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?

(N/A) Although the image size is larger than the object,the angular size of the image is equal to the angular size of the object. $A$ magnifying glass allows us to see objects placed closer than the least distance of distinct vision $(25\, cm)$. $A$ closer object subtends a larger angle at the eye. Thus,the magnifying glass provides angular magnification by allowing the object to be placed much closer to the eye than the naked eye can resolve.
$(b)$ Yes,the angular magnification changes. When the distance between the eye and the magnifying glass is increased,the angular magnification decreases slightly because the angle subtended at the eye becomes less than the angle subtended at the lens.
$(c)$ We cannot indefinitely decrease the focal length of a convex lens because manufacturing lenses with extremely small focal lengths is technically difficult. Furthermore,lenses with very small focal lengths suffer from severe spherical and chromatic aberrations,which distort the image quality.
$(d)$ The angular magnification of the eyepiece is given by $m_e = (1 + D/f_e)$. Thus,a smaller $f_e$ results in higher magnification. For the objective lens,the magnification is $m_o \approx L/f_o$. To achieve high total magnification $(M = m_o \times m_e)$,both $f_o$ and $f_e$ must be small.
$(e)$ If the eye is placed too close to the eyepiece,the field of view is restricted,and we cannot collect all the refracted light,leading to a blurred image. The eye should be placed at the 'eye-ring' (the position of the exit pupil) to capture the maximum light and obtain the best field of view.