If the focal length of the objective lens and the eye lens are $4 \, mm$ and $25 \, mm$ respectively in a compound microscope,and the length of the tube is $16 \, cm$,find its magnifying power for the relaxed eye position.

In a Ramsden eyepiece, two plano-convex lenses, each of focal length $f$, are separated by a distance of $d = 2f/3$. If the separation distance is $12 \ cm$, what is the equivalent focal length (in $cm$) of the eyepiece?

$A$ compound microscope has a magnifying power of $30$. The focal length of its eyepiece is $5 \, cm$. Assuming the final image is formed at the least distance of distinct vision,the magnification produced by the objective is:

The image formed by an objective of a compound microscope is

$A$ wire mesh consisting of very small squares is viewed at a distance of $8 \, cm$ through a magnifying converging lens of focal length $10 \, cm$,kept close to the eye. The magnification produced by the lens is

$A$ compound microscope consists of an objective lens of focal length $2 \ cm$ and an eyepiece of focal length $6.25 \ cm$ separated by a distance of $15 \ cm$. How far from the objective should an object be placed in order to obtain the final image at the least distance of distinct vision $(25 \ cm)$ (in $cm$)?