1. Image formed by the lens in the first pass

For the convex lens, $f=+0.5\,\text{m}, \qquad u=-1\,\text{m}$ (using Cartesian sign convention, object on the left gives negative object distance)

Lens formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Substitute values: $\frac{1}{0.5}=\frac{1}{v}-\left(-\frac{1}{1}\right)$ $2=\frac{1}{v}+1$ $\frac{1}{v}=1$ $v=1\,\text{m}$

So the lens forms a real image $1\,\text{m}$ to the right of the lens.

2. Position of this image relative to the mirror

The mirror is $2\,\text{m}$ to the right of the lens. So the image formed by the lens is $1\,\text{m}$ in front of the mirror.

Thus for the plane mirror, the object is at a distance of $1\,\text{m}$ in front of it. A plane mirror forms an image at the same distance behind it.

Hence the mirror forms an image $1\,\text{m}$ behind the mirror.

3. This mirror image acts as object for the lens on return path

Now light reflects back and passes through the lens again. For the lens, the object is now on its right side.

Distance of mirror image from lens: $2+1=3\,\text{m}$

So for the second pass through the lens, the object is at $u=+3\,\text{m}$ (on the right side of the lens, so positive in Cartesian convention)

Again use lens formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ $\frac{1}{0.5}=\frac{1}{v}-\frac{1}{3}$ $2=\frac{1}{v}-\frac{1}{3}$ $\frac{1}{v}=2+\frac{1}{3}=\frac{7}{3}$ $v=\frac{3}{7}\,\text{m}$

Since $v>0$, the final image is formed on the left side of the lens? Let us interpret carefully.

Using the Cartesian convention with light now traveling from right to left, positive $v$ means image on the left of the lens, i.e. on the side from which refracted rays emerge. Therefore the final image is real.

4. Distance of final image from the mirror

Mirror is $2\,\text{m}$ to the right of the lens, and final image is $\frac{3}{7}\,\text{m}$ to the left of the lens.

Therefore distance from mirror: $2+\frac{3}{7}=\frac{14+3}{7}=\frac{17}{7}\,\text{m}$ $\frac{17}{7}\approx 2.43\,\text{m}$

So the physically correct result is:

• distance from mirror $=\dfrac{17}{7}\,\text{m}\approx 2.43\,\text{m}$
• nature: real

5. Comparison with options

None of the given options matches exactly. The closest in nature is option A, but the distance should be about $2.43\,\text{m}$, not $2.6\,\text{m}$.

Therefore, the stored answer appears inconsistent with the calculation.