Two plane mirrors of length $L$ are separated by distance $L$ and a man $M_2$ is standing at distance $L$ from the connecting line of mirrors as shown in figure. A man $M_1$ is walking in a straight line at distance $2L$ parallel to mirrors at speed $U$. The time for which man $M_2$ at O will be able to see image of $M_1$ is

The correct option is C $\frac{6L}{U}$
The set-up is symmetric about the line passing through O. So we need to consider only the upper half.

Consider the following ray coming from $M_1$ (farthest point)


By AAA criterion, $\Delta ABC$ and $\Delta COD$ are similar.

i.e $\frac{OD}{BC}=\frac{CD}{M_{1}B}$
i.e $\frac{L}{2L}=\frac{3L}{2x}$
i.e $x=3L$

So net perpendicular distance from $M_1$ to horizontal line passing through $OD$

$=3L+\frac{3L}{2}=\frac{9L}{2}$

Now, the nearest point from mirror


Again by similarity of triangles

$\frac{L}{2L}=\frac{L}{2x}$
i.e $x=L$ and.
Perpendicular distance to the horizontal line passing through OD
$=L+\frac{L}{2}=\frac{3L}{2}$

So difference in the distances $=\frac{9L}{2}-\frac{3L}{2}=3L$

By symmetry, distance covered by $M_1=6L$

Total time taken $=\frac{6L}{U}$