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

Explanation

Consider two coherent sources `S_1 and S_2` separated by a distance d.. Let a screen be placed at a distance .D. from the coherent sources.
The point .O. on the screen is equidistant from `S_1 and S_2` . So that path difference between the two light waves from `S_1 and S_2` reaching O is zero. Thus the point O has maximum intensity. Consider point .P. at a distance .x. from .O.. The path difference between the light waves from `S_1 and S_2` reaching point P is
`delta=S_(2)P-S_(1)P`
From the figure `Delta^(le)S_(2)PF`
`(S_(2)P)^(2)=(S_(2)F)^(2)+(FP)^(2)`
`D^(2)+(x+d/2)^(2)`
Similarly , `Delta^(le) S_(1)PE`
`(S_(1)P)^(2)=(S_(1)E)^(2)+(EP)^2`
`D^(2)+(x-d/2)^(2)`
`(S_(2)P)^(2)-(S_(1)P)^(2)=[D^(2)+(x+d/2)^(2)]-[D^(2)+(x-d/2)^(2)]`
`[D^(2)+x^(2)+d^2/4+2(x)(d/2)]-[D^(2)+x^2+(d^2)/4-2(x)(d/2)]`
`(S_(2)P)^(2)-(S_(1)P)^(2) =2xd`
or `S_(2)P-S_(1)P=(2xd)/((S_(2)P+S_(1)P))`
Since .P. is very close to O and `d lt lt D`
`S_(2)P + S_(1)P ~~2D`
`:.` Path difference : `S_(2) P - S_(1) P = (2xd)/(2D) = (xd)/D " "....(1)`
Equation (1) represents the path difference between light waves from `S_1 and S_2` superposing at point P.
For bright fringe or maximum intensity at P
`S_(2)P - S_(1)P = nlamdan = 0,1,2,3......`
eqn(1) `implies (xd)/(D) = nlamda`
or `x = (nlamdaD)/d`
The distance of the `n^(th)` bright fringe from the centre .O. of the screen is
`x_(n) = n((lamdaD)/(d))`
The distance of the `(n+1)^(th)` bright fringe from the centre .O. of the screen is
`x_(n) =n((lamdaD)/d)`
The distance of the `(n+1)^(th)` bright frings from the centre .O. of the screen is
`x_(n+1)=(n+1)((lamdaD)/d)`
By definition of fringe width
`Beta=(x_(n+1))-x_(n)`
`(n+1)((lamdaD)/d)-n((lamdaD)/d)`
`= (lamdaD)/d(n+1-n)`
`beta = (lamdaD)/d`