The focal lengths of the objective and the eyepiece of a compound microscope are $2 \,cm$ and $3 \,cm$ respectively, and the distance between them is $15 \,cm$. The final image formed by the eyepiece is at infinity. The distances of the object and the image produced by the objective lens from the objective lens are respectively:

The focal lengths of the objective and eyepiece of a compound microscope are $1 \, cm$ and $5 \, cm$ respectively. If the magnification is $45$,what is the length of the tube in $cm$?

The least distance of distinct vision is $25 \, cm$. The magnifying power of a simple microscope with a focal length of $5 \, cm$ is:

$(a)$ $A$ 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. Where should the lens be held in order to view the squares distinctly with the maximum possible magnifying power?
$(b)$ What is the magnification in this case?
$(c)$ Is the magnification equal to the magnifying power in this case? Explain.

Which of the following statements are true in the context of a Compound Microscope?

The magnifying power of a simple microscope is $6$. The focal length of its lens in metres will be,if the least distance of distinct vision is $25\,cm$.