Question

A Cassegrain telescope uses two mirrors as shown in Fig. Such a telescope is built with the mirrors 2omm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

Answer 1

Given,
radius of curvature of objective mirror , R1 = 220mm
so, focal length of objective mirror, f1 = 110mm
radius of curvature of secondary mirror, R2 = 140mm
so, focal length of secondary mirror, f2 = 70mm
distance between two mirrors , d = 20mm

when object is placed at infinity , parallel rays fall on objective mirror, after reflection would collect at its focus f1 .
in turn they fall on secondary mirror which is at distance of 20mm from objective mirror.
so, for secondary mirror,
object distance, u = f1 - d = 110mn - 20mm = 90mm
now use formula,
1/v + 1/u = 1/f2

1/v = 1/f2 - 1/u

= 1/70 - 1/90

= (9 - 7)/630

= 1/315

v = 315mm

hence, Image is formed 315mm to the right of secondary mirror.

Answer 2

Given in the question :-

Here the Curvature of objective mirror R₁ = 220 mm

curvature of secondary mirror R₂ = 140 mm

The distance between these mirror , d = 20 mm

Now ,

focal length of the objective mirror , f₁ = R₁ /2 = 110 mm

focal length of the secondary mirror , f₂ = R₂/2 = 70 mm .

Secondary mirror , u = (100 - 20) = 90 mm.

Now , 1/v + 1/u = 1/f₂

1/v = 1/70 - 1/90

1/ v = (2) / 630

v = 315 mm

v = 31.5 cm

The final image will be at 31.5 cm

Hope it helps :-)