A beam of light consisting of two wavelengths 650nm and 520nm is used to obtain interference fringes in Young's double slit experiment. The distance between slits is 2 mm and between the plane of slits and screen is 120 cm. The least distance from the central maximum where the bright hinges due to both wavelength coincide is

let the n^th bright fringe of wavelength ${\lambda }_m$ and m ^th Bright fringe of wavelength ${\mathrm{\lambda }}_{\mathrm{m}}$ coincide at a distance x _n from central maximum .then 
$$\begin{gathered} {\mathrm{x}}_{\mathrm{n}}=\frac{{\mathrm{n\lambda }}_{\mathrm{n}}\mathrm{D}}{2\mathrm{d}}=\frac{{\mathrm{m\lambda }}_{\mathrm{m}}\mathrm{D}}{2\mathrm{d}} \\ \mathrm{or} \frac{\mathrm{m}}{\mathrm{n}}=\frac{{\mathrm{\lambda }}_{\mathrm{n}}}{{\mathrm{\lambda }}_{\mathrm{m}}}=\frac{650\mathrm{nm}}{520\mathrm{nm}}=\frac{5}{4} \\ \mathrm{So}, \mathrm{least} \mathrm{integral} \mathrm{values} \mathrm{of} \mathrm{m} \mathrm{and} \mathrm{n} \mathrm{satisfying} \mathrm{above} \\ \mathrm{requhemeDt} \mathrm{are} \mathrm{m} = 5\mathrm{andn}= 4 \\ \mathrm{Smallest} \mathrm{value} \mathrm{of} {\mathrm{x}}_{\mathrm{n}}, \mathrm{is} \mathrm{given} \mathrm{by} \\ {\left({\mathrm{x}}_{\mathrm{n}}\right)}_{\min }=\frac{{\mathrm{n\lambda }}_{\mathrm{n}}\mathrm{D}}{\mathrm{d}}=\frac{4\times \left(650\times {10}^{-9}\right)\times 1.20}{2\times {10}^{-3}} = 1.56 \mathrm{mm} \end{gathered}$$