Two plane mirrors of lingth L are separated by distance L and a man `M_(2)` is standing at distance L from the connecting line of mirrors as shown in figure. A man `M_(1)` , is walking in a straight line at distance 2 L parallel to mirrors at speed u,then man at O will be able to see image of `M_(1)` ,for total time:

Options

A. `(4L)/(u)`
B. `(3L)/(u)`
C. `(6L)/(u)`
D. `(9L)/(u)`

Answer

C

Explanation

`tantheta=(PR)/(0R)=(AC)/(PC) rArr (3L//2)/(L)=(AC)/(2L) rArr AC=3L`
`tanalpha=(QR)/(OR)=(BD)/(DQ) rArr BD=(L//2)/(L)xx2L=L`
`therefore AB=AD-BD=(3L+L)-L=3L`.
For `A` to `B` observ can observe `M_(1)`
Making similar diagram for lower mirror
Total visible distance `=3L+3L=6L`
Total time `=(6L)/(u)` .