Q. A fish looking up through the water sees the outside world, contained in a circular horizon. $\frac{4}{6}$ and the fish is 12 cm below the water surface, the radius of this circle in cm is:
AIEEEAIEEE 2005Ray Optics and Optical Instruments
Solution:
The situation is shown in figure.
$sin\,\theta_{C}=\frac{1}{\mu}$
$tan\,\theta_{C}=\frac{AB}{AO}$
$AB=QA\,tan\,\theta_{C}$
$\therefore AB=OA\,tan\,\theta_{C}$
or $AB=\frac{OA}{\sqrt{\mu^{2}-1}}$
$=\frac{12}{\sqrt{\left(\frac{4}{3}\right)^{2}-1}}=\frac{36}{\sqrt{7}}$
