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

Explanation

The light rays from the two (coherent) slits, reaching a point .P. on the screen, have a path difference `(S_(2)P-S_(1)P)`. The point .P. would ,therefore be a
(i) Point of maxima (bright fringe), if `S_(2)P-S_(1)P=nlambda`
(ii) Point of minima (dark fringle), if `S_(2)P-S_(1)P=(2n+1)lambda//2`

We have
`(S_(2)P)^(2)-(S_(1)P)^(2)={D^(2)-(x+(d)/(2))^(2)}-{D^(2)+(x-(d)/(2))^(2)}`
`=2xd`
`S_(2)P-S_(1)P=(2xd)/(S_(2)P+S_(1)P)=(2xd)/(2D)=(xd)/(D)`
`:.` We have maxima at point, where
`(xd)/(D)=nlambda`
and minima at points where
`(xd)/(D)=((2n+1)/(2))lambda`
Now, fringe width `beta=` separation between two successive maxima (or two successive minima)`=x_(n)-x_(n-1)`
`:.beta=(lambdaD)/(d)`