Question

A Cassegrain telescope uses two mirrors as shown in figure. 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?

• A. 235 mm away from small mirror
• B. 285 mm away from small mirror
• C. 305 mm away from small mirror
• D. 315 mm away from small mirror

Answer 1

Answer: (D)

Given, distance between objective mirror and another mirror
$d=20 \,mm$
Radius of curvature of objective mirror $=R_{1}=220 \,mm$
$\therefore $ Focal length of objective mirror, $f_{1}=\frac{220}{2}=110 \,mm$
Radius of curvature of small mirror $=R_{2}=140\, mm$
$\therefore $ Focal length of small mirror, $f_{2}=\frac{140}{2}=70\, mm$
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for small mirror.
So, the object distance for small mirror $u=f_{1}-d$
i.e., $u=110-20=90 \,mm\, v=315\, mm$ of $v =31.5\, cm$
Using mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f_{2}}$
$\frac{1}{v}=\frac{1}{f_{2}}-\frac{1}{u}=\frac{1}{70}-\frac{1}{90}$
$=\frac{9-7}{630}=\frac{2}{630}$
Thus, the final image is formed at $315\,mm$ away from small mirror.