JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Geometrical Optics

Only the mirror strip of width $d$ reflects light, and the reflected rays look as though they diverge from the image $S^{\prime }$, which lies as far behind the mirror as $S$ is in front, a distance $L$ behind the mirror plane. The man walks along a line $2L$ in front of the mirror, so he is $L+2L=3L$ from the image. The two edges of the mirror cut out a diverging beam from $S^{\prime }$; the width of this beam where it crosses the man's path is scaled up from the mirror width in the ratio of the distances from the image, namely $\frac{3L}{L}=3$. So the band over which the man keeps seeing the reflected source is $3d$ wide, and that is the length over which he can see the image.