Answer
(a) $109.8m$
(b) real
(c) $-5.50$
Work Step by Step
(a) We know that
$\frac{1}{f}=\frac{1}{d_{\circ}}+\frac{1}{d_i}$
This can be rearranged as:
$d_i=\frac{fd_{\circ}}{d_{\circ}-f}$
We plug in the known values to obtain:
$d_i=\frac{(16.9m)(20.0m)}{20.0m-16.9m}=109.8m$
Thus, the astronomer's image is located in front of the mirror.
(b) Since the distance of the image $d_i$ is positive, the image is real.
(c) We know that
$m=-\frac{d_i}{d_{\circ}}$
We plug in the known values to obtain:
$m=-\frac{110m}{20.0m}$
$m=-5.50$
