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Question

In Young's double slit interference experiment the wavelength of light used is 6000A. If the path difference between waves reaching a ponit P on the screen is 1.5 microns, then at that point P:

A
second bright band occurs
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B
second dark band occurs
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C
third dark band occurs
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D
thirdbright band occurs
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Solution

The correct option is C third dark band occurs
Given λ=6000A=6000×1010m=6×107mand =1.5 microns=1.5×106m. For bright fringes: =nλ; where n is an integer.
n=λ=1.5×1066×107=52,which is not an integer.
hence, path difference of 1.5×106m does not correspond to a bright fringe. For dark fringes, we have
=(n12)λor 1.5×106=(n12)×(6×107)
which gives n12=52 or n=3., Hence a path difference of 1.5×106m corresponds to the third dark fringe.

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