The length of the compound microscope is $14 \, cm$. The magnifying power for a relaxed eye is $25$. If the focal length of the eye lens is $5 \, cm$,then the object distance for the objective lens will be.......$ cm$.

The magnification power of a compound microscope is given in terms of the magnification of the objective $m_0$ and the magnification power of the eyepiece $m_E$. The total magnification is:

In a compound microscope,the magnified virtual image is formed at a distance of $25 \, cm$ from the eye-piece. The focal length of its objective lens is $1 \, cm$. If the magnification is $100$ and the tube length of the microscope is $20 \, cm$,then the focal length of the eye-piece lens (in $cm$) is

$A$ microscope consists of an objective of focal length $1.9 \,cm$ and an eyepiece of focal length $5 \,cm$. The two lenses are kept at a distance of $10.5 \,cm$. If the image is to be formed at the least distance of distinct vision, the distance at which the object is to be placed before the objective is (least distance of distinct vision is $25 \,cm$). (in $\,cm$)

When can more magnification of a compound microscope be obtained?

What is a simple microscope? Obtain the equation of magnification for the image formed at the near point (normal vision distance).