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

Explanation

` S_(1) & S_(2)to Slits`
d `to` Slit separation
D `to` Distance of screen from the plane of slits From the figure,
`S_(2)P^(2)-S_(1)p^(2)=[D^(2)+(x+(d)/(2))^(2)]-[D^(2)+(x-(d)/(2))^(2)]=2xd`
`impliesS_(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`
`therefore` Path difference :`S_(2)P-S_(1)P=(2xd)/(2D)=(xd)/(D)` ........(1)
Let `lambda` be the wavelength of light ,for a bright fringe at P.
Path difference :`S_(2)P-S_(1)P=nlambda` ........(2)
Thus,from (1) and (2), `(xd)/(D)=nlambda`
`implies` For `n^(th)` bright fringe :`x_(n)=n((lambdaD)/(d))`
For `(n+1)^(th)` bright fringe :`x_(n+1)=(n+1)((lambdaD)/(d))`
Fringe width :`beta=_(n+1)-x_(n)=(lambdaD)/(d)`
Note:Alternate method using sin `theta` and `tantheta` should be considered.