A stationary observer O looking at a fish F in water `(mu_(w)=4//3)` through a converging lens of focal length `90.0cm` . The lens is allowed to fall frelly from a height `62.0cm` with its axis vertical. The fish and the observer are on the principal axis of the lens. The fish moves up with constant velocity `100cm//s`. Initially it was at depth of `44.0cm` . Find the velocity (in cm//s) with which the fish appears to move with respect to lens to the observer at `t=0.2 s`. (take `g=10m//s^(2))`

Answer

22

Explanation

At `t=0.2 sec` , velocity of lens
`V_(l)=gt=2m//s` (downwards)
`therefore` For lens the fish appears to approach with a speed of `2+(1xx(3)/(4))=(11)/(4)m//s`
at a distance of `(42+(24)/((4/3)))=60 cm `

`therefore` distance of image from lens, `V=(-60xx90)/(-60+90)=-180 cm` .
`therefore` Velocity of image `w.r.t` . lens `V_(i)=(v^(2)/(u^(2)))(du)/(dt)=((-180)/(-60))^(2)xx(11)/(4)=(99)/(4)m//s`
velocity of image `w.r.t`. observer `=V_(i)-2=(99)/(4)-2(91)/(4)m//s=22.75 m//s` (upwards)