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

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(A) The man can see the image of the source $S$ in the mirror as long as he remains within the field of view of the image.Let the mirror be $AB$ with

Vedclass · Accepted Answer

$6$

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(A) The man can see the image of the source $S$ in the mirror as long as he remains within the field of view of the image.Let the mirror be $AB$ with length $L = 20\,cm$. The source $S$ is at a distance $d_1 = 10\,cm$ from the mirror.The man walks along a line at a distance $d_2 = 20\,cm$ from the mirror. The total distance of the man from the source $S$ is $d_1 + d_2 = 10\,cm + 20\,cm = 30\,cm$.By using similar triangles,the width of the field of view $PQ$ at a distance $d_1 + d_2$ from the mirror is given by:$\frac{PQ}{L} = \frac{d_1 + d_2}{d_1} = \frac{30\,cm}{10\,cm} = 3$$PQ = 3 	imes L = 3 	imes 20\,cm = 60\,cm$.The man walks at a speed $v = 10\,cm/s$.The time $t$ for which he can see the image is $t = \frac{PQ}{v} = \frac{60\,cm}{10\,cm/s} = 6\,s$.

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