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: (Final image is at infinity)

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 $....... \,cm$.

If the tube-length $(L)$ of a compound microscope increases,then its magnification . . . . . . .

The power of two convex lenses $A$ and $B$ are $8$ diopters and $4$ diopters respectively. If they are to be used as a simple microscope,the magnification of

$A$ compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically,creating two identical plano-convex lenses. One of them is used in place of the original objective lens. To retain the same magnification while keeping the object distance unchanged,the tube length has to be

$A$ person with a normal near point $(25 \;cm)$ using a compound microscope with an objective of focal length $8.0 \;mm$ and an eyepiece of focal length $2.5 \;cm$ can bring an object placed at $9.0 \;mm$ from the objective into sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.