If the tube-length $(L)$ of a compound microscope increases,then its magnification . . . . . . .

$A$ microscope has an objective of focal length $1.5\, cm$ and an eye-piece of focal length $2.5\, cm$. If the distance between objective and eye-piece is $25\, cm$,the approximate value of magnification produced for a relaxed eye is?

In a simple microscope,if the final image is located at infinity,then its magnifying power is:

$A$ compound microscope produces a magnification of $24$. The focal length of the eyepiece is $5 \ cm$. The final image is formed at the least distance of distinct vision. The magnification produced by the objective is

If the focal length of the objective lens and the eye lens are $4 \, mm$ and $25 \, mm$ respectively in a compound microscope,and the length of the tube is $16 \, cm$,find its magnifying power for the relaxed eye position.

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