A) 5
B) 30
C) 56.5
D) 125
Correct Answer: C
Solution :
Magnification of microscope for minimum distance of distinct vision is \[M=\frac{-{{v}_{0}}}{{{u}_{0}}}\left( 1+\frac{D}{{{f}_{e}}} \right)\] where, \[{{v}_{o}}+{{u}_{e}}=25,\]\[{{f}_{o}}=2\,cm,\]\[{{f}_{e}}=5\,cm,\] \[{{v}_{e}}=-\,25\,cm\] For eye-lens, \[\frac{1}{{{f}_{e}}}=\frac{1}{{{v}_{e}}}-\frac{1}{{{u}_{e}}}\] \[\frac{1}{{{u}_{e}}}=\frac{1}{{{v}_{e}}}-\frac{1}{{{f}_{e}}}\] \[=\frac{-1}{25}-\frac{1}{5}\] \[{{u}_{e}}=-\frac{25}{6}\] \[{{v}_{o}}=25-\frac{25}{6}=\frac{125}{6}\] For objective, \[\frac{1}{{{f}_{o}}}=\frac{1}{{{v}_{o}}}-\frac{1}{{{u}_{o}}}\] \[\frac{1}{{{u}_{o}}}=\frac{1}{{{v}_{o}}}-\frac{1}{{{f}_{o}}}\] \[=\frac{6}{125}-\frac{1}{2}\] \[{{u}_{o}}=-\frac{250}{113}\] \[\therefore \] \[M=\left( \frac{125/6}{250/113} \right)\left( 1+\frac{25}{5} \right)\] \[=56.5\]
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