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, the radius of this circle in cm is (a) (36)/(√(7)) (b) 36√(7) (c) 4√(5) (d) 36√(5)

Answer: (a). surfacefishh = 12 cmrC Light from outside reaches the fish only inside a cone of half-angle C, the critical angle: sin C = 1/μ = 3/4. Radius of the circle: r = htan C = (h)/(√(μ² - 1)) r = (12)/(√(16/9 - 1)) = (12)/(√(7)/3) = (36)/(√(7)) cm So the answer is (a) (36)/(√(7)).

A fish looking up through the water sees the outside world contained… | JEE Main 2005 (AIEEE)

Q57JEE Main 2005 · AIEEEMedium

Total Internal Reflection

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 $12\text{ cm}$ below the surface, the radius of this circle in $\text{cm}$ is

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Answer: (a)

Light from outside reaches the fish only inside a cone of half-angle $C$ , the critical angle: $\sin C = \frac{1}{\mu} = \frac{3}{4}$ .
Radius of the circle: $r = h\tan C = \frac{h}{\sqrt{\mu^2 - 1}}$
$r = \frac{12}{\sqrt{\frac{16}{9} - 1}} = \frac{12}{\sqrt{7}/3} = \frac{36}{\sqrt{7}}\text{ cm}$
So the answer is (a) $\frac{36}{\sqrt{7}}$ .

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