In a compound microscope,if the objective produces an image $I_o$ and the eyepiece produces an image $I_e$,then

$A$ 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.

If the tube-length $(L)$ of a compound microscope increases, then its magnification . . . . . . .

In a compound microscope,the focal lengths of two lenses are $1.5 \text{ cm}$ and $6.25 \text{ cm}$. An object is placed at $2 \text{ cm}$ from the objective and the final image is formed at $25 \text{ cm}$ from the eye lens. The distance between the two lenses is .............. (in $\text{cm}$).

The length of the tube of a compound microscope is $10 \, cm$. The focal lengths of the objective and eye lenses are $0.5 \, cm$ and $1.0 \, cm$ respectively. The magnifying power of the microscope is about:

$A$ 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: