$A$ 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

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?

$A$ simple magnifying lens is used in such a way that an image is formed at $25 \, cm$ away from the eye. In order to have $10$ times magnification,the focal length of the lens should be

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 objective of a compound microscope is essentially

If in a compound microscope, the objective has a focal length of $1.0 \text{ cm}$, the eyepiece has a focal length of $2.0 \text{ cm}$, the tube length is $20 \text{ cm}$, and the near point for an observer is $25 \text{ cm}$, then the value of the magnification of the compound microscope will be . . . . . . .