In a compound microscope,the focal lengths of the objective and eye lens are $2.5 \, cm$ and $5 \, cm$ respectively. An object is placed at $3.75 \, cm$ in front of the objective. If the final image is formed at the least distance of distinct vision,then the distance between the two lenses (length of the microscope tube) is.......$cm$.

$A$ 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 $10 \, cm$) held close to the eye.
$(a)$ What is the magnification produced by the lens? How much is the area of each square in the virtual image?
$(b)$ What is the angular magnification (magnifying power) of the lens?
$(c)$ Is the magnification in $(a)$ equal to the magnifying power in $(b)$? Explain.

$A$ compound microscope consists of an objective lens of focal length $2.0 \, cm$ and an eyepiece of focal length $6.25 \, cm$ separated by a distance of $15 \, cm$. What is the magnifying power when the final image is formed at the least distance of distinct vision $(25 \, cm)$?

When the length of a microscope tube increases,its magnifying power

The power of two convex lenses $A$ and $B$ are $8$ diopters and $4$ diopters respectively. If they are to be used as a simple microscope,the magnification of

$A$ 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.