Question

In a compound microscope, magnification will be large, if the focal length of the eye piece is

• A. Large
• B. Smaller
• C. Equal to that of objective
• D. Less than that of objective

Answer 1

The correct option is B Smaller

The maximum magnifying power of the compound microscope is,

$M_{max}=\frac{-v}{u}(1+\frac{D}{f_e})$

Where, $D=\text{Least distance of distinct vision}{f}_e=\text{Focal length of eyepiece}u=\text{The object distance for the objective lens}v=\text{The image distance for the objective lens}$

From the above relation we get, in a compound microscope magnification will be large, if the focal length of the eye piece is smaller.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.