A) 9.75 mm
B) 15.6 mm
C) 1.56 mm
D) 7.8 mm
Correct Answer: D
Solution :
[d] For common maxima, \[{{n}_{1}}{{\lambda }_{1}}={{n}_{2}}{{\lambda }_{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{520\times {{10}^{-9}}}{650\times {{10}^{-9}}}=\frac{4}{5}\] For \[{{\lambda }_{1}}\] \[y=\frac{{{n}_{1}}{{\lambda }_{1}}D}{d}\], \[{{\lambda }_{1}}=650nm\] \[y=\frac{4\times 650\times {{10}^{-9}}\times 1.5}{0.5\times {{10}^{-3}}}\] or \[y=7.8mn\]You need to login to perform this action.
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