Question

A small fish 0.4 m below the surface of a lake, is viewed through a simple converging lens of focal length 2m. 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 $({\mu }_{water}=\frac{4}{3})$

• A. A distance of 1.2 m from the water surface
• B. A distance of 0.46m from the water surface
• C. A distance of 0.3 m from the water surface
• D. The same location of fish

Answer 1

The correct option is B A distance of 0.46m from the water surface

Apparent distance of fish from lens $u=0.2+\frac{h}{\mu }$
$=0.2+\frac{0.4}{\frac{4}{3}}=0.5m$
From $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\Rightarrow \frac{1}{(+2)}=\frac{1}{v}-\frac{1}{(-0.5)}$
$v=-0.66m$
The image of the fish is still where the fish is 0.46m below the water surface.