Physics · Ch 6 — Optics
Radius of Illumination (Snell's Window)
Radius of Illumination (Snell's Window)
When a light source such as an electric bulb is placed inside a water tank, light travels outward from it in every direction inside the water. Light striking the water surface at an angle less than the critical angle refracts through and escapes into the air above; light striking at an angle greater than the critical angle undergoes total internal reflection and stays trapped inside the water. Because at every point of the surface some ray from the source manages to arrive at less than the critical angle, the entire surface, viewed from outside, appears illuminated. Looking at it from the reverse direction -- an observer's eye at depth underwater, looking upward -- only light arriving from within a cone of half-angle equal to the critical angle can reach the eye directly from outside air; this restricted, bright circular patch of view is called Snell's window, with radius (derived from the right-triangle geometry relating the depth , the radius , and the crit …
What this figure shows. A small light source such as an electric bulb sits inside a water tank, sending rays outward in every direction inside the water. Rays that strike the surface at an angle below the critical angle refract through and escape into the air; rays that strike beyond the critical angle undergo total internal reflection back into the water instead. From outside, this means the entire water surface appears lit up -- because at every point on the surface some ray from the source manages to arrive at less than the critical angle, righ …
What this figure shows. An observer's eye sits at depth d below a water surface, looking upward. Only light arriving from within a cone of half-angle equal to the critical angle ic can reach the eye directly from outside air (anything beyond that angle undergoes total internal reflection at the surface and cannot enter); the diagram uses the right triangle formed by the depth d, the horizontal radius R of this bright circular patch and the slant distance from the eye to the patch's edge to derive R = d tan(ic), which is then rewritten purely in terms of the refractive indices as R = d/sqrt(n^2-1) -- this bright circular patch, containing the entire co …
Worked out. An observer 10 m deep in a swimming pool of water (refractive index n = 4/3) looks upward on a sunny day. Using R = d/sqrt(n^2-1) with d = 10 m and n = 4/3 gives R = 10/sqrt((4/3)^2 - 1) = 10/sqrt(16/9 - 1) = 10 divided by the square root of 7/9, working out to R = 30/sqrt7 = 11.32 m -- the radius of the bright circular Snell's window patch seen overhead. The half-angle of the viewing cone is the critical angle itself, ic = sin inverse(1/n) = sin inverse(3/4) = 48.6 degrees, so the total angular width of the cone within which the observer can see the entire outside world is 2 times …