A diverging lens with a focal length of 20 cm and a converging...

For the final image to form at the location of the object itself, the light rays must retrace their path after reflection from the mirror.

This condition is satisfied if the rays from the lens strike the mirror normally, which requires the image formed by the lens to be at the center of curvature of the mirror.

The focal length of the converging mirror is $f_M = 10 \text{ cm}$, so its radius of curvature is $R = 20 \text{ cm}$. The center of curvature is $20 \text{ cm}$ in front of the mirror.

Since the distance between the lens and the mirror is $5 \text{ cm}$, the center of curvature is located $20 - 5 = 15 \text{ cm}$ in front of the lens.

The diverging lens must form a virtual image of the object at this position, so the image distance for the lens is $v = -15 \text{ cm}$.

Using the lens formula $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f_L}$ with $f_L = -20 \text{ cm}$:

$\dfrac{1}{-15} - \dfrac{1}{u} = \dfrac{1}{-20}$

$\dfrac{1}{u} = \dfrac{1}{20} - \dfrac{1}{15}$

$\dfrac{1}{u} = \dfrac{3 - 4}{60} = -\dfrac{1}{60}$

$u = -60 \text{ cm}$

The object should be placed $60 \text{ cm}$ from the lens on the side further away from the mirror.

Answer: $60 \text{ cm}$ from the lens on the side further away from the mirror