A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?

Given:


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


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


Therefore, focal length of the objective mirror, f1 = 0.5 × R1 = 110 mm


Radius of curvature of the secondary mirror, R2 = 140 mm


Therefore, focal length of the secondary mirror, f2 = 0.5 × R2 = 70 mm


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


Let the virtual object distance for the secondary mirror be u,


u = f1 – d …(1)


u = 110 – 20


u = 90 mm


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


…(2)


By putting the values in equation (2), we get,



v = 315 mm


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


NOTE: Cassegrainian telescopes are one of the most widely used telescope of current times. It is a combination of a parabolic primary mirror and secondary hyperbolic mirror.


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