Two plane mirrors of length L are separated b | vedclass

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(A) The light from man $M_1$ hits the mirrors and then reaches man $M_2$. The positions of the moving man $M_1$ need to be found when the light from h

Vedclass · Accepted Answer

$\frac{4L}{u}$

Vedclass · Answer

(A) The light from man $M_1$ hits the mirrors and then reaches man $M_2$. The positions of the moving man $M_1$ need to be found when the light from him hits the edges of the mirrors to reach the standing man $M_2$ at $O$.Let the line of motion of $M_1$ be at a distance $2L$ from the plane of the mirrors. Let the mirror edges be at $E, F, G, H$ as shown in the figure. The observer $O$ is at a distance $L$ from the gap between the mirrors.Using similar triangles for the upper mirror (edges $E$ and $G$):For edge $E$ (closer to $O$): The distance of $M_1$ from the line passing through $O$ parallel to the mirrors is $x_E$. By similar triangles,$\frac{x_E}{L} = \frac{2L}{L} \implies x_E = 2L$.For edge $G$ (further from $O$): The distance of $M_1$ from the line passing through $O$ is $x_G$. By similar triangles,$\frac{x_G}{L+L} = \frac{2L}{L} \implies x_G = 4L$.The length of the path visible through the upper mirror is $AB = x_G - x_E = 4L - 2L = 2L$.Similarly,for the lower mirror,the visible path length $CD$ is also $2L$.Total visible length $= AB + CD = 2L + 2L = 4L$.Total time $t = \frac{	ext{Total distance}}{u} = \frac{4L}{u}$.

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