Problem — The fish's strange sky

Problem

The fish’s strange sky. A fish looks upward from beneath the smooth surface of a pond. Its effective refractive index is that of water, $n \approx 1{,}33$, against air with $n \approx 1$. (a) Calculate the critical angle beyond which rays coming from outside can no longer enter the water. (b) Sketch the ray diagram the fish sees: what is the shape of the “window” through which the objects of the outside world reach it? What does it see instead when looking outside that circular window? (c) Estimate the radius of that window if the fish is at a depth of $1{,}0\;\text{m}$.

Solution

(a) Critical angle. By reversibility of rays, the water↔air critical angle is the same critical angle: $\sin\theta_c = n_\text{aria}/n_\text{acqua}$:

$\sin\theta_c = \frac{1}{1{,}33} = 0{,}752 \quad\Rightarrow\quad \ev{\theta_c \approx 48{,}8^\circ}$

(b) Snell’s window. All the light coming from the hemisphere of the sky (180°) is compressed, by refraction, within a cone of half-angle $\theta_c \approx 48{,}8^\circ$ about the vertical. The fish therefore sees the entire outside world within a bright circle (the “Snell’s window”) above itself. Outside that circle the angle exceeds $\theta_c$: total internal reflection occurs and the fish sees the bottom of the pond reflected, as in a mirror.

(c) Radius of the window. At depth $h = 1{,}0\;\text{m}$, the edge of the cone intersects the surface at a distance

$r = h\,\tan\theta_c = 1{,}0\cdot\tan 48{,}8^\circ = 1{,}0\cdot 1{,}14 \approx \ev{1{,}15\;\mathrm{m}}$

Adapted from (Povey 2015, §11.5).

Linked atoms

Argomenti: Optics Concetti: Refraction and Snell’s law Competenze: Limiting-case and thought-experiment analysis