The focal lengths of the objective and eye-lens of a microscope are $1 \,cm$ and $5 \,cm$ respectively. If the magnifying power for the relaxed eye is $45$,then the length of the tube is $....... \,cm$.

$A$ 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?

We wish to make a microscope with the help of two positive lenses,both with a focal length of $20\, mm$ each,and the object is positioned $25\, mm$ from the objective lens. How far apart should the lenses be so that the final image is formed at infinity?......$mm$

The focal lengths of the objective and the eye-piece of a compound microscope are $2.0 \, cm$ and $3.0 \, cm$ respectively. The distance between the objective and the eye-piece is $15.0 \, cm$. The final image formed by the eye-piece is at infinity. The two lenses are thin. The distances in $cm$ of the object and the image produced by the objective measured from the objective lens are respectively:

An angular magnification (magnifying power) of $30X$ is desired using an objective of focal length $1.25 \, cm$ and an eyepiece of focal length $5 \, cm$. How will you set up the compound microscope?

$A$ 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?