If the focal length of the objective and eye lens are $1.2 \, cm$ and $3 \, cm$ respectively,and the object is placed $1.25 \, cm$ away from the objective lens,and the final image is formed at infinity,the magnifying power of the microscope is:

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:

$A$ microscope has an objective of focal length $1 \ cm$ and an eye-piece of focal length $6 \ cm$. If the tube length is $30 \ cm$ and the image is formed at the least distance of distinct vision,what is the magnification produced by the microscope? Take $D = 25 \ cm$.

In order to increase the magnifying power of a compound microscope:

For a compound microscope,the focal lengths of the objective lens and the eye lens are ${f_o}$ and ${f_e}$ respectively. The microscope provides magnification when:

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: