$A$ 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 $(20 \, cm)$.

Write the equation of magnification for a simple microscope when the image is formed at the near point.

In a compound microscope,the objective lens and eyepiece have focal lengths of $0.95 \ cm$ and $5 \ cm$ respectively,and are kept at a distance of $20 \ cm$. The final image is formed at a distance of $25 \ cm$ from the eyepiece. Calculate the magnifying power.

In a compound microscope,the focal length and the aperture of the objective lens used are respectively:

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:

If the focal length of the objective lens and the eye lens are $4 \, mm$ and $25 \, mm$ respectively in a compound microscope,and the length of the tube is $16 \, cm$,find its magnifying power for the relaxed eye position.