Obtain the equation for bandwidth in Young's double slit experimeet.
Conditon for bright fringe (or) maxima

Explanation

The condition for the constructive interference or the point P to the have a bright fringe is, path difference, `delta = nlambda`
whre, n = 0, 1,2…
`therefore(dy)/(D)=nlambda`
`y=n(lambdaD)/(d)(or)y_(n)=n(lambdaD)/(d)`
This is the condition for the point P to be a bright fringe. The distance is the distance of the `n^(th)` bright fringe from the point O.
Condition for dark fringe (or) minima
The condition for the destructive interference or the poit P to be have a dark fringe is,
Path difference, `delta = (2n - 1) (lambda)/(2)`
where, n 1,2,3...
`therefore (dy)/(D) = (2n - 1) (lambda)/(2)`
`y=((2n-1))/(2)(lambdaD)/(d)(or)y_(n)=((2n-1))/(2)(lambdaD)/(d)`
Equation for bandwidth
The bandwidth `(beta)` is defined as the distance between any two consecutive bright or dark fringes.
The distance between `(n + 1)^(th) and n^(th)` consecutive bright fringes from O is given by,
`beta=y_(n+1)-y_(n)=((n+1)(lambdaD)/(D))-(n(lambdaD)/(d))`
`beta = (lambdaD)/(d)`
Similarly, the distance between `(n+1)^th and n^(th)` consecutive dark fringes from O is given by, `beta=y_((n+1))-y_(n)=((2(n+1)-1)/(2)(lamdaD)/(d))-(((2n-1))/(2)(lambdaD)/(d))`
`beta = (lambdaD)/(d)`
The above equation show that the bright dark fringes are of same width equally spaced on either side of central bright fringe.