A stationary observer O looking at a fish (in water of `mu=4//3` ) through a converging lens of focal length `90.0cm. ` the lens is allowed to falll freely from a height of `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^(-1)` . Initially, it was at a depth of `44.00cm. ` Finally the velocity with which the fish appears to move to the observer at `t=0.2sec. `

Answer

`2475cm//s`

Explanation

At `t-0.2s, ` velocity of lens `V_(i)=g t=2ms^(-1)` (downwards)
Therefore, for lens, the fish appears to approach with a speed of
`2xx(1xx(3)/(4))=(11)/(4)ms^(-1)`
At a distance of `[42+(24)/(((4)/(3)))]=60cm`
`:. ` Image of fish from lens, `V=(-60xx90)/(-60+90)=-180cm`
`:. ` Velocity of image w.r.t. lens
`V_(1)=((v^(2))/(u^(2)))(du)/(dt)=((-180)/(-60))^(2)xx(11)/(4)=99/4 m//s -2475cms^(-1)`