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_2P - S_1P`
From the figure `Delta^("le") S_2 PF`
`(S_2P)^2 = (S_2F)^2 + (FP)^2`
` = D^2 + (x + d/2)^2`
Similarly `Delta^("le")S_1PE`
`(S_1P)^2 = (S_1E)^2 + (EP)^2`
`=D^2 + (x - d/2)^2`
`(S_2P)^2 - (S_1P)^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_2P)^2 - (S_1P)^2 - 2xd`
or `S_2P - S_1P = (2xd)/((S_2P + S_1P))`
Since .P. is very close to O and d < < D
`S_2P + S_1P ~~2D`
`therefore` Path difference : `S_2P - S_1P = (2xd)/(2D) = (xd)/(D) to (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_2P - S_1P = n lamda " " n = 0,1,2,3.......`
From eqn(1) `(xd)/(D) = n lamda`
or `x = (n lamdaD)/(d)`
The distance of the nth bright fringe from the centre .O. of the screen is
`x_n = n ((lamda D)/(d))`
The distance of the `(n + 1)^(th)` bright fringe from the centre .O. of the screen is
`x_(n+1) = (n+1) ((lamda D)/(d))`
By definition of fringe width
`beta = x_(n+1) - x_n`
` = (n+1) ((lamda D)/(d)) - n ((lamda D)/(d))`
`(lamda D)/(d) ( n +1 -n)`
`beta = (lamda D)/(d)`