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Q. In double slit experiment, the distance between two slits is $0.6$ mm and these are illumunated with light of wavelength $ 4800\, \mathring{A} $ . The angular width of first dark fringe on the screen distant $120\, cm$ from slits will be :

BHUBHU 2006Wave Optics

Solution:

Angular fringe width of first minima is
$ \frac{2x}{D}=2(2n-1)\frac{A}{2d} $
$=(2n-1)\frac{\lambda }{d} $
Given, $ d=0.6\text{ }mm=0.6\times {{10}^{-3}}m $
$ \lambda =4800{AA}=4.8\times {{10}^{-7}}m,\text{ }n=1 $
$ \therefore $ $ \frac{2x}{D}=\frac{(2\times 1-1)\times 4.8\times {{10}^{-7}}}{0.6\times {{10}^{-3}}} $
$=8\times {{10}^{-4}}rad. $