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)

32

Answer 2

For objective lens

$\frac{1}{f_{o}} = \frac{1}{v_{o}} - \frac{1}{u_{o}}$

$\Rightarrow \frac{1}{( + 4)} = \frac{1}{v_{o}} - \frac{1}{( - 4.5)} \Rightarrow v_{o} = 36\mspace{6mu} cm$

$\therefore|m_{D} ⥂ ⥂ ⥂ |\mspace{6mu} = \frac{v_{o}}{u_{o}}\left( 1 + \frac{D}{f_{e}} \right) = \frac{36}{4.5}\left( 1 + \frac{24}{8} \right) = 32$