A beam of light consisting of two wavelengths ${\lambda }_1$$=750 \mathrm{nm}$ and ${\lambda }_2=450 \mathrm{nm}$ is used in Young's double-slit experiment. The separation between the slits is $2 \mathrm{mm}$ and the distance of the screen from the plane of the slits is $100 \mathrm{cm}$. What is the minimum distance between two successive regions of complete darkness?

In this case, for dark fringes,
$$\begin{gathered} y_1=\left(n-\frac{1}{2}\right)\frac{{\lambda }_{1}D}{d} \\ {y}_2=\left(m-\frac{1}{2}\right)\frac{{\lambda }_{2}D}{d} \end{gathered}$$
The waves will produce complete darkness at a point on the screen where $y_1=y_2$, i.e.,
$\left(n-\frac{1}{2}\right)\frac{{\lambda }_{1}D}{d}=\left(m-\frac{1}{2}\right)\frac{{\lambda }_{2}D}{d}$
$$\begin{gathered} \Rightarrow \frac{\left(n-\frac{1}{2}\right)}{\left(m-\frac{1}{2}\right)}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{450 \mathrm{nm}}{750 \mathrm{nm}}=\frac{3}{5} \\ \Rightarrow \frac{2n-1}{2m-1}=\frac{3}{5}=\frac{9}{15}=\frac{15}{25}=...... \end{gathered}$$
So, the minimum integral values of n and m which satisfy this condition are $m_1=3$ and $n_1=2$. The next values are $m_2=8$ and $n_2=5$. So we get the first region of complete darkness when 2^nd dark fringe of ${\lambda }_1$ falls on the 3^rd dark fringe of ${\lambda }_2$,
$(\Delta y{)}_1=\left(2-\frac{1}{2}\right)\frac{{\lambda }_{1}D}{d}$
The next region of complete darkness occurs when the 5^th dark fringe of ${\lambda }_1$ falls on 8^th dark fringe of ${\lambda }_2$,
$$\begin{gathered} (\Delta y{)}_2=\left(5-\frac{1}{2}\right)\frac{{\lambda }_{1}D}{d} \\ \therefore (\Delta y{)}_{\min }=\left(5-\frac{1}{2}\right)\frac{{\lambda }_{1}D}{d}-\left(2-\frac{1}{2}\right)\frac{{\lambda }_{1}D}{d} \\ =3\frac{{\lambda }_{1}D}{d} \\ =\frac{3\times \left(750\times {10}^{-9}\right)\times 1.0}{2\times {10}^{-3}} \\ =1.125 \mathrm{mm} \end{gathered}$$