An optical arrangement consists of two concave mirrors M$$_1$$ and M$$_2$$, 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 M$$_1$$ and M$$_2$$ are 20 cm and 24 cm, respectively. The distance between L and M$$_2$$ is 20 cm. A point object S is placed at the mid-point between L and M$$_2$$ on the axis. When the distance between L and M$$_1$$ is $$n/7$$ cm, one of the images coincides with S. The value of $$n$$ is ______.

For reflection from $$M_2$$:

$$\frac{1}{v} + \frac{1}{(-10)} = \frac{1}{(-12)}$$

$$\frac{1}{v} = \frac{1}{10} - \frac{1}{12}$$

$$v = +60\text{ cm}$$ (for $$I_1$$)

For refraction from $$L$$:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} - \frac{1}{(-80)} = \frac{1}{10}$$

$$v = +\frac{80}{7}$$ (For $$I_2$$)

This image should be at focus of $$\text{M}_1$$ $$\implies$$ $$\therefore \frac{20}{2} + \frac{80}{7} = \frac{n}{7}$$ $$\implies$$ $$n = 150$$

Also, If $$I_2$$ is formed at pole of $$M_1$$ then $$\frac{n}{7} = \frac{80}{7}$$ $$\implies$$ $$n = 80$$

And further if $$\text{I}_2$$ is formed at centre of curvature of $$\text{M}_1$$ then $$\frac{n}{7} = \frac{80}{7} + 20$$ $$\implies$$ $$\therefore n = 220$$