Question

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

Answer 1

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

x = ?

x_{n =} (2n-1) D$\lambda$/2d

x_{n =} (2n-1) $\beta$/2 {$\beta$= D$\lambda$/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