$A$ 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 $cm$)?

To improve the resolving power of a compound microscope, we should:

In a Ramsden eyepiece,two planoconvex lenses,each of focal length $f$,are separated by a distance of $12 \ cm$. The equivalent focal length (in $cm$) of the eyepiece is:

If in a compound microscope $m_1$ and $m_2$ are the linear magnifications of the objective lens and the eye lens respectively,then the magnifying power of the compound microscope will be:

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: For a simple microscope,the angular size of the object equals the angular size of the image.
Reason $R$: Magnification is achieved as the small object can be kept much closer to the eye than $25\, cm$ and hence it subtends a large angle.
In the light of the above statements,choose the most appropriate answer from the options given below:

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 $.... \, cm$.