The least distance of distinct vision is $25 \, cm$. The magnifying power of a simple microscope with a focal length of $5 \, cm$ is:

The focal length of the objective and eye lens of a microscope are $4 \, cm$ and $8 \, cm$ respectively. If the least distance of distinct vision is $24 \, cm$ and the object distance is $4.5 \, cm$ from the objective lens, then the magnifying power of the microscope will be:

$A$ 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 $cm$)?

The magnification of a compound microscope is $30$. If the focal length of the eyepiece is $5 \, cm$,what is the magnification of the objective lens?

The focal length of the objective and eye lens of a microscope are $4 \ cm$ and $8 \ cm$ respectively. If the least distance of distinct vision is $24 \ cm$ and the object distance is $4.5 \ cm$ from the objective lens,then the magnifying power of the microscope will be: (Final image is at infinity)

The focal lengths of the objective and the eyepiece of a compound microscope are $2 \,cm$ and $3 \,cm$ respectively, and the distance between them is $15 \,cm$. The final image formed by the eyepiece is at infinity. The distances of the object and the image produced by the objective lens from the objective lens are respectively: