The magnification power of a compound microscope is given in terms of the magnification of the objective $m_0$ and the magnification power of the eyepiece $m_E$. The total magnification 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?

The near vision of an average person is $25 \, cm$. To view an object with an angular magnification of $10$,what should be the power of the microscope?

For observing a cricket match,a binocular is preferred to a terrestrial telescope because

Which of the following statements are true in the context of a Compound Microscope?
$(A)$ Each lens produces a virtual and inverted image.
$(B)$ The objective has a very short focal length.
$(C)$ The eyepiece is used as a simple magnifying glass.
$(D)$ The objective and eyepiece are convex and concave lenses respectively.

$A$ compound microscope has a magnifying power of $30$. The focal length of its eyepiece is $5 \, cm$. Assuming the final image is formed at the least distance of distinct vision,the magnification produced by the objective is: