A) \[60\times {{10}^{-4}}cm\]
B) \[10\times {{10}^{-4}}cm\]
C) \[10\times {{10}^{-5}}cm\]
D) \[6\times {{10}^{-5}}cm\]
Correct Answer: D
Solution :
Distance of \[{{n}^{th}}\] dark fringe from central fringe \[{{x}_{n}}=\frac{(2n-1)\lambda \,D}{2d}\] \[\therefore \,\,\,{{x}_{2}}=\frac{(2\times 2-1)\lambda \,D}{2d}=\frac{3\lambda \,D}{2d}\] \[\Rightarrow 1\times {{10}^{-3}}=\frac{3\times \lambda \times 1}{2\times 0.9\times {{10}^{-3}}}\Rightarrow \lambda =6\times {{10}^{-5}}\,cm\]
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