Question

Let the focal length of the objective and eye lens be $1.2\,cm$ and $3\,cm$ , respectively. Let the object is put $1.25\,cm$ away from the objective lens and the final image is formed at infinity. What is the magnifying power of the microscope?

• A. $150$
• B. $200$
• C. $250$
• D. $400$

Answer 1

Answer: (B)

The microscope here, is a compound microscope which is a combination of objective and eyepiece.
For the objective lens,
$\frac{1}{f_{o}}=\frac{1}{v_{o}}-\frac{1}{u_{o}}$
Objective lens is convex, so $f_{o}=1.2\,cm$
$\frac{1}{+ 1 . 2}=\frac{1}{v_{o}}-\frac{1}{\left(\right. - 1 . 25 \left.\right)}\frac{1}{+ 1 . 2}=\frac{1}{v_{o}}+\frac{1}{1 . 25}\frac{1}{v_{o}}=\frac{1}{1 . 2}-\frac{1}{1 . 25}v_{o}=30\,cm$
When final image is at infinity,
Net magnification, $m=m_{o}\times m_{e}$
$m=\frac{v_{o}}{u_{o}}\times \frac{D}{f_{e}}$
$m=\frac{30}{- 1 . 25}\times \frac{25}{3}\left|m\right|=200$