Answer –

Below is the diagram of a Cassegrain telescope. It has a concave and a convex mirror.

We are given –

Distance between the secondary mirror and the objective mirror is d = 20 mm

Radius of curvature of the objective mirror is R₁ = 220 mm

Thus, the focal length of the objective mirror is f₁ = $\frac{R_{1}}{2}$ = 110 mm

Radius of curvature of the secondary mirror is R₂ = 140 mm

Therefore, focal length of the secondary mirror is f₂ = $\frac{R_{2}}{2}$

= $\frac{140}{2}$

= 70 mm

The image , formed by the objective mirror of an object placed at infinity will act as a virtual object for the secondary mirror.

Therefore, the virtual object distance for the secondary mirror is given by –

u = f₁ – d = 110 – 20

= 90 mm

According to the mirror formula for the secondary mirror, we can determine the image distance (v) as:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f_{2}}$

$\frac{1}{v}$ = $\frac{1}{f_{2}}$ – $\frac{1}{u}$

= $\frac{1}{70}$ – $\frac{1}{90}$ = $\frac{2}{630}$

Therefore, we have –

v = $\frac{630}{2}$ = 315 mm

Hence, the distance between the final image and the secondary mirror is 315mm.