In a compound microscope,let $u_0$ and $v_0$ be the object distance and image distance respectively. The objective of focal length $f_0$ magnifies a tiny object into a real,inverted image. The linear magnification of the objective is

Which of the following statements are true in the context of a Compound Microscope?

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 $cm$)

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?

The image formed by an objective lens of a compound microscope is

An angular magnification (magnifying power) of $30X$ is desired using an objective of focal length $1.25 \, cm$ and an eyepiece of focal length $5 \, cm$. How will you set up the compound microscope?