$A$ simple microscope is used to see an object first in blue light and then in red light. Due to the change from blue to red light, its magnifying power

In a compound microscope,the focal length of the objective lens is $1.2 \, cm$ and the focal length of the eyepiece is $3.0 \, cm$. When the object is kept at $1.25 \, cm$ in front of the objective,the final image is formed at infinity. The magnifying power of the compound microscope is:

In a Ramsden eyepiece, two plano-convex lenses, each of focal length $f$, are separated by a distance of $d = 2f/3$. If the separation distance is $12 \ cm$, what is the equivalent focal length (in $cm$) of the eyepiece?

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$.

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.

$A$ 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.