A beam of light consisting of two wavelengths, 650 nm and 520 n m is used to obtain interference fringes in Young's double-slit experiment. What is the least distance (in m ) from a central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is 3 m m and the distance between the plane of the slits and the screen is 150 c m .

Step-by-step solution

Let "y' be the linear distance between the centre of screen and the point at which the bright fringes due to both wavelength coincides. Let n 1 th number of bright fringe with wavelength λ 1 coincides with n 2 m number of bright fringe with wavelength λ 2 . ∴ n 1 β 1 = n 2 β 2 ⇒ n 1 λ 1 D d = n 2 λ 2 D d ⇒ n 1 λ 1 = n 2 λ 2 …(i) Also at first position, the n th bright fringe of one will coincide with ( n + 1 ) th bright of other. If λ 2 < λ 1 So then n 2 > n 1 ⇒ n 2 = n 1 + 1 …(ii) Using equation (ii) and equation