Question

In a compound microscope, the magnified virtual image is formed at a distance of $25\text{cm}$ from the eye-piece. The final image formed at a distance lens $1\text{cm}.$ If the magnification is $100$ and the tube length of the microscope is $20\text{cm},$ then the focal length of the eye-piece lens (in $\text{cm}$) is close to (up to two decimal places)

Answer 1

Given, $v=25\text{cm};L=20\text{cm}$
$f_0=1\text{cm}$, magnification, $M=100$

For the near point, the magnification is

$M=\frac{L}{f_0}(1+\frac{D}{f_e})$
$\Rightarrow 100=\frac{20}{1}(1+\frac{25}{f_e})$
$\Rightarrow f_e=\frac{25}{4}=6.25\text{cm}$