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

`22.75 cm//s` (upwards)

Explanation

At `t=0.2 sec, ` velocity of lens
`V_(1)=g t=2ms^(-1)` (downward)
For lens, the fish appers to approach with a speed of
`2+(1xx(3)/(4))=(11)/(4)ms^(-1)`
at distance of `(42+(24)/((4)/(3)))=60cm`
`:.` Image of the fish from the lens,
`V=(-60xx90)/(-60+90)=-180cm`
`:.` Velocity of image w.r.t. the lens,
`V_(i)=((v^(2))/u^(2))(du)/(dv)=((-180^(2))/(-60))xx(11)/(4)=(99)/(4)ms^(-1)`
Velocity of image w.r.t. the observer is
`V_(1)-2=(99)/(4)-2(91)/(4)ms^(-1)`
`=22.75 cms^(-1)` (upward)