Answer
(a) diverging
(b) converging
(c) distant object: -0.20 diopters; near objects: +2.3 diopters
Work Step by Step
(a) We know that the diverging lens can produce an image of a distant object at a Librarian's far point distance; therefore, the Librarian should wear glasses with diverging lenses.
(b) We know that a converging lens in front of the eye can correct farsightedness. The convex lens focuses light from an object inside the near point to produce an image that is beyond the near point and hence the eye can focus on the image of the object.
(c) We know that
$d_i=5.0m-0.020m=4.98m$
and $\frac{1}{f}=\frac{1}{\infty}+\frac{1}{-4.98m}$
$\implies \frac{1}{f}=-0.2diopters$
Now the object distance from the lens is
$d_{\circ}=0.25m-0.020m=0.23m$
and the image distance from the lens is
$d_i=-(0.50m-0.020m)$
$\implies d_i=-0.48m$
Now we can find the refractive power of eyeglass lenses for the near point object as
$\frac{1}{f}=\frac{1}{0.23m}-\frac{1}{0.48m}=2.3diopters$
