A point source of light,S,is placed at a dist | vedclass

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(C) Let the mirror be placed along the $y$-axis from $y = -d/2$ to $y = d/2$. The source $S$ is at $(L, 0)$. The image $S'$ of the source $S$ is forme

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$3d$

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(C) Let the mirror be placed along the $y$-axis from $y = -d/2$ to $y = d/2$. The source $S$ is at $(L, 0)$. The image $S'$ of the source $S$ is formed at $(-L, 0)$.The man walks along the line $x = 2L$. The rays from the image $S'$ that reach the man must pass through the edges of the mirror.The rays from $S'$ passing through the top edge $(0, d/2)$ and bottom edge $(0, -d/2)$ of the mirror define the field of view.Using similar triangles,the height $h$ of the field of view at a distance $x = 2L$ from the mirror is given by the ratio of distances.The distance from the image $S'$ to the mirror is $L$,and the distance from the image $S'$ to the man is $L + 2L = 3L$.By similar triangles,the width of the field of view $h$ is given by $\frac{h}{d} = \frac{3L}{L} = 3$.Therefore,$h = 3d$.

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