Answer
$-15\times$.
Work Step by Step
The focal length of the eyepiece is found using the definition of the power of a lens, equation 23–7.
$$f_e=\frac{1}{P_e}=\frac{1}{19D}=0.0526m=5.26cm$$
As explained on page 724, when the viewing eye is relaxed, and assuming a faraway object and image, the distance between the lenses of a telescope is the sum of the focal lengths of the objective and of the eyepiece.
$$\mathcal{l}=f_o+f_e$$
Solve for the focal length of the objective.
$$f_o=\mathcal{l}-f_e=85cm-5.26cm=79.7cm$$
Finally, solve for the magnification of the telescope using equation 25–3.
$$M=-\frac{f_o}{f_e}=-\frac{79.7cm}{5.3cm}=-15\times$$
