In a Young's double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength λ0=750nm  and  λ=900nm. The minimum distance from the common central bright fringe on a screen 2m  from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is

From the given data, note that the fringe width β1 for λ1=900 nm  is greater than fringe width β2 for   λ2=750 nm.This means that at though the central maxima of the two coincide, but first maximum for  λ1=900 nm will be further away from the first maxima for  λ2=750 nm, and so on. A stage may come when this mismatch equals β2,, then again maxima of λ1=900 nm,  will coincide with a maxima of λ2=750 nm,  let this correspond to nth order fringe for λ1. Then it will correspond to  (n+1)th order fringe for λ2.Therefore  nλ1Dd=(n+1)λ2Dd⇒n×900×10−9=(n+1)750×10−9 ⇒n=5Minimum distance fromCentral maxima  =nλ1Dd=5×900×10−9×22×10−3=45×10−4m=4.5 mm