A Cassegrain telescope uses two mirrors as shown in figure below. Such a telescope is built with the mirrors 20 mm 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?

Fig. Schematic diagram of a reflecting telescope (Cassegrain)

Solution

Here given,
Radius of curvature of objective mirror,R₁ = 220 mm
Radius of curvature of secondary mirror, R₂= 140 mm
Distance between the two mirror, d = 20 mm
We can calculate the focal length of both mirrors.

∴

Focal length of objective mirror, f₁ =

\frac{R_{1}}{2} = \frac{220}{2} = 110 m m

Focal length of secondary mirror,

f_{2} = \frac{R_{2}}{2} = \frac{140}{2} = 70 mm

When object is at infinity, parallel rays falling on objective mirror, on reflection, would collect at its focus at f₁.
In turn, they fall on secondary mirror which is at a distance of 20 mm from objective mirror.

For secondary mirror,
Object distance, u = f₁ – d = 110 – 20 = 90 mm

Now, using formula,

\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{9 - 7}{630} = \frac{2}{630}

v = \frac{630}{2} = 315 mm = 31.5 cm

to the right of secondary mirror.

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