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
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