The magnification of a compound microscope is $30$. The focal length of the eyepiece is $5 \ cm$ and the image is formed at the near point (distance of distinct vision) of $25 \ cm$. The magnification of the objective lens is:

In a compound microscope,the focal length of the objective and eyepiece lens is $4\,cm$ and $8\,cm$ respectively. If the length of the tube is $30\,cm$,then the magnifying power for normal adjustment is:

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 at the least distance of distinct vision $(25 \ cm)$,the magnification produced by the objective is:

$(a)$ $A$ card sheet divided into squares each of size $1 \; mm^2$ is being viewed through a magnifying glass (a converging lens of focal length $9 \; cm$) held close to the eye. Where should the lens be held in order to view the squares distinctly with the maximum possible magnifying power?
$(b)$ What is the magnification in this case?
$(c)$ Is the magnification equal to the magnifying power in this case? Explain.

$A$ compound microscope has two lenses. The magnifying power of one is $5$ and the combined magnifying power is $100$. The magnifying power of the other lens is