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  4. In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:

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Q. In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:

NEETNEET 2020Wave Optics

Solution:

Fringe width, $\beta=\frac{\lambda\,D}{d}$
$\beta' = \frac{\lambda\,D'}{d'}$
Now, $d'=\frac{d}{2}$ and
$D'=2\,D$
So, $\beta'=\frac{\lambda \times 2 D}{d / 2}$
$=\frac{4 \,\lambda\,D}{d}$
$\beta'=4\,\beta$
Fringe width becomes 4 times