In Young's double-slit experiment, the fringe width with the light of wavelength $6000 \mathrm{\AA }$ is found to be $4.0 \mathrm{mm}$. What will be the fringe width if the light of wavelength $4800 \mathrm{\AA }$ is used?

Given $\mathrm{\beta }=4.0 \mathrm{mm}$ and $\lambda =6000 \mathrm{\AA }$.
We know that the fringe width is given by:
$\mathrm{\beta }=\frac{\mathrm{\lambda D}}{\mathrm{d}}$            (i)
for $\lambda \text{'}=4800\stackrel{\circ }{A}$, the fringe width will be:
$\beta \text{'}=\frac{\lambda \text{'}D}{d}$                         (ii)
From (i) and (ii), we have:
$\beta \text{'}=\beta \frac{\lambda \text{'}}{\lambda }=\frac{4.0 \mathrm{mm}\times 4800 \mathrm{\AA }}{6000 \mathrm{\AA }}=3.2 \mathrm{mm}$