Question

A fish in a lake (refractive index of water is $\frac{4}{3}$) is viewed through a convex lens. From water surface, the lens is placed in air at half of the distance of the fish from the water surface, so that the image is formed at the fish itself. The focal length of the lens is how many times the depth of fish in water?

Answer 1

plane water surface,
$\frac{1}{v}-\frac{4/3}{(-u)}=\frac{1-4/3}{\infty }=0$
$v=-0.75u$
for convex lens,
$u_1=-(0.75+0.5)u=-1.25u$
$v_1=-(u+0.5u)=-1.5u$
rays are coming from air and are viewed through air.
from $\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f}\Rightarrow -\frac{1}{1.5u}-\frac{1}{(-1.25u)}=\frac{1}{f}$
$f=7.5u$