In YDSE distance of second bright fringe from central bright fringe is x. What is the distance of third dark band from second bright fringe?

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2x3

3x4

answer is D.

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Detailed Solution

x=2β=2.λDd⇒β=λDd=x2 (where β is fringe width)Distance of dark band from second bright fringe β2=12×Fringe width =12.λDd=12.x2=x4

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