The least distance of distinct vision is $25 \ cm$. Find the magnifying power of a simple microscope of focal length $5 \ cm$ if the final image is formed at the least distance of distinct vision.

$A$ compound microscope consists of an objective lens of focal length $1 \, cm$ and an eyepiece of focal length $5 \, cm$ with a separation of $10 \, cm$. The distance between an object and the objective lens,at which the strain on the eye is minimum,is $\frac{n}{40} \, cm$. The value of $n$ is $....$

Magnifying power of a simple microscope is (when final image is formed at $D = 25 \ cm$ from eye)

$A$ card sheet divided into squares each of size $1 \; mm^{2}$ is being viewed through a magnifying glass (a converging lens of focal length $9 \; cm$) held close to the eye. What should be the distance between the object and the magnifying glass if the virtual image of each square in the figure is to have an area of $6.25 \; mm^{2}$? Would you be able to see the squares distinctly with your eyes very close to the magnifier?

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:

The magnification of a compound microscope is $30$. The focal length of the eyepiece is $5 \ cm$ and the image is formed at the near point (distance of distinct vision) of $25 \ cm$. The magnification of the objective lens is: