A small fish, 0.4m below the surface of a lake, is viewed through a simple converging lens of focal length 3m. The lens in kept at 0.2m above the water surface such that the fish lies on the optical axis of the lens. Find the image of the fish seen by the observer. The refractive index of water is `4//3`.

Explanation

`"Apparent depth of water "=("real depth")/(mu)=(0.4)/(4//3)=0.3 m` Since lens is kept at a height of 0.2 m above the surface of water hence distance of bulb from lens becomes `0.3 m + 0.2 m = 0.5 m`.
Now we can use lens formula as follows :
`(1)/(v)-(1)/(u)=(1)/(f)`
`rArr" "(1)/(v)-(1)/(-0.5)=(1)/(3)`
`rArr" "(1)/(v)=(1)/(3)-2`
`rArr" "v=-(3)/(5)=-0.6m`
Hence, location of the final image of bulb is 0.6 m below the lens and this happens to be the actual position of object.