Question

A small fish $0.4m$ 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 $\left({\mu }_{\text{water }}=\frac{4}{3}\right)$

• 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

Answer: (D)

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.