$A$ 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.

Which type of electromagnetic waves are used in an optical microscope? Why?

The length of the compound microscope is $15 \ cm$. The magnifying power for a relaxed eye is $25$. If the focal length of the eye lens is $6 \ cm$,then the object distance for the objective lens will be: (in $cm$)

$A$ thin convex lens of focal length $5 \, cm$ is used as a simple microscope by a person with a normal near point of $25 \, cm$. What is the magnifying power of the microscope?

$A$ microscope has an objective of focal length $1.5\, cm$ and an eye-piece of focal length $2.5\, cm$. If the distance between objective and eye-piece is $25\, cm$,the approximate value of magnification produced for a relaxed eye is?

Answer the following questions:
$(a)$ The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
$(b)$ In viewing through a magnifying glass,one usually positions one's eyes very close to the lens. Does angular magnification change if the eye is moved back?
$(c)$ Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
$(d)$ Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
$(e)$ When viewing through a compound microscope,our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?