In a microscope of tube length $10 \ cm$, two convex lenses are arranged with focal lengths of $2 \ cm$ and $5 \ cm$. The total magnification obtained with this system for normal adjustment is $(5)^{k}$. The value of $k$ is . . . . . . .

In a compound microscope,the intermediate image 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.

The focal length of the objective and eye lens of a microscope are $4 \, cm$ and $8 \, cm$ respectively. If the least distance of distinct vision is $24 \, cm$ and the object distance is $4.5 \, cm$ from the objective lens, then the magnifying power of the microscope will be:

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 $cm$?

In a compound microscope,the focal length and aperture of the objective lens used are,respectively: