A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart.

Solution:

The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.

Distance between the objective mirror and the secondary mirror, d = 20 mm

Radius of curvature of the objective mirror, R1 = 220 mm

Hence, focal length of the objective mirror, $f_{1}=\frac{R_{1}}{2}=110 \mathrm{~mm}$

Radius of curvature of the secondary mirror, R1 = 140 mm

Hence, focal length of the secondary mirror, $f_{2}=\frac{R_{2}}{2}=\frac{140}{2}=70 \mathrm{~mm}$

The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.

Hence, the virtual object distance for the secondary mirror, $u=f_{1}-d$

$=110-20$

$=90 \mathrm{~mm}$

Applying the mirror formula for the secondary mirror, we can calculate image distance (v)as:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f_{2}}$

$\frac{1}{v}=\frac{1}{f_{2}}-\frac{1}{u}$

$=\frac{1}{70}-\frac{1}{90}=\frac{9-7}{630}=\frac{2}{630}$

$\therefore v=\frac{630}{2}=315 \mathrm{~mm}$

Hence, the final image will be formed 315 mm away from the secondary mirror.