Question

A stationary observer $O$ locking at a fish (in water of $\mu =4/3$) through a converging lens of focal length $90.0cm$. The lens is allowed to fall 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 $44.0cm$. Find the velocity with which the fish appears to move to the observer at $t=0.2\sec$.

Answer 1

At $t=0.2\sec$, velocity of lens
$V_l=gt=2m{s}^{-1}\left(downward\right)$
For lens, the fish appears to approach with a speed of $2+(1\times \frac{3}{4})=\frac{11}{4}m{s}^{-1}$
at distance of $(42+\frac{34}{\left(\frac{4}{3}\right)})=60cm$
$\therefore$ Image of the fish from the lens,
$V=\frac{-60\times 90}{-60+90}=-80cm$
$\therefore$ Velocity of image w.r.t the lens,
$V_i=\left(\frac{v^2}{u^2}\right)\frac{du}{dt}=\left(\frac{-{180}^2}{-60}\right)\times \frac{11}{4}=\frac{99}{4}m{s}^{-1}$
Velocity of image w.r.t the observer is
$V_1-2=\frac{99}{4}-2=\frac{91}{4}m{s}^{-1}$
$=22.75cm{s}^{-1}\left(upward\right)$.