A converging mirror M_1 , a point source S , and a diverging mirror...

For the converging (concave) mirror $M_1$, the focal length is $f_1 = -20 \text{ cm}$.
For the diverging (convex) mirror $M_2$, the focal length is $f_2 = +20 \text{ cm}$.
Let the distance between the two mirrors be $d$.

Consider the first reflection at $M_1$:
The object (source $S$) is at a distance of $30 \text{ cm}$ from $M_1$, so $u_1 = -30 \text{ cm}$.
Using the mirror formula for $M_1$:
$\dfrac{1}{v_1} + \dfrac{1}{u_1} = \dfrac{1}{f_1}$
$\dfrac{1}{v_1} - \dfrac{1}{30} = -\dfrac{1}{20}$
$\dfrac{1}{v_1} = \dfrac{1}{30} - \dfrac{1}{20} = -\dfrac{1}{60}$
$v_1 = -60 \text{ cm}$
The first image $I_1$ is formed $60 \text{ cm}$ in front of $M_1$.

Now, consider the second reflection at $M_2$:
The image $I_1$ acts as a virtual object for $M_2$. Since $I_1$ is $60 \text{ cm}$ from $M_1$ and the mirrors are separated by $d$, $I_1$ is located at a distance of $(60 - d)$ behind $M_2$.
Thus, the object distance for $M_2$ is $u_2 = +(60 - d)$.

The final image is formed at the position of the source $S$ itself, which is at a distance of $(d - 30)$ in front of $M_2$.
Thus, the image distance for $M_2$ is $v_2 = -(d - 30)$.

Using the mirror formula for $M_2$:
$\dfrac{1}{v_2} + \dfrac{1}{u_2} = \dfrac{1}{f_2}$
$\dfrac{1}{-(d - 30)} + \dfrac{1}{60 - d} = \dfrac{1}{20}$
$\dfrac{-(60 - d) + (d - 30)}{(d - 30)(60 - d)} = \dfrac{1}{20}$
$\dfrac{2d - 90}{90d - d^2 - 1800} = \dfrac{1}{20}$

Cross-multiplying yields:
$40d - 1800 = 90d - d^2 - 1800$
$d^2 - 50d = 0$
$d(d - 50) = 0$

Since $d \neq 0$, we get $d = 50 \text{ cm}$.
The distance separating the two mirrors is $50 \text{ cm}$.

Answer: $50 \text{ cm}$