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 2 m 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=900nm 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=750nm, 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=750nm letthis 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=4.5mm