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