question_answer

A beam of light consisting of two wavelengths 650 nm and 520 nm is used to obtain interference fringes in a Young's double slit experiment. Distance between slits and screen is 1m and that between slits is 0.1 mm. What is the least distance from the central maximum, where the bright fringes due to both the wavelengths coincide?

Answer:

            Given, \[{{\lambda }_{1}}=650nm=650\times {{10}^{-9}}m,\]             \[{{\lambda }_{2}}=520nm=520\times {{10}^{-9}}m,\] D =1 m, d = 0.1 mm \[=1\times {{10}^{-4m}}\] Let the nth fringe due to \[{{\lambda }_{2}}\]  coincide with \[(n-1)\] th bright fringe due to \[{{\lambda }_{1}},\] then we have             \[n{{\lambda }_{2}}=(n-1){{\lambda }_{1}}\] \[n\times 520\times {{10}^{-9}}=(n-1)\times 650\times {{10}^{-9}}\] \[\Rightarrow 4n=(n-1)5\Rightarrow n=5\] \[\therefore \] The least distance required, \[y=n{{\lambda }_{2}}\frac{D}{d}\] \[=\frac{5\times 520\times {{10}^{-9}}\times 1}{{{10}^{-4}}}=2600\times {{10}^{-5}}\] \[\therefore \] \[y=2.6\times {{10}^{-2}}m\]