A Cassegrainian telescope uses two mirrors as shown in Fig. Such a telescope is built with the mirrors `20 mm` apart. If the radius of curvature of large mirror is `220 mm` and the small mirror is `40 mm`, where will the final image of an object at infinity be ?

Explanation

Here, radius of curvature of objective mirror
`R_(1) = 220 mm` ltbr. Radius of curvature of secondary mirror
`R_(2) = 140 mm` ,
`f_(2) = (R_(2))/(2) = (140)/(2) = 70 mm`
Distance between the two mirrors, `d = 20 mm`.
When object is at infinity, parallel rays falling on objective mirror, on reflection, would collect at its focus at
`f_(1) = (R_(1))/(2) = (220)/(2) = 110 mm`
Instead, they fall on secondary mirror at `20 mm` from objective mirror.
`:.` For secondary mirror, `u = f_(1) - d = 110 - 20 = 90 mm`
From `(1)/(v) + (1)/(u) = (1)/(f_(2))`
`(1)/(v) = (1)/(f_(2)) - (1)/(u) = (1)/(70) - (1)/(90) = (9 - 7)/(630) = (2)/(630)`
`v = (630)/(2) = 315 mm = 31.5 cm` to the right of secondary mirror.