A Cassegrain telescope uses two mirrors as shown in above figure. Such a telescope is built with the mirrors $20 m m$ apart. If the radius of curvature of the large mirror is $220 m m$ and the small mirror is $140 m m$ , where will the final image of an object at infinity be?

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Solution

Distance between the objective mirror and the secondary mirror, $d = 20 m m$
Radius of curvature of the objective mirror, $R_{1} = 220 m m$
Hence, focal length of the objective mirror, $f_{1} = R_{1} / 2 = 110 m m$
Radius of curvature of the secondary mirror, $R_{2} = 140 m m$
Hence, focal length of the secondary mirror, $f_{2} = R_{2} / 2 = 70 m m$
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.
Hence, the virtual object distance for the secondary mirror, $u = f_{1} - d = 110 - 20 = 90 m m$
Applying the mirror formula for the secondary mirror,
$1 / v + 1 / u = 1 / f_{2}$
$\Rightarrow v = 315 m m$
Hence, the final image will be formed $315 m m$ away from the secondary mirror.

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A Cassegrain telescope uses two mirrors as shown in above figure. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?

A Cassegrain telescope uses two mirrors as shown in above figure. Such a telescope is built with the mirrors $20 m m$ apart. If the radius of curvature of the large mirror is $220 m m$ and the small mirror is $140 m m$ , where will the final image of an object at infinity be?