1. Set up coordinates

Take the mirror to lie in the plane $x=0$, with its center at $y=0$ and vertical width ignored since motion is along the horizontal direction shown.

• Mirror extends from $y=-\dfrac d2$ to $y=+\dfrac d2$.
• Source $S$ is at distance $L$ in front of the mirror, so let $S=(L,0).$
• Its virtual image in the plane mirror is at $S'=(-L,0).$
• The man walks along a line parallel to the mirror at distance $2L$, so his eye position is $M=(2L,y).$

The man can see the source in the mirror iff the straight line joining $M$ to the virtual image $S'$ meets the mirror segment.

2. Equation of the line joining $M(2L,y)$ and $S'(-L,0)$

We need the point where this line cuts the mirror plane $x=0$.

Using section formula / linear interpolation:

• From $x=-L$ to $x=2L$, total change in $x$ is $3L$.
• The mirror plane $x=0$ is at a distance $L$ from $S'$ toward $M$.
• So it divides the segment in the ratio $\frac{L}{3L}=\frac13$ from $S'$ to $M$.

Hence the corresponding $y$-coordinate at $x=0$ is $y_{\text{hit}}=0+\frac13(y-0)=\frac y3.$

3. Condition for visibility

For the ray to strike the mirror, this intersection point must lie within the mirror width: $-\frac d2 \le \frac y3 \le \frac d2.$

Multiplying by $3$, $-\frac{3d}{2} \le y \le \frac{3d}{2}.$

Therefore, the total distance over which the man can walk and still see the image is $\frac{3d}{2}-\left(-\frac{3d}{2}\right)=3d.$

4. Check options

• A: $d$ ❌
• B: $\dfrac d2$ ❌
• C: $3d$ ✅
• D: $2d$ ❌

So the correct answer is Option C.