The objective lens of a compound microscope produces a magnification of $10$. In order to get an overall magnification of $100$ when the image is formed at $25 \, cm$ from the eye,the focal length of the eye lens should be:

The magnifying power of a microscope with an objective of $5\, mm$ focal length is $400$. The length of its tube is $20\, cm$. Then the focal length of the eye-piece is.....$cm$

The magnifying power of a simple microscope is $6$. The focal length of its lens in metres will be,if the least distance of distinct vision is $25\,cm$.

The focal length of the objective lens and the eyepiece of a microscope are $2 \, cm$ and $5 \, cm$ respectively,and the distance between them is $20 \, cm$. Find the distance of the object from the objective lens when the final image is seen by the eye at $25 \, cm$ from the eyepiece.

$A$ simple microscope is a combination of two lenses in contact. Their powers are $+20$ $D$ and $-4$ $D$. The distance of distinct vision is $25$ cm. When seen through the microscope, the size of an object $3$ mm high is (in $cm$)

Obtain the magnification for the image formed at infinity for a simple microscope.