$A$ compound microscope has a magnifying power of $30$. The focal length of its eyepiece is $5 \, cm$. Assuming the final image is formed at the least distance of distinct vision,the magnification produced by the objective is:

$A$ card sheet divided into squares each of size $1\,mm^2$ is being viewed at a distance of $9\,cm$ through a magnifying glass (a converging lens of focal length $9\,cm$) held close to the eye. What is the angular magnification (magnifying power) of the lens?

$A$ compound microscope has a magnifying power of $30$. The focal length of its eyepiece is $5 \ cm$. Assuming the final image is at the least distance of distinct vision $(25 \ cm)$,the magnification produced by the objective is:

$A$ person with a normal near point $(25 \;cm)$ using a compound microscope with an objective of focal length $8.0 \;mm$ and an eyepiece of focal length $2.5 \;cm$ can bring an object placed at $9.0 \;mm$ from the objective into sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

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:

The objective of a compound microscope is essentially