1. Image formed by the lens in the first pass

The object is placed $40\text{ cm}$ in front of a converging lens of focal length $20\text{ cm}$.

Using the lens formula:

For a convex lens, using Cartesian sign convention:

So,

So the lens forms a real image $40\text{ cm}$ to the right of the lens.

2. This image acts as object for the mirror

The mirror is placed $60\text{ cm}$ to the right of the lens. So the image formed by the lens is at a point:

in front of the converging mirror.

The converging mirror has focal length $10\text{ cm}$, so this object is at:

from the mirror.

For a concave (converging) mirror, an object at $2f$ forms its image at $2f$, same size.

Hence the mirror forms the image $20\text{ cm}$ in front of the mirror, i.e. again at the same point where the lens had formed the first image.

So after reflection, rays retrace back toward the lens as if coming from a point $40\text{ cm}$ to the right of the lens.

3. Second pass through the lens

Now this image acts as an object for the lens on its right side, at a distance $40\text{ cm}$ from the lens.

For the lens, light is now incident from the right, and the object for this pass is at $40\text{ cm}$ on the right side. Since this is again at $2f$ of the lens, a convex lens will form the image at $2f$ on the opposite side, with same size.

Therefore, the final image is formed $40\text{ cm}$ to the left of the lens.

Thus, the final image is:

• $40\text{ cm}$ from the convergent lens
• same size as the object

4. Check options

• A: $20\text{ cm}$ from the convergent mirror, same size as the object — this is the intermediate mirror image, not the final image. ❌
• B: $40\text{ cm}$ from the convergent mirror, same size as the object — incorrect. ❌
• C: $40\text{ cm}$ from the convergent lens, same size as the object — correct. ✅
• D: $20\text{ cm}$ from the convergent mirror, twice the size of the object — incorrect. ❌

So the correct answer is C.