An optical arrangement consists of two concave mirrors M₁ and M₂ and a convex lens L with a common principal axis, as shown in the figure. The focal length of L is 10 cm. The radii of curvature of M₁ and M₂ are 20cm and 24cm, respectively. The distance between L and M₂ is 20cm. A point object S is placed at the mid-point between L and M₂ on the axis. When the distance between L and M₁ is $\frac{\mathrm{n}}{7}\mathrm{cm}$, one of the images coincides with S. The value of n is _______.

$$\begin{gathered} \frac{1}{\mathrm{V}}+\frac{1}{-10}=\frac{1}{-12} \frac{1}{\mathrm{V}}=\frac{1}{10}-\frac{1}{12} \\ \mathrm{V}=+60{\mathrm{U}}^{\mathrm{\prime }}=60+20 \\ {\mathrm{U}}^{†}=80\frac{1}{{\mathrm{V}}^{\mid }}-\frac{1}{-80}=\frac{1}{10}\mathrm{V}=\frac{80}{7} \end{gathered}$$
This image should be at centre of curvature of M₁ (or) at pole of M₁ for the image to be formed at object itself 
$\therefore$distance between M & L can be $\frac{80}{7}\text{ Or }20+\frac{80}{7}=\frac{220}{7}$
$\therefore$n = 80 or 220
If lens forms first image then it becomes parallel light beam and falls on mirror which forms image at 10cm from M₁ then it again falls on lens and then again it M₂ and finally image is formed at object.  
$\frac{1}{80}+\frac{1}{\mathrm{x}-10}=\frac{1}{10}\frac{1}{\mathrm{x}-10}=\frac{8}{80}-\frac{1}{80}\mathrm{x}-10\frac{80}{7}\mathrm{x}=\frac{150}{7}$