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, \[y=n{{\lambda }_{2}}\frac{D}{d}\] \[=\frac{5\times 520\times {{10}^{-9}}\times 1}{{{10}^{-4}}}=2600\times {{10}^{-5}}\] \[\therefore \] \[y=2.6\times {{10}^{-2}}m\]
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