$A$ compound microscope consists of an eyepiece of focal length $6.25 \, cm$ and an objective lens of focal length $2.0 \, cm$,separated by a distance of $15 \, cm$. What is the magnifying power when the final image is formed at infinity?

The magnifying power of a simple microscope is $6$. The focal length of its lens in metres will be,if the least distance of distinct vision is $25\,cm$.

The magnifying power of a microscope with an objective of $5\, mm$ focal length is $400$. The length of its tube is $20\, cm$. Then the focal length of the eye-piece is.....$cm$

To obtain a magnified image at the distance of distinct vision $(D)$ using a simple microscope,the object should be placed:

$A$ 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.

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