A small fish 0.4 m below the surface of a lake is viewed through a simple converging lens of focal length 3m. The lens is kept at 0.2m above the water surface such that fish lies on the optical axis of the lens. The image of the fish seen by observer will be at $(μ_{water} = \frac{4}{3})$

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A distance of 0.2 m from the water surface

A distance of 0.6m from the water surface

A distance of 0.3 m from the water surface

The same location of fish

answer is D.

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Detailed Solution

Apparent distance of fish from lens $u = 0 .2 + \frac{h}{μ}$
$= 0 .2 + \frac{0 .4}{\frac{4}{3}} = 0 .5 m$ From $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{(+ 3)} = \frac{1}{v} - \frac{1}{(- 0 .5)}$

$\Rightarrow v = - 0 .6 m$
The image of the fish is still where the fish is 0.4 m below the water surface.

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