The objective and the eyepiece of a microscope have focal lengths of $4 \, mm$ and $25 \, mm$,respectively. The objective produces a real image $30$ times the size of the object. The final image is viewed at infinity. The near point of the microscope user is at $25 \, cm$. The overall magnification of the microscope is

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

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

Obtain the equation of magnification for a compound microscope.

If the focal length of the objective and eye lens are $1.2 \, cm$ and $3 \, cm$ respectively,and the object is placed $1.25 \, cm$ away from the objective lens,and the final image is formed at infinity,the magnifying power of the microscope 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.