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

Explanation

A and B represented two coherent sources separated by a distance d.
Let a screen be placed at a distance D from the sources. The point `theta` on the screen is equidistant from A and B .
It is given
BP-AP=`delta`
eqn.(1)-(2)
Then `delta=BP^2-AP^2`
In the
`triangle^"le"` BRP, `BP^2=BR^2+PR^2=D^2+(PO+OR)^2`
`BP^2=D^2+(x + d//2)^2`...(1)
In the `triangle^"le"` APQ,
`AP^2 =AQ^2 +QP^2 =D^2 + (OP-OQ)^2`
`AP^2 =D^2 +(x - d//2)^2` ....(2)
`BP^2-AP^2=D^2+(x+d//2)^2 -D^2 +(x-d//2)^2`
`=[D^2+x^2+d^2/4+2xd//2] -[D^2+x^2+d^2/4-2xd//2]`
`=cancelD^2+cancelx^2+cancel(d^2/4)+xd -cancelD^2-x^2-cancel(d^2/4)+xd`
=xd+xd=2xd
`(BP^2-AP^2)=2xd`
(BP-AP)(BP+AP)=2xd
`BP-AP=(2xd)/((BP+AP))`
BP-AP=`(cancel2xd)/(cancel2D)` P is very close to D
Path Difference `Delta=(xd)/D " " Delta` is not shown in drawing .
For constructive interference `delta=nlambda`
`nlambda=(xd)/D rArr x=(Dnlambda)/d`
`x_1=(Dlambda)/d rArr x_2=(2Dlambda)/d`
The distance between the centers of any two consecutive bright fringes is called the fringe width.
`beta=x_2-x_1=(2Dlambda)/d-(Dlambda)/d`
`beta=(Dlambda)/d`