Question

The focal length of objective and eye lens of a microscope are $4\, cm$ and $8 \,cm$ respectively. If the least distance of distinct vision is $24 \, cm$ and object distance is $4.5 \,cm$ from the objective lens, then the magnifying power of the microscope will be

• A. 18
• B. 32
• C. 64
• D. 20

Answer 1

Answer: (B)

For objective lens $\frac{1}{f_{o}}=\frac{1}{v_{o}}-\frac{1}{u_{o}}$
$\Rightarrow \frac{1}{(+4)}+\frac{1}{v_{o}}-\frac{1}{(-4.5)}$
$ \Rightarrow v_{o}=36 \,cm$
$\therefore \left|m_{D}\right|=\frac{v_{o}}{u_{o}}\left(1+\frac{D}{f_{e}}\right)$
$=\frac{36}{4.5}\left(1+\frac{24}{8}\right)=32$