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`
radius of curvature of secondary mirror
`R_1 = 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.