Question

A stationary observer O looking at a fish (in water of $\mu =4/3)$ through a converging lens of focal length 90.0 cm. The lens is allowed to fall freely from a height of 62.0 cm with its axis vertical. The fish and the observer are on the principal axis of the lens. The fish moves up with constant velocity $100m{s}^{-1}$. Initially, it was at a depth of 44.0 cm. Find the velocity with which the fish appears to move to the observer at $t=0.2sec$

• A. 12.75 cm/s (upwards)
  
• B. 22.75 cm/s (upwards)
  
• C. 29.75 cm/s (upwards)
  
• D. 52.75 cm/s (upwards)
  

Answer 1

The correct option is B 22.75 cm/s (upwards)

Given: A stationary observer O looking at a fish (in water of $\mu =\frac{4}{3}$) 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 $100m/s$. Initially, it was at a depth of $44.0cm$.

To find the velocity with which the fish appears to move to the observer at $t=0.2sec$

Solution:

At $t=0.2sec$, velocity of lens

$V_1=gt=2m/s$ (downward)

For lens, the fish appears to approach with a speed of

$2+(1\times \frac{3}{4})=\frac{11}{4}m/s$

at distance of $(42+\frac{24}{\frac{4}{3}})=60cm$

Therefore image of the fish from the lens,

$V=\frac{-60\times 90}{-60+90}=-180cm$

Therefore the velocity of image w.r.t the lens

$V_1=\left(\frac{v^2}{u^2}\right)\frac{du}{dt}=\left(\frac{-{180}^2}{-60}\right)\times \frac{11}{4}=\frac{99}{4}m/s$

Velocity of image w.r.t the observer is

$V_1-2=\frac{99}{4}-2=\frac{91}{4}m/s=22.75cm/s$ (upward)

is the velocity with which the fish appears to move to the observer