In a Ramsden eyepiece,two plano-convex lenses,each of focal length $f$,are separated by a distance of $d = 2f/3$. If the separation distance is $12 \ cm$,what is the equivalent focal length (in $cm$) of the eyepiece?

We wish to make a microscope with the help of two positive lenses,both with a focal length of $20\, mm$ each,and the object is positioned $25\, mm$ from the objective lens. How far apart should the lenses be so that the final image is formed at infinity?......$mm$

The focal lengths of the objective and eye-lens of a microscope are $1 \,cm$ and $5 \,cm$ respectively. If the magnifying power for the relaxed eye is $45$,then the length of the tube is $....... \,cm$.

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

$A$ 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 $....$

If we need a magnification of $375$ from a compound microscope of tube length $150\; mm$ and an objective of focal length $5\; mm$,the focal length of the eye-piece should be close to.....$mm$.