A cassegrain telescope is built with the mirrors 20 mm apart. If the radius of curvature of the larger mirror is 220 mm and the small mirror is 140 mm. where will the final image of an object at infinity be ?

Explanation

Here, radius of curvature of objective mirror `R_(1)=220mm` radius of curvature of secondary mirror `R_(2)=140mm`,
`f_(2)=(R_(2))/(2)=(140)/(2)=70mm`
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)=110mm`
Instead, they fall on secondary mirror at 20 mm from objective mirror.
`therefore" For secondary mirror, u"=f_(1)-d=110-20="90 mm"`
From `(1)/(upsilon)+(1)/(u)=(1)/(f_(2))`
`(1)/(upsilon)=(1)/(f_(2))(-1)/(u)=(1)/(70)-(1)/(90)=(9-7)/(630)=(2)/(630)`
`upsilon=(630)/(2)="315 mm = 31.5 cm to the right of secondary mirror."`