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Step-by-Step Solution
Step 1: Identify the given data
• Object distance from the convex lens, $u = -1\,\text{m}$ (using the convention that light initially travels from left to right).
• Focal length of the convex lens, $f = +0.5\,\text{m}$.
• A plane mirror is placed $2\,\text{m}$ behind the lens.
Step 2: Find the image formed by the lens on the first pass
Using the lens formula:
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$
Substituting $f = +0.5\,\text{m}$ and $u = -1\,\text{m}$:
$$
\frac{1}{0.5}
= \frac{1}{v}
- \left(\frac{1}{-1}\right)
\quad\Longrightarrow\quad
2
= \frac{1}{v} + 1
$$
Hence,
$$
\frac{1}{v} = 1
\quad\Longrightarrow\quad
v = 1\,\text{m}
$$
So the lens forms the first image $1\,\text{m}$ behind the lens.
Step 3: Consider the plane mirror’s effect
The plane mirror is $2\,\text{m}$ behind the lens, so the first image (at $1\,\text{m}$ behind the lens) is $1\,\text{m}$ in front of the mirror. A plane mirror creates its image behind itself at the same distance as the object is in front. Therefore, the mirror forms an image:
$$
1\,\text{m} \text{ (behind the mirror)}
= \bigl(2 + 1\bigr)\,\text{m}
= 3\,\text{m} \text{ behind the lens.}
$$
This “mirror image” now serves as a new object for the lens but from the right side (i.e., at $+3\,\text{m}$ from the lens center).
Step 4: Find the final image formed by the lens with the mirror image as the object
When the lens sees this object at $3\,\text{m}$ on its right, the light effectively travels right-to-left for this second pass. In the usual sign convention, an object on the right side of the lens (with light incident from the right) is assigned $u = -3\,\text{m}$. Again, using the lens formula:
$$
\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
\quad\Longrightarrow\quad
\frac{1}{0.5} = \frac{1}{v} - \frac{1}{-3}
$$
$$
2 = \frac{1}{v} + \frac{1}{3}
\quad\Longrightarrow\quad
\frac{1}{v} = 2 - \frac{1}{3} = \frac{5}{3}
\quad\Longrightarrow\quad
v = \frac{3}{5} = 0.6\,\text{m}.
$$
A positive $v$ in this right-to-left scenario means the final image is formed $0.6\,\text{m}$ to the left of the lens (the same side as the original object).
Step 5: Calculate the final image distance from the mirror
Since the plane mirror is $2\,\text{m}$ to the right of the lens, the distance from the final image (at $0.6\,\text{m}$ to the left of the lens) to the mirror is:
$$
2.0\,\text{m} + 0.6\,\text{m} = 2.6\,\text{m}.
$$
This final image is real (it appears on the same side as the original object and can be obtained on a screen).
Answer
The final image is located $2.6\,\text{m}$ from the mirror and is real, matching Option (1).
(Reference figure)
