This is not the cool picture. I made that with a LEGO minifig based on this awesome image of a guy in a pool – http://i.imgur.com/im37Rwn.jpg. Yes. You must click that link and see the awesome picture.
What’s going on? In short, the light coming from the underwater part of his body refracts as it goes from water to air. This makes an image appear to the right of where the body actually is located. I did the same thing above with the LEGO figure and a shallow container of water. It’s a cool trick.
But where is the guy actually at? Sure, he’s right where his head is – but can I get an estimate for the pool depth? First, the physics. When light is in a material like water, it has a slower apparent speed. This speed change causes the light to bend – we call this refraction. The amount of light bending depends on the type of material and the incident angle. This can be described with Snell’s Law:
Here the n‘s represent the index of refraction for the two materials. This is just the ratio of the speed of light in a vacuum to the apparent speed of light in that material. For the angles, this diagram will help.
It doesn’t matter which direction the light is traveling, the lines and angles will still follow Snell’s Law.
Now let’s look at the guy in the pool. Here is a top view along with the camera.
Why does the image appear at that location? It’s because both cameras and humans are stupid (not really). All we can do is trace the light back to where we think it comes from. Even though the path of light bends, our eyes just make the assumption that it doesn’t.
How do you get the largest separation between head and body? The angular difference between these two objects (the image and the object) would be θ1 – θ2. Of course, Snell’s law also says there is a relationship between these two angles. Let me assume that the index of refraction in air is 1. This means I can write:
As θ1 increases, so does the angular separation between the head and body. Let me just go ahead and make a plot of the angular separation as function of incident angle. That’s just what I like to do. Oh, let’s say the index of refraction for water is 1.33 – that’s important for the plot below.
If you look at the pool wall straight on – the incident angle would be zero degrees and there would be no angular shift (duh). As you look more at an angle, the angular difference between increases. But what happens as you move the person closer to the boundary of water-glass? In the diagram, this distance is labeled as s. Well, you can see above that the solution for the angular size does not depend on s. However, something would change. The angular distance between the head and the body would stay the same but the angular size of both the head and increase. So, the head and body would stay in their relative locations but they would get bigger. The bigger they get, the less the angular shift seems to be.
There is something that changes as the object (or person) moves further back away from the wall – the apparent depth. Look at this diagram of the light after it reflects off an object in the water.
If you look at the light rays from the edges of the object, they both refract away from the object at the boundary. Tracing these rays back puts the image closer to the boundary than the actual object. It isn’t too hard to derive, but you can show that this apparent depth is:
If you are really careful, you might be able to use this to measure the actual depth of the person in the pool. Measure the angular size of both the part of his neck out of the water and under the water. Since they are both actually his real neck, in real life they should be the same width. This will give you a ratio of the image size and the object size. Well, that won’t give you the depth, but it could give you the index of refraction.
You could also estimate the index of refraction by measuring the angular distance between the head and body. All you would need to do is to estimate the incident angle for the camera looking at the water.