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Addition of Roman Numerals

A fringe of c...

Question

A fringe of certain interference pattern is 0.002 cm.what is the distance of 5th dark fringe

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Solution

Given : fringe width, $β$ = 0.002cm = 2 x 10^-5m
x = ?

x_n₌(2n-1) D $λ$ /2d

x_n₌(2n-1) $β$ /2 { $β$ = D $λ$ /d) }
x_n = (2x5-1) 2 x 10^-5m /2
= 18 x 10^-5m/2
= 9x10^-5m
the distance of 5th dark fringe is 9x10^-5m = 0.009cm

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