Question

A microscope is having objective of focal length $1\text{cm}$ and eye-piece of focal length $6\text{cm}$. If tube length is $30\text{cm}$ and image is formed at the least distance of distinct vision, what is the magnification produced by the microscope ? Take $D=25\text{cm}$.

Answer 1

By compound microscope for image formed at the least distance of distinct vision,
$m=\frac{L}{f_0}(1+\frac{D}{f_e})$ (where, length of tube $L=30\text{cm}$, focal length of objective lens $f_0=1\text{cm}$, focal length of eye-piece $f_e=6\text{cm},D=25\text{cm})$
$=\frac{30}{1}(1+\frac{25}{6})=30\times \frac{(6+25)}{6}$
$=5\times 31=155$