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 $cm$?

$A$ microscope consists of an objective of focal length $1.9 \,cm$ and an eyepiece of focal length $5 \,cm$. The two lenses are kept at a distance of $10.5 \,cm$. If the image is to be formed at the least distance of distinct vision, the distance at which the object is to be placed before the objective is (least distance of distinct vision is $25 \,cm$). (in $\,cm$)

$A$ 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. What should be the distance between the object and the magnifying glass if the virtual image of each square in the figure is to have an area of $6.25 \; mm^{2}$? Would you be able to see the squares distinctly with your eyes very close to the magnifier?

If the tube-length $(L)$ of a compound microscope increases, then its magnification . . . . . . .

What are the objective and the eye-piece in an optical instrument?

The maximum magnification that can be obtained with a convex lens of focal length $2.5\, cm$ is (the least distance of distinct vision is $25\, cm$).