
i.e. `beta=(lambda D)/(d)`
Fringe width `beta prop D, beta prop lambda and beta prop (1)/(d).`
We can show that fringe width between any two dark fringes is also `beta =(lambda D)/(d)`.