$A$ Cassegrain telescope uses two mirrors as shown in the figure. Such a telescope is built with the mirrors $20 \; mm$ apart. If the radius of curvature of the large mirror is $220 \; mm$ and the small mirror is $140 \; mm$,where will the final image of an object at infinity be?

(N/A) The following figure shows a Cassegrain telescope consisting of a concave mirror (objective) and a convex mirror (secondary).
Distance between the objective mirror and the secondary mirror,$d = 20 \; mm$.
Radius of curvature of the objective mirror,$R_1 = 220 \; mm$.
Hence,the focal length of the objective mirror,$f_1 = \frac{R_1}{2} = 110 \; mm$.
Radius of curvature of the secondary mirror,$R_2 = 140 \; mm$.
Hence,the focal length of the secondary mirror,$f_2 = \frac{R_2}{2} = \frac{140}{2} = 70 \; mm$.
The image of an object placed at infinity,formed by the objective mirror,acts as a virtual object for the secondary mirror. The distance of this virtual object from the secondary mirror is $u = f_1 - d = 110 - 20 = 90 \; mm$.
Applying the mirror formula for the secondary mirror,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f_2}$,where $f_2$ is positive for a convex mirror:
$\frac{1}{v} + \frac{1}{90} = \frac{1}{70}$
$\frac{1}{v} = \frac{1}{70} - \frac{1}{90} = \frac{9 - 7}{630} = \frac{2}{630}$
$v = \frac{630}{2} = 315 \; mm$.
Thus,the final image is formed $315 \; mm$ away from the secondary mirror.