(a)
Hint: Recall the properties of a magnifying glass

Step: Find the nature of the image produced by the magnifying glass.
A magnifying glass helps one see the objects placed closer than the least distance of distinct vision. When an object is placed closer than the least distance of distinct vision, it magnifies it and forms its magnified image at the near point. In that case, the angles subtended by the object and the image are equal.  A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.

(b)
Hint: Angular magnification is independent of image distance.

Step: Find the angular magnification when the distance between the eye and a magnifying glass is increased.
When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended by the object at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.

(c)
Hint: \(\text{focal length}\propto \frac{1}{\text{Magnifying power}}\)

Step: Find the cons of very small focal length.
The lenses with very small focal lengths are very difficult to make. The focal length of a convex lens cannot be decreased by a greater amount. A convex lens having a very small focal length produces spherical and chromatic aberrations.

(d)
Hint: \(m=-\frac{v_{0}D}{u_{0}f_{e}}\)

Step: Find the magnifying power of a compound microscope.
The magnifying power of a compound microscope is given by:

$\mathrm{m}=-\frac{{\mathrm{v}}_0}{{\mathrm{u}}_0}\left(\frac{\mathrm{D}}{{\mathrm{f}}_{\mathrm{e}}}\right) \mathrm{for} \mathrm{normal} \mathrm{adjustment}$
$\mathrm{m}=\mathrm{m}=-\frac{{\mathrm{v}}_0}{{\mathrm{u}}_0}\left(1+\frac{\mathrm{D}}{{\mathrm{f}}_{\mathrm{e}}}\right) \mathrm{for} \mathrm{final} \mathrm{image} \mathrm{at} \mathrm{D}.$

For large magnification, both ${\mathrm{f}}_{\mathrm{e}} \mathrm{and} {\mathrm{u}}_0$ should be small. An object is placed at a distance slightly more than the focal length of the objective. So ${\mathrm{f}}_{\mathrm{o}}$ should also be small as ${\mathrm{u}}_0$ is small.

(e)
Hint: Eye-ring is the best position for viewing.

Step: Find the position of eyes for viewing the compound microscope.
When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much-refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred. The best position of the eye for viewing through a compound microscope is at the eye ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.