The magnification of a compound microscope is $30$. If the focal length of the eyepiece is $5 \, cm$,what is the magnification of the objective lens?

The magnification produced by the objective lens and the eye lens of a compound microscope are $25$ and $6$ respectively. The magnifying power of this microscope is

$A$ compound microscope consists of an objective lens of focal length $2.0 \, cm$ and an eyepiece of focal length $6.25 \, cm$ separated by a distance of $15 \, cm$. What is the magnifying power when the final image is formed at the least distance of distinct vision $(25 \, cm)$?

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

$A$ microscope is an optical instrument which:

In a Ramsden eyepiece, two plano-convex lenses, each of focal length $f$, are separated by a distance of $d = 2f/3$. If the separation distance is $12 \ cm$, what is the equivalent focal length (in $cm$) of the eyepiece?