The length of the compound microscope is $15 \ cm$. The magnifying power for a relaxed eye is $25$. If the focal length of the eye lens is $6 \ cm$,then the object distance for the objective lens will be: (in $cm$)

In a simple microscope,if the final image is located at infinity,then its magnifying power is:

The 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:

The objective of a compound microscope is essentially

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.

$A$ 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 $cm$)?