Question

In Young’s double slit experiment, the slits are 2mm apart and are illuminated by photons of two wavelengths λ1=12000Å and λ2=10000Å. At what minimum distance from the common central bright fringe on the screen 2m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other ?

Answer 1

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Answer 2

the answer is 4.5mm.From the given data, note that the fringe width (b1) for λ1=900nm is greater than fringe width (b2) for λ2=750nm. This means that at though the central maxima of the two coincide, but first maximum for λ1=900nm will be further away from the first maxima for λ2=750nm, and so on.

A stage may come when this mismatch equals b2, then again maxima of λ1=900nm, will coincide with a maxima of λ2=750nm, let this correspond to nth order fringe for l1.

Then it will correspond to (n+1)th order fringe for l2.

Therefore nλ1Dd=(n+1)λ2Dd ⇒n×900×10−9=(n+1)750×10−9

⇒n=5 Minimum distance from Central maxima =nλ1Dd=5×900×10−9×22×10−3 =45×10−4m=4.5mm