The objective of a compound microscope is essentially

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

To improve the resolving power of a compound microscope, we should:

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 $.... \, cm$.

When the final image is formed at the least distance of distinct vision $(D)$ from the eye,the magnifying power of a simple microscope is ($f$ is the focal length of the lens):

$A$ card sheet divided into squares each of size $1 \; mm^{2}$ is being viewed through a magnifying glass (a converging lens of focal length $9 \; cm$) held close to the eye. What should be the distance between the object and the magnifying glass if the virtual image of each square in the figure is to have an area of $6.25 \; mm^{2}$? Would you be able to see the squares distinctly with your eyes very close to the magnifier?