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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

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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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