Q. A fish, looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface of water, the radius of the circle in centimetre is
A) $\frac{12 \times 3}{\sqrt{5}}$
B) $12 \times 3 \times \sqrt{5}$
C) $\frac{12 \times 3}{\sqrt{7}}$
D) $12 \times 3 \times \sqrt{7}$
Explanation
(c) From figure, $\tan{C = \frac{r}{12}}$
$or\ r = 12\tan C$
$or\ r = \frac{12\sin C}{\sqrt{1 - \sin^{2}C}}$
$r = \frac{12 \times \frac{1}{\mu}}{\sqrt{1 - \frac{1}{\mu^{2}}}} = \frac{12}{\sqrt{\mu^{2} - 1}} = \frac{12}{\sqrt{\left( \frac{4}{3} \right)^{2} - 1}\ }$
$ie,r = \frac{12 \times 3}{\sqrt{7}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.