# Total Internal Reflection and snell's law

1. Aug 1, 2012

### schaefera

When we talk about Snell's law, and total internal reflection in particular, we usually will draw diagrams as if light is coming off a point in a single, straight line (that bends at an interface, of course).

My question is, though, how does this light behave when it's coming off an extended object? For example, if I stand at the side of a pool and there is a person, say, under the surface, then how do I know whether or not I will see him? If the light that hits my eyes comes from beyond the critical angle, then I shouldn't be able to see anything from under the water there. But isn't it possible that some of the light he emits (well, reflects but it's as if he emits it) will come from above the critical angle while some will come below it?

As in, shouldn't the diagram we draw of light coming from this person really include lines going at all angles up to surface, some from every angle, and then once they hit the interface they will bend according to Snell's law, but rays from seemingly many different points will reach my eyes, not just from one as we usually draw in diagrams?

This is also what often confused me about deriving Snell's law from Fermat's principle-- how can we single out an individual line of light and ignore all the others? Not all light reaching my eyes goes through a single point on the interface, does it?

2. Aug 1, 2012

### Staff: Mentor

You can consider every point on that person to be emitting light in every direction. This would be extremely difficult to draw and would be a big mess. Instead you could draw lines from several points on the person to your eye and see if they will reflect off the surface of the water or not. You could stand at a certain point where part of the person is visible and part reflects.

You could do this, but generally we only care about the rays entering our eyes that were emitted or reflected from the object, not ALL rays entering our eyes.

We have to single it out in order to calculate it. We can usually take one ray and calculate whether it reflects or not and then apply it to all the nearby rays as long as the difference in the angles are small enough. This of course isn't an exact answer, so you wouldn't do this if you were designing lenses or something.

The light emitted from different points will take different paths to your eye, so no, they do not all travel through the same point on the interface.

3. Aug 1, 2012

### schaefera

So really, in this situation, I could actually imagine all the light going the opposite direction (as if drawing arrows FROM my eye TO the object), and then use Snell's law to figure the angles out, keeping in mind that the critical angle cannot be surpassed. Even though this is backwards from what is happening when I see someone it the water, it's a good way to figure out just where the boundary of what I can see lies?

4. Aug 1, 2012

### Staff: Mentor

No, total internal reflection happens when you go from a medium with a higher index of refraction to a medium with a lower index of refraction. If you do it the other way you will not get any total internal reflection, which would be incorrect.

5. Aug 2, 2012

### schaefera

But I'm saying the paths they take are symmetric as long as you remember that at a cretain angle you hit total reflection-- so we can treat the light as going either way keeping this in mind as a way to think about which light reaches our eyes (not for anything more than a mnemonic, though)

6. Aug 2, 2012

### jbriggs444

There is no "blind spot" (assuming that there is no shoreline to block line of sight).

Light from underwater that hits just shy of the critical angle will emerge just a bit above the horizontal. Pick just the right angle and the ray will refract just right to hit the observers eye, no matter how low that eye is.

Light from underwater that hits at or past the critical angle will be reflected, but that's OK. That just means that the image that the underwater guy sees will be concentrated in a limited viewing angle. He can see everything above-water. But he sees it all as if by looking through a "fish-eye" lens -- mapped to a finite viewing area.

[Hence the name "fish-eye", I would expect]

7. Aug 2, 2012

### schaefera

That would mean there is a blind spot, though? You say that past the critical angle all light will be reflected.

8. Aug 3, 2012

### jbriggs444

That's not a blind spot. That's an angle where you can look and not see out.

But the entire outside world is nonetheless visible to you -- compressed down to fit within angles at which you can look and see out.

9. Aug 3, 2012

### Staff: Mentor

You can think of it however you like, but you will run into problems if you try to trace the rays from your eye to the object. I don't know of any reason to either, it doesn't seem to make anything simpler and if you forget that you have that critical angle you're screwed.