The total magnification produced by a compound microscope is $24$ when the final image is formed at the least distance of distinct vision. If the focal length of the eyepiece is $5 \ cm$,the magnification produced by the objective is

An object viewed from a near point distance of $25 \, cm$,using a microscopic lens with magnification $6$,gives an unresolved image. $A$ resolved image is observed at infinite distance with a total magnification double the earlier,using an eyepiece along with the given lens and a tube of length $0.6 \, m$. The focal length of the eyepiece is equal to $.... \, cm$.

$A$ 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 $D = 25 \ cm$.

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 with an objective lens of focal length $1 \, cm$ and an eyepiece of $2 \, cm$ focal length has a tube length of $20 \, cm$. Calculate the magnifying power of the microscope,if the final image is formed at the least distance of distinct vision.

In a compound microscope,cross-wires are fixed at the point: