$A$ microscope has an objective of focal length $1.5\, cm$ and an eye-piece of focal length $2.5\, cm$. If the distance between objective and eye-piece is $25\, cm$,the approximate value of magnification produced for a relaxed eye is?

The near vision of an average person is $25 \, cm$. To view an object with an angular magnification of $10$,what should be the power of the microscope?

The focal lengths of the objective and the eye-piece of a compound microscope are $2.0 \, cm$ and $3.0 \, cm$ respectively. The distance between the objective and the eye-piece is $15.0 \, cm$. The final image formed by the eye-piece is at infinity. The two lenses are thin. The distances in $cm$ of the object and the image produced by the objective measured from the objective lens are respectively:

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

$A$ 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 $(25 \, cm)$?

In a compound microscope,the focal length and the aperture of the objective lens used are respectively: