Question

A smallfish, 0.4m below the surface of a lake is viewed through a simple converging lens of focal length 3m. The lens 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. Refractive index of the water is 4/3.

Answer 1

Hi
As apparant distance is= 0.2+h/u(refractive index)
=0.2+0.4/4/3=0.5
from lens formula focal length is 3cm +ve and u is -ve 0.5
then putting in formula
1/f=1/v-1/u
we get
1/3=1/v-1/(-0.5)
= -0.6m
The image of the fish is still where the fish 0.4m below the water surface
Ty
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Answer 2

Answer:

At the actual position of the fish

Explanation:

after refraction, it is formed at 30 m below water

after refraction it is formed 60 m below the lens

also the lens is 60 m above water