A beam of light consisting of two wavelengths 4500 $\stackrel{0}{A}$ and 7500 $\stackrel{0}{A}$ is used to obtain interference fringes in Young’s double slit experiment. The distance between the slits is 1 mm and the distance between the plane of the slits and the screen is 120cm. What is the minimum distance (in mm) between two successive regions of complete darkness on the screen ?

$$\begin{gathered} Let the nth dark fringe of wavelength {\lambda }_n and \\ the m^{th} dark fringe of wavelength {\lambda }_m coincide at a distance y from \\ the centre of the screen, then \\ y=\left(n-\frac{1}{2}\right)\frac{{\lambda }_{n}D}{d}=\left(m-\frac{1}{2}\right)\frac{{\lambda }_{m}D}{d}.....\left(1\right) \\ At this position there is complete darkness on the screen eq. \left(1\right) gives : \\ \frac{n-\frac{1}{2}}{m-\frac{1}{2}}=\frac{{\lambda }_m}{{\lambda }_n}=\frac{7500A^0}{4500A^0}=\frac{5}{3} \\ Which gives m=\frac{6n+2}{10}=\frac{3n+1}{5}...\left(2\right) \\ Integral values of n and m which satisfy eq. \left(2\right) are \\ n = 3, m = 2, n =8, m = 5 and so on. \\ Let n_1=3 be the centre of the screen of the first and n_2=8 the second regions of darkness are given by : \\ {y}_1=\left(n_1-\frac{1}{2}\right)\frac{{\lambda }_{n}D}{d} and y_2=\left(n_2-\frac{1}{2}\right)\frac{{\lambda }_{n}D}{d} \\ \therefore \Delta y=y_2-y_1=(n_2-n_1)\times \frac{{\lambda }_{n}D}{d} \\ = (8-3)\times \frac{4500\times {10}^{-10}\times 1.2}{1\times {10}^{-3}}=2.7\times {10}^{-3}m=2.7mm \end{gathered}$$