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 water surface, the radius of this circle, in cm, is

Question

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 water surface, the radius of this circle, in cm, is (A) 36 √7 (B) (36/√7) (C) 36 √5 (D) 4 √5

Tardigrade Admin · Accepted Answer

The situation is shown in figure. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e0a74ac08fedb3e90c84c013b287272c-.png" /> sin θC=(1/μ), hence, tan θC=(A B/A O) ∴ A B=O A tan θC or A B=(O A/√μ2-1)=(12/√((4)3)2-1)=(36/√7)

Q. A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of water is $\frac{4}{3}$ and the fish is $12 c m$ below the water surface, the radius of this circle, in $c m$ , is

$367$

$\frac{36}{7}$

$365$

$45$

The situation is shown in figure.

$sin θ_{C} = \frac{1}{μ}$ , hence, $tan θ_{C} = \frac{A B}{A O}$
$∴ A B = O A tan θ_{C}$
or $A B = \frac{O A}{μ ^{2} - 1} = \frac{12}{( \frac{4}{3} ) ^{2} - 1} = \frac{36}{7}$

Q. A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of water is 34​ and the fish is 12cm below the water surface, the radius of this circle, in cm, is

367​

7​36​

365​

45​