Please wait...
Contact Us
Need assistance? Contact us on below numbers

For Enquiry

9:00am - 9:00pm IST all days.

Business Inquiry (North)

Business Inquiry (West / East / South)



Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number


Mon to Sat - 10 AM to 7 PM


Asked by sarveshvibrantacademy 18th May 2019, 11:13 AM
Answered by Expert
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 Expert 18th May 2019, 12:44 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp