If in a compound microscope,the objective has a focal length of $1.0 \text{ cm}$,the eyepiece has a focal length of $2.0 \text{ cm}$,the tube length is $20 \text{ cm}$,and the near point for an observer is $25 \text{ cm}$,then the value of the magnification of the compound microscope will be . . . . . . .

The focal length of the objective and eye lens of a microscope are $4 \, cm$ and $8 \, cm$ respectively. If the least distance of distinct vision is $24 \, cm$ and the object distance is $4.5 \, cm$ from the objective lens, then the magnifying power of the microscope will be:

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 $25\,cm$.

$A$ 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

The image formed by an objective of a compound microscope 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.