In Young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths${\lambda }_1=12000\stackrel{0}{A}$and ${\lambda }_2=10000\stackrel{0}{A}$. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Distance between slits, d = 2 mm
Distance between screen and slits,
D = 2 m
Let ‘y’ be the distance on the screen from the centre, where the two bright bands coincide.
Let n₁, n₂ be the no. of bands of sources $\lambda$₁ , $\lambda$₂ in distance ‘y’.
$\begin{array}{l}\therefore y=n_1{\beta }_1=n_2{\beta }_2\\ n_1\frac{D{\lambda }_1}{d}=n_2\frac{D{\lambda }_2}{d}\\ n_1{\lambda }_1=n_2{\lambda }_2\\ \frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{10000}{12000}=\frac{5}{6}\end{array}$
$\therefore$5 bands of first source are accommodated in distance y.
 $\begin{array}{l}\therefore y=5\times {\beta }_1=5\times \frac{D{\lambda }_1}{d}\\ y=\frac{5\times 12000\times {10}^{-10}\times 2}{2\times {10}^{-3}}\\ =60\times {10}^{-4}m\\ =6mm.\end{array}$