The magnifying power of a simple microscope can be increased if we use an eyepiece of:

$A$ 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

$A$ compound microscope has a magnifying power of $30$. The focal length of its eyepiece is $5 \ cm$. Assuming the final image is at the least distance of distinct vision $(25 \ cm)$,the magnification produced by the objective is:

An object viewed from a near point distance of $25 \, cm$,using a microscopic lens with magnification $6$,gives an unresolved image. $A$ resolved image is observed at infinite distance with a total magnification double the earlier,using an eyepiece along with the given lens and a tube of length $0.6 \, m$. The focal length of the eyepiece is equal to $.... \, cm$.

The length of the compound microscope is $15 \ cm$. The magnifying power for a relaxed eye is $25$. If the focal length of the eye lens is $6 \ cm$,then the object distance for the objective lens will be: (in $cm$)

The focal length of the objective and eye lenses of a microscope are $1.6 \, cm$ and $2.5 \, cm$ respectively. The distance between the two lenses is $21.7 \, cm$. If the final image is formed at infinity,what is the linear magnification?