A Cassegrain telescope uses two mirrors as shown in fig. Such a telescope is built with the mirros 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, find the value of image distance of small mirror in cm.

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answer is 31.5.

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Detailed Solution

$\begin{array}{l} u = - \infty, \\ R = - 220 m m = - 22 c m \end{array}$

$\begin{array}{l} f = - 11 c m \\ \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \\ \frac{1}{- \infty} + \frac{1}{v} = - \frac{1}{11} o r v = - 11 c m \\ R e a l i m a g e f o r m e d b y o b j e c t i v e m i r r o r a c t s a s v i r t u a l o b j e c t f o r t h e s e c o n d a r y m i r r o r . \\ O b j e c t d i s t a n c e f o r s e c o n d a r y m i r r o r = (11 - 2) c m = 9 c m \\ N o w, u = + 9 c m, f = + 7 c m \\ \frac{1}{v} + \frac{1}{9} = \frac{1}{7} \Rightarrow v = \frac{9 \times 7}{7 - 9} = - 31.5 c m \end{array}$