$A$ microscope has an objective of focal length $2 \ cm$,an eyepiece of focal length $4 \ cm$,and a tube length of $40 \ cm$. If the distance of distinct vision is $25 \ cm$,the magnification of the 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.

$A$ microscope has an objective of focal length $1 \ cm$ and an eye-piece of focal length $6 \ cm$. If the tube length is $30 \ cm$ and the image is formed at the least distance of distinct vision,what is the magnification produced by the microscope? Take $D = 25 \ cm$.

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

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 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 should an object be placed in order to obtain the final image at the least distance of distinct vision $(25 \ cm)$ (in $cm$)?