A converging lens and a diverging mirror are positioned with a...

For the light to emerge parallel to the principal axis after passing through the converging lens, the virtual object for the lens (which is the image formed by the mirror) must lie at the focal point of the lens.

Since the light travels from the mirror to the lens, the focal point of the lens for this incident light is at a distance of $25 \text{ cm}$ from the lens, on the side of the mirror.

The separation between the lens and the mirror is $15 \text{ cm}$. Therefore, the image formed by the mirror must be located $25 - 15 = 10 \text{ cm}$ behind the mirror.

For the diverging (convex) mirror, the focal length is $f = +40 \text{ cm}$ and the image distance is $v = +10 \text{ cm}$.

Using the mirror formula:

$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{1}{10} + \dfrac{1}{u} = \dfrac{1}{40}$

$\dfrac{1}{u} = \dfrac{1}{40} - \dfrac{1}{10} = \dfrac{1 - 4}{40} = -\dfrac{3}{40}$

$u = -\dfrac{40}{3} \approx -13.33 \text{ cm}$

The point source must be placed at a distance of $13.33 \text{ cm}$ in front of the mirror.

The distance of the point source from the lens is:

$15 - 13.33 = 1.67 \text{ cm}$