The length of the compound microscope is $15 \ cm$. The magnifying power for a relaxed eye is $25$. If the focal length of the eye lens is $6 \ cm$,then the object distance for the objective lens will be: (in $cm$)

$A$ card sheet divided into squares each of size $1 \, mm^{2}$ is being viewed at a distance of $9 \, cm$ through a magnifying glass (a converging lens of focal length $10 \, cm$) held close to the eye.
$(a)$ What is the magnification produced by the lens? How much is the area of each square in the virtual image?
$(b)$ What is the angular magnification (magnifying power) of the lens?
$(c)$ Is the magnification in $(a)$ equal to the magnifying power in $(b)$? Explain.

Maximum magnification produced by a simple microscope of focal length $f=5 \,cm$ is

In a compound microscope,the objective of focal length ${f_o}$ and the eyepiece of focal length ${f_e}$ are placed at a distance $L$ such that $L$ equals:

$A$ compound microscope has an eyepiece of focal length $10 \, cm$ and an objective of focal length $4 \, cm$. Calculate the magnification,if an object is kept at a distance of $5 \, cm$ from the objective so that the final image is formed at the least distance of distinct vision $(20 \, cm)$.

$A$ microscope has an objective of focal length $1 \ cm$ and an eye-piece of focal length $6 \ cm$. If the tube length is $30 \ cm$ and the image is formed at the least distance of distinct vision,what is the magnification produced by the microscope? Take $D = 25 \ cm$.