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Asked by sarveshvibrantacademy | 18 May, 2019, 11:13: AM
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Figure describes the question.  Man-2 is standing at a distance L before the line joining the mirror.
Each mirror is of length L and they are separated by a distance L. Man-2 is watching the Man-1 moving along the line CEFD
which is at a distance 2L from mirrors. Speed of Man-2 is u m/s.
 
As shown in figure, Man-2 starts seeing Man-1, when he is C.  At that instant light ray is reflected at A with angle of incidence θ.
Man-1 will  see Man-2 till Man-2 reaches the point D. At that instant light ray is reflected at B with angle of incidence θ.
 
By considering reflected ray AO and normal AE, we get tanθ = 3/2.
 
Now if we consider tanθ = 3/2 in ΔAEC, we get CE = 3L. Similarly FD = 3L.
 
Hence total length CD = 3L+3L+3L = 9L.
 
Hence time of viewing Man-1 by Man-2 = 9L/u
Answered by Thiyagarajan K | 18 May, 2019, 12:44: PM

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