If three fringes are missing in front of the double slit then express `'lambda'` in terms of `'d' and 'D'`

Explanation

`x=(2n+1)(lambda)/(2)(D)/(d)` for D.I.
`x=(2.2+1)(lambda)/(2)(D)/(d)` where `x=d`
i.e., `d=5 (lambda)/(2)(D)/(d)`
`therefore lambda=(2d^2)/(5D)`