In a compound microscope,the magnification will be large if the focal length of the eyepiece is:

In a compound microscope,the objective lens and eyepiece have focal lengths of $0.95 \ cm$ and $5 \ cm$ respectively,and are kept at a distance of $20 \ cm$. The final image is formed at a distance of $25 \ cm$ from the eyepiece. Calculate the magnifying power.

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

In a compound microscope,the focal length and the aperture of the objective lens used are respectively:

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:

$A$ 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: