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Question

In a Young's double slit experiment, the slits are 2mm apart and are illuminated with a mixture of two wavelength λ0=750 nm and λ=900 nm . 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

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Solution

The correct option is **D** 4.5mm

From the given data, note that the fringe width (β1) for λ1900 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 λ2750 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+)750×10−9⇒n=5

Minimum distance from

Central maxima =nλ1Dd=5×900×10−9×22×10−3

=45×10−4m=4.5 mm

From the given data, note that the fringe width (β1) for λ1900 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 λ2750 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+)750×10−9⇒n=5

Minimum distance from

Central maxima =nλ1Dd=5×900×10−9×22×10−3

=45×10−4m=4.5 mm

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