Snell's Window
You may or may not have seen the effect before, but it's there, over your head, every
time you dive. Look at the surface, and you'll probably see it: Snell's Window. It
manifests itself as a bright circle directly overhead, while all the water outside the
circle is darker. But there's another effect, one you'll see most when the surface
water is calm: inside the circle, you can see objects on the surface clearly. Outside the
circle, you can't.
So why does Snell's Window exist, and exactly how big is it?
Let's start with the equation that governs the bending (or "refraction") of
light when it passes from one medium to another. Look at the diagram below, which
shows a light beam going from air into water. When doing optics calculations, the
convention is to measure angles from the "normal,"an imaginary line
perpendicular to the surface in question. In the diagram, you'll see the line
between the light beam and the normal in air marked as Q_{i}
("i" is for "incident"), and the same thing on the water side as Q_{r} ("r" is for "refracted").
There is one other piece of information we need: the index of refraction of both air and
water. The index of refraction basically measures how much the medium in question
bends light. To be precise, it's the ratio of the speed of light in that medium to
the speed of light in a vacuum, but that's not important for this discussion.
Suffice it to say that denser materials have higher index of refraction numbers.
The equation we want is called "Snell's Law," after the mathematician who
discovered the law in the 17th century.Snell's Law will let us explain and
quantify Snell's Window, which is why the effect is named after the same person as the
law. Anyway, here is a sketch that shows the geometries.
Using the geometries above, here is Snell's Law:
n_{i} Sin Q_{i} = n_{r} Sin Q_{r}
Remember, n is the index of refraction, and Q is the angle
measured from the normal.
So how does this help us explain Snell's Window? Let's start by making a few
observations. First off, if a ray of light hits with zero angle (that is, perfectly
perpendicular to the water), it will pass through with no bending. However, if it
hits with some angle, the water will bend the light towards the
normal. That is, for whatever angle it hits with, after it passes into the water the
light will be closer to the normal than it had been. So if a ray of light hits at 30
degrees from vertical, it will end up in the water at a smaller angle, maybe 20 degrees
from the vertical. This is an important observation.
Why? Because the largest angle light can hit the water with is 90 degrees; at
that point, it's not hitting the water, but rather is parallelling it. So if we
plug 90 degrees into Snell's Law above, we can solve for the maximum angle from the
vertical that light can ever have underwater! Think about that: at an
incident angle of 90 degrees from the normal, light will still be bent towards the normal
by the water, so it will enter the water at an angle less than 90 degrees!
Here's the implication of that last statement:no direct light will enter your
eyes at an angle larger than this. That's not to say that you don't get any
light from greater angles, just that any light from greater angles must be scattering
rather than refracting. It won't come in the form of an image you can see, but as a random
and diffuse glow. Here's a sketch showing what happens:
You have to pay close attention to this sketch to understand it. Each ray
of light is numbered, on both the air and water sides. Incident ray #1 becomes
refracted ray #1, #2 becomes #2, etc. Notice how ray #1 is close to the
normal, and stays close to the normal. As the incident rays get farther from the
normal, so do the refracted raysbut the refracted rays get farther away
"slower" than the incident rays do. Notice how ray #4, which strikes the water
at an almost grazing angle in air, does not end up at anything like a grazing angle under
water!
So let's do some algebra, and get a number. We want to solve Snell's Law for Q_{r}. We are also interested in the special case of
light that just grazes the surface of the water, so plug90 degrees in for Q_{i}. Then divide both sides by n_{r}.
Also note than Sin(90) = 1, so the Sin Q_{i} factor
goes away. Then our equation is:
Sin Q_{r} = n_{i} / n_{r}
Now we solve for Q_{r} by taking the inverse sine of
both sides:
Q_{r} = sin^{1} (n_{i}/n_{r}).
For practical purposes, the index of refraction of air is 1.0, and the index of
refraction of water is 1.33. Then we get
Q_{r} = sin^{1} (1/1.33) = sin^{1}
(0.75) = 48.6 deg ~ 50 deg.
So the magic angle is around 50 degrees. Since that's measured from the vertical, it
means you get a cone of vision roughly 100 degrees wide over your head.
There are two things worth noting here. (1) Light will never actually strike at
an incident angle of 90 degrees, because it will reflect off the surface first. Thus
the 48.6 degrees we arrived at earlier is probably more like 45 or even 42 degrees.
Also, wave action at the surface will tend to 'smear' the actual incident angle, so
there will be a fringe area close to the edges of Snell's window where some light makes it
through and some doesn't. You can see this in pictures with Snell's Window in them; just
look near the edges of the Window, and you will see patterns of light and dark that come
from the waves. (2) Snell's Window has a corresponding inverse effect for light
travelling from water into air (or from an medium with high index of refraction to one
with low index). Just think about what would happen if the incident light ray were
from the water at the 48.6 degree angle we derived above: it would "exit" the
water at 90 degreesthat is, it would end up parallel to the surface of the water.
At higher angles, it would actually reflect back into the water. This effect,
known as "total internal reflection," is what makes fiber optic cables work.
