A converging mirror  $M_1$, a point source S and a diverging mirror  $M_2$ are arranged as shown in figure. The source is placed at a distance of 30 cm from $M_1$ . The focal length of each of the mirrors is 20 cm. Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself, then

For mirror  $M_1:$
 $\begin{array}{l}u=-30cm\\ f=-20cm\end{array}$
By mirror formula,      $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$      , we have
   $\frac{1}{-30}+\frac{1}{v}=\frac{1}{-30}$
Which on solving gives v = -60 cm.
For mirror $M_2:$ The image formed by mirror  $M_1$ behaves like virtual object for mirror $M_2$ .  Let it is at a distance x from the pole of mirror $M_2$ . Thus 
    $\begin{array}{l}{u}_2=+x\\ v_2=-(30-x)\end{array}$
Again by mirror formula, we have 
    $\frac{1}{x}+\frac{1}{-(30-x)}=\frac{1}{20}$
Which on solving gives x = 10 cm or 60 cm ,
X = 60 cm is not possible.  $\therefore x=10cm$
(a)  Thus the separation between the mirrors = 60-10 = 50 cm
(b)  The image formed by mirror  $M_2$ is at a distance 10 cm.