The magnification of a compound microscope is $30$. The focal length of the eyepiece is $5 \ cm$ and the image is formed at the near point (distance of distinct vision) of $25 \ cm$. The magnification of the objective lens is:

The length of the compound microscope is $14 \, cm$. The magnifying power for a relaxed eye is $25$. If the focal length of the eye lens is $5 \, cm$,then the object distance for the objective lens will be.......$ cm$.

Obtain the equation of magnification for a compound microscope.

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:

We wish to make a microscope with the help of two positive lenses,both with a focal length of $20\, mm$ each,and the object is positioned $25\, mm$ from the objective lens. How far apart should the lenses be so that the final image is formed at infinity?......$mm$

$A$ 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