A light source, which emits two wavelengths  ${\lambda }_1=400\text{\hspace{0.17em}nm\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}{\lambda }_2=600\text{\hspace{0.17em}nm,}$ is used in Young’s double site experiment. If recorded fringe widths for ${\lambda }_1$ and ${\lambda }_2$ are ${\beta }_1$ and ${\beta }_2$ and the number of fringes for them within a distance y on one side of the central maximum are $m_1$ and  $m_2$, respectively, then

The figure width in Young’s double slit experiment is given by $\beta =\lambda D/d.$ Thus, the ratio of fringe width for the two wavelengths is ${\beta }_1/{\beta }_2={\lambda }_1/{\lambda }_2=400/600=2/3.$
  
The number of fringes in a distance y are  $m=y/\beta .$ Hence, the ratio of number of fringes of two wavelengths in a given distance y is 
$m_1/m_2={\beta }_2/{\beta }_1=3/2.$
The third maximum for ${\lambda }_2$ and fifth minimum of  ${\lambda }_1$ will occur at
 $$\begin{gathered} y_2=3{\beta }_2=3\times 600D/d=1800D/d, \\ {y}_1=(5-1/2){\beta }_1=4.5\times 400D/d=1800D/d. \end{gathered}$$