question_answer

A beam of light consisting of two wavelengths 650 nm and 520 nm, is used to obtain interference fringes in Young's double slit experiment.

(i) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 650 nm.

(ii) What is the least distance from the central maximum where the bright fringes due to both wavelengths coincide?

The distance between the two slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm.

Answer:

Given, separation between two slits, \[d=2mm=2\times {{10}^{-3}}m\] Distance of screen from slit, D = 120 cm = 1.2 m Wavelength,       \[{{\lambda }_{2}}=520\,nm=5.2\times {{10}^{-7}}m\] Wavelength,       \[{{\lambda }_{2}}=520\,nm=5.2\times {{10}^{-7}}m\] (i) Since, separation of nth order bright fringe from centre fringe is given by, \[{{Y}_{n}}=\frac{nD\lambda }{d}\] \[\therefore \] For \[{{3}^{rd}}\] order bright fringe, \[{{Y}_{3}}=\frac{3D{{\lambda }_{1}}}{d}\] where,   \[\lambda ={{\lambda }_{1}}=6.5\times {{10}^{-7}}m\] \[\therefore \]      \[{{Y}_{3}}=\frac{3\times 1.2\times 6.5\times {{10}^{-7}}}{2\times {{10}^{-3}}}\]             \[=1.17\times {{10}^{-3}}m=1.17\,mm\] (ii) Let nth order bright fringe of wavelength \[{{\lambda }_{1}}\]  coincides with \[(n+1)\,th\] order bright fringe of wavelength \[{{\lambda }_{2}}.\] \[\therefore \]      \[\frac{nD{{\lambda }_{1}}}{d}=\frac{(n+1)D{{\lambda }_{2}}}{d}\] or         \[n{{\lambda }_{1}}=(n+1){{\lambda }_{2}}\] or         \[\frac{n+1}{n}=\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{6.5\times {{10}^{-7}}}{5.2\times {{10}^{-7}}}\] or         \[\left( 1+\frac{1}{n} \right)=\frac{5}{4}\] or \[\frac{1}{n}=\frac{5}{4}-1=\frac{1}{4}\] \[\Rightarrow \]   n = 4 \[\therefore \] the required least distance             \[=\frac{nD{{\lambda }_{1}}}{d}=\frac{4\times 1.2\times 6.5\times {{10}^{-7}}}{2\times {{10}^{-3}}}\]             \[=1.56\times {{10}^{-3}}\,m\]             = 1.56 mm