The magnifying power of a microscope with an objective of $5\, mm$ focal length is $400$. The length of its tube is $20\, cm$. Then the focal length of the eye-piece is.....$cm$

$A$ compound microscope has an eyepiece of focal length $10 \, cm$ and an objective of focal length $4 \, cm$. Calculate the magnification,if an object is kept at a distance of $5 \, cm$ from the objective so that the final image is formed at the least distance of distinct vision $(20 \, cm)$.

$A$ thin convex lens of focal length $5 \, cm$ is used as a simple microscope by a person with a normal near point of $25 \, cm$. What is the magnifying power of the microscope?

In a compound microscope,the focal length of the objective and eyepiece lens is $4\,cm$ and $8\,cm$ respectively. If the length of the tube is $30\,cm$,then the magnifying power for normal adjustment is:

If the focal length of the objective and eye lens are $1.2 \, cm$ and $3 \, cm$ respectively,and the object is placed $1.25 \, cm$ away from the objective lens,and the final image is formed at infinity,the magnifying power of the microscope is:

Which of the following statements are true in the context of a Compound Microscope?
$(A)$ The objective lens forms a virtual and magnified image.
$(B)$ The objective lens has a very short focal length.
$(C)$ The eyepiece acts as a simple magnifying glass.
$(D)$ The eyepiece is a concave lens.