The focal lengths of the objective and the eyepiece of a compound microscope are $1 \, cm$ and $5 \, cm$ respectively. An object is placed at a distance of $1.1 \, cm$ from the objective. If 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?
$(A)$ Each lens produces a virtual and inverted image.
$(B)$ The objective has a very short focal length.
$(C)$ The eyepiece is used as a simple magnifying glass.
$(D)$ The objective and eyepiece are convex and concave lenses respectively.

$A$ card sheet divided into squares each of size $1 \; mm^{2}$ is being viewed through a magnifying glass (a converging lens of focal length $9 \; cm$) held close to the eye. What should be the distance between the object and the magnifying glass if the virtual image of each square in the figure is to have an area of $6.25 \; mm^{2}$? Would you be able to see the squares distinctly with your eyes very close to the magnifier?

In which of the following is the final image erect?

An object viewed from a near point distance of $25 \, cm$,using a microscopic lens with magnification $6$,gives an unresolved image. $A$ resolved image is observed at infinite distance with a total magnification double the earlier,using an eyepiece along with the given lens and a tube of length $0.6 \, m$. The focal length of the eyepiece is equal to $.... \, cm$.

$A$ compound microscope has two lenses. The magnifying power of one is $5$ and the combined magnifying power is $100$. The magnifying power of the other lens is