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  4. A small fish 0.4 m below the surface of a lake, is viewed through a simple converging lens of focal length 3 m. The lens is kept at 0.2 m 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 text water =(4/3))

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Q. A small fish $0.4\, m$ below the surface of a lake, is viewed through a simple converging lens of focal length $3\, m$. The lens is kept at $0.2 \,m$ 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 $\left(\mu_{\text {water }}=\frac{4}{3}\right)$Physics Question Image

Ray Optics and Optical Instruments

Solution:

Apparent distance of fish from lens
$u =0.2+\frac{h}{\mu} $
$=0.2+\frac{0.4}{4 / 3}=0.5\, m $
From $ \frac{1}{f} =\frac{1}{v}-\frac{1}{u}$
$\Rightarrow \frac{1}{(+3)} =\frac{1}{v}-\frac{1}{(-0.5)} v=-0.6\, m$
The image of the fish is still where the fish is $0.4\, m$ below the water surface.