Question

# Two plane mirrors of length L are separated by distance L and a man M2 is standing at distance L from the connecting line of mirrors as shown in figure. A man M1 is walking in a straight line at distance 2L parallel to mirrors at speed U. The time for which man M2 at O will be able to see image of M1 is

A
8LU
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B
3LU
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C
6LU
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D
nLU
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Solution

## The correct option is C 6LUThe 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 M1 (farthest point) By AAA criterion, ΔABC and ΔCOD are similar. i.e ODBC=CDM1B i.e L2L=3L2x i.e x=3L So net perpendicular distance from M1 to horizontal line passing through OD =3L+3L2=9L2 Now, the nearest point from mirror Again by similarity of triangles L2L=L2x i.e x=L and. Perpendicular distance to the horizontal line passing through OD =L+L2=3L2 So difference in the distances =9L2−3L2=3L By symmetry, distance covered by M1=6L Total time taken =6LU

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