In a Young's double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength λ0=750mm and λ=900mm. 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=900mm is greater than fringe width ((β2) for λ2=750mm. This means that at though the central maxima of the two coincide, but first maximum for λ1=900mm will be further away from the first maxima for λ2=750mm and so on. A stage may come when this mismatch equals β2, then again maxima of λ1=900mm will coincide with a maxima of λ2=750mm,  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 from Central maxima =nλ1Dd=5×900×10−9×22×10−3 =45×10−4m=4.5mm