The magnification produced by the objective lens and the eye lens of a compound microscope are $25$ and $6$ respectively. The magnifying power of this microscope 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 magnification will be large if the focal length of the eyepiece is:

$A$ compound microscope consists of an objective lens of focal length $2 \ cm$ and an eyepiece of focal length $6.25 \ cm$ separated by a distance of $15 \ cm$. How far from the objective lens should an object be placed in order to obtain the final image at the least distance of distinct vision (in $cm$)? (Take $D = 25 \ cm$).

$A$ microscope consists of an objective of focal length $1.9 \,cm$ and an eyepiece of focal length $5 \,cm$. The two lenses are kept at a distance of $10.5 \,cm$. If the image is to be formed at the least distance of distinct vision, the distance at which the object is to be placed before the objective is (least distance of distinct vision is $25 \,cm$). (in $\,cm$)

Magnifying power of a simple microscope is (when final image is formed at $D = 25 \ cm$ from eye)