Question

A small fish $0.4m$ below the surface of a lake, is viewed through a simple convering 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. Find the distance of image of fish as seen by observer.
$({\mu }_{water}=\frac{4}{3})$

• A. A distance of $0.2m$ from the water surface
• B. A distance of $0.6m$ from the water surface
• C. A distance of $0.3m$ from the water surface
• D. The same location of fish

Answer 1

The correct option is D The same location of fish

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