A converging lens with a focal length of 15 cm and a converging...

The light from the point source can form an image in two ways: by directly refracting through the lens, and by first reflecting from the mirror and then refracting through the lens.

For the first image (formed directly by the lens):
The source is at a distance of $40 \text{ cm}$ from the lens.
Using the lens formula $\dfrac{1}{v_1} - \dfrac{1}{u} = \dfrac{1}{f}$:
$\dfrac{1}{v_1} - \dfrac{1}{-40} = \dfrac{1}{15}$
$\dfrac{1}{v_1} = \dfrac{1}{15} - \dfrac{1}{40} = \dfrac{8 - 3}{120} = \dfrac{1}{24}$
$v_1 = +24 \text{ cm}$
The positive sign indicates that the image is formed on the side opposite to the object, which is away from the mirror.

For the second image (formed after reflection from the mirror):
The distance between the lens and the mirror is $50 \text{ cm}$. The source is $40 \text{ cm}$ from the lens, so it is $50 - 40 = 10 \text{ cm}$ from the mirror.
For the converging (concave) mirror, $u_m = -10 \text{ cm}$ and $f_m = -10 \text{ cm}$.
Using the mirror formula $\dfrac{1}{v_m} + \dfrac{1}{u_m} = \dfrac{1}{f_m}$:
$\dfrac{1}{v_m} - \dfrac{1}{10} = -\dfrac{1}{10} \Rightarrow \dfrac{1}{v_m} = 0 \Rightarrow v_m = \infty$
The reflected rays become parallel to the principal axis and travel towards the lens.
For the lens, these parallel rays act as an object at infinity ($u_2 = -\infty$).
Using the lens formula:
$\dfrac{1}{v_2} - \dfrac{1}{-\infty} = \dfrac{1}{15} \Rightarrow v_2 = +15 \text{ cm}$
Since the rays are coming from the mirror, they pass through the lens and converge on the other side, which is away from the mirror.

Thus, the two images are formed at $15 \text{ cm}$ and $24 \text{ cm}$ from the lens, both on the side away from the mirror.