As shown in the figure, a point source S is positioned midway...

For a single image to be formed, the images produced by both mirrors must coincide.

Let the distance between the mirrors be $d$. The point source $S$ is placed midway, so the object distance for each mirror is $u = -\dfrac{d}{2}$.

Using the mirror formula for one of the concave mirrors (with focal length $F = -f$):
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{F}$
$\dfrac{1}{v} - \dfrac{2}{d} = -\dfrac{1}{f}$
$\dfrac{1}{v} = \dfrac{2}{d} - \dfrac{1}{f} = \dfrac{2f - d}{df}$
$v = -\dfrac{df}{d - 2f}$

The negative sign indicates the image is formed at a distance $\dfrac{df}{d - 2f}$ in front of the mirror. By symmetry, the second mirror forms its image at the same distance.

Case 1: The images coincide at a finite distance.
For the two images to coincide, they must meet at the same point in space. By symmetry, they must meet at the midpoint. Thus, the distance of each image from its respective mirror must be $\dfrac{d}{2}$.
$\dfrac{df}{d - 2f} = \dfrac{d}{2}$
$\dfrac{2f}{d - 2f} = 1 \Rightarrow 2f = d - 2f \Rightarrow d = 4f$
Here, the images coincide exactly at the position of the source $S$.

Case 2: The images coincide at infinity.
This occurs when the reflected rays are parallel, meaning the object is placed at the principal focus of both mirrors.
$\dfrac{d}{2} = f \Rightarrow d = 2f$

Thus, the values of $d$ for which only a single image is formed are $2f$ and $4f$.

Answer: $2f, 4f$