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

The length of the compound microscope is $14 \, cm$. The magnifying power for a relaxed eye is $25$. If the focal length of the eye lens is $5 \, cm$,then the object distance for the objective lens will be.......$ cm$.

$(a)$ $A$ 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.

$A$ 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 lens should an object be placed in order to obtain the final image at the least distance of distinct vision (in $cm$)? (Take $D = 25 \ cm$).

The focal length of the objective lens and the eyepiece of a microscope are $2 \, cm$ and $5 \, cm$ respectively,and the distance between them is $20 \, cm$. Find the distance of the object from the objective lens when the final image is seen by the eye at $25 \, cm$ from the eyepiece.

To obtain a magnified image at the distance of distinct vision $(D)$ using a simple microscope,the object should be placed: