In Young's double slit experiment, if the separation between coherent sources is halved and the distance of th

Question

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 :

Accepted Answer

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

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