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Started by Hasan Fakhruddin
Mon, 12 Dec 1994 23:09
Re: ACTIVITY 12
Author: Hasan Fakhruddin
Date: Mon, 12 Dec 1994 23:09
Date: Mon, 12 Dec 1994 23:09
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1242 bytes
For the diffraction experiment using the hair/wire what equation is applicable
to determine the width of the hair/wire? Is it,
d sin(theta) = n (lambda)
where, d = width of the hair/wire
theta = angular position of the nth dark fringe
lambda = wavelength ?
If yes, then what is the justification for this equation? Are you regarding
the edges of the obstacle as two sources as in the Young's double slit exp?
-Hasan Fakhruddin
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