A Cassegrain telescope uses two mirrors of radii of curvature 220 mm and 140 mm. The distance b/w the two mirrors is 20 mm. Where will the final image of an object at infinity be?

Explanation

By using the formula for mirrors,
`(1)/(f)=(1)/(u)+(1)/(v)` we get
Since, `f=r//2=(220)/(2)=110` mm
hence, `(1)/(110)=(1)/(oo)+(1)/(v)`
`therefore v=110`mm
virtual object distance for the second mirror `=(110-20)=90` mm
For the second mirror,
Hence `(1)/(-70)=(1)/(-90)+(1)/(v)`
`therefore (1)/(v)=(1)/(-70)+(1)/(90)=(-90+70)/(6300)`
or `v= -(6300)/(20)= -315` mm
The image is formed at 315 mm from the smaller mirror in the direction of light.