In a compound microscope,the focal length and the aperture of the objective lens used are respectively:

$A$ compound microscope consists of an objective lens of focal length $2.0 \;cm$ 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
$(a)$ the least distance of distinct vision $(25\;cm)$,and
$(b)$ at infinity? What is the magnifying power of the microscope in each case?

If we need a magnification of $375$ from a compound microscope of tube length $150\; mm$ and an objective of focal length $5\; mm$,the focal length of the eye-piece should be close to.....$mm$.

The focal lengths of the objective and eyepiece of a compound microscope are $1 \, cm$ and $5 \, cm$ respectively. If the magnification is $45$,what is the length of the tube in $cm$?

$A$ card sheet divided into squares each of size $1 \, mm^{2}$ is being viewed at a distance of $9 \, cm$ through a magnifying glass (a converging lens of focal length $10 \, cm$) held close to the eye.
$(a)$ What is the magnification produced by the lens? How much is the area of each square in the virtual image?
$(b)$ What is the angular magnification (magnifying power) of the lens?
$(c)$ Is the magnification in $(a)$ equal to the magnifying power in $(b)$? Explain.

The image formed by an objective of a compound microscope is