In a compound microscope,the magnified virtual image is formed at a distance of $25 \, cm$ from the eye-piece. The focal length of its objective lens is $1 \, cm$. If the magnification is $100$ and the tube length of the microscope is $20 \, cm$,then the focal length of the eye-piece lens (in $cm$) is

The power of two convex lenses $A$ and $B$ are $8$ diopters and $4$ diopters respectively. If they are to be used as a simple microscope,the magnification of

$(a)$ $A$ 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 is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of the original objective lens. To retain the same magnification while keeping the object distance unchanged, the tube length has to be

$A$ compound microscope consists of an objective lens of focal length $2.0 \, cm$ and an eyepiece of focal length $6.25 \, cm$ separated by a distance of $15 \, cm$. What is the magnifying power when the final image is formed at the least distance of distinct vision $(25 \, cm)$?

$A$ simple microscope is used to see an object first in blue light and then in red light. Due to the change from blue to red light, what is the effect on its magnifying power?