In Young’s double slit experiment the screen is at a distance D =1m from the plane of the slit and slits are illuminated by plane monochromatic light of wavelength $\mathrm{\lambda }$ = 5000A°. P is a point on the screen at a distance y = 10 mm from the central maximum. If, by some special arrangement, the slits be moved symmetrically apart with a relative velocity v = 1 mm / s , the number of fringes crossing the point P per unit time will be ……..

Let l be the distance between the two slits, at any instant, then 
It is given that $\mathrm{v}=\frac{\mathrm{dl}}{\mathrm{dt}} .....\left(\mathrm{i}\right)$
The path difference reaching the point P from the slits is evidently $\mathrm{\Delta x}=\mathrm{y}\frac{\mathrm{l}}{\mathrm{D}}$
Differentiating both sides with respect to time t, we get $\frac{\mathrm{d\Delta x}}{\mathrm{dt}}=\frac{\mathrm{y}}{\mathrm{D}}\cdot \frac{\mathrm{dl}}{\mathrm{dt}}=\frac{\mathrm{yv}}{\mathrm{D}} ....\left(\mathrm{ii}\right)$
Since, a change in optical path difference of $\mathrm{\lambda }$, corresponds to one fringe, the number of fringes crossing point P per unit time will be
$\left(\frac{\mathrm{d\Delta x}}{\mathrm{dt}}\right)\frac{1}{\mathrm{\lambda }}=\frac{\mathrm{yv}}{\mathrm{\lambda D}}=\frac{10\times {10}^{-3}\times 1\times {10}^{-3}}{5\times {10}^{-7}\times 1}=20$