$A$ compound microscope consists of an objective lens of focal length $2 \ cm$ and an eyepiece of focal length $6.25 \ cm$ separated by a distance of $15 \ cm$. How far from the objective lens should an object be placed in order to obtain the final image at the least distance of distinct vision (in $cm$)? (Take $D = 25 \ cm$).

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:

The least distance of distinct vision is $25 \ cm$. Find the magnifying power of a simple microscope of focal length $5 \ cm$ if the final image is formed at the least distance of distinct vision.

The focal length of the objective lens and the eyepiece of a microscope are $2 \, cm$ and $5 \, cm$ respectively,and the distance between them is $20 \, cm$. Find the distance of the object from the objective lens when the final image is seen by the eye at $25 \, cm$ from the eyepiece.

In a compound microscope,the focal length of the objective and eyepiece lens is $4\,cm$ and $8\,cm$ respectively. If the length of the tube is $30\,cm$,then the magnifying power for normal adjustment is:

$A$ 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.