A beam of light consisting of two wavelengths 4500 A0 and 7500 A0 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 ?

Let the nth dark fringe of wavelength λn  and  the mth dark fringe of wavelength λm  coincide at a distance y from  the centre of the screen, then  y=n−12λnDd=m−12λmDd.....(1) At this position there is complete darkness on the screen eq. (1) gives :  n−12m−12=λmλn=7500A04500A0=53  Which gives  m=6n+210=3n+15...(2) Integral values of n and m which satisfy eq. (2) are  n = 3, m = 2, n =8, m = 5 and so on.  Let n1=3  be the centre of the screen of the first and n2=8 the second regions of darkness are given by : y1=n1−12λnDd  and  y2=n2−12λnDd  ∴Δy=y2−y1=(n2−n1)×λnDd                          = (8−3)×4500×10−10×1.21×10−3=2.7×10−3m=2.7mm