# Why can't we see around corners?

1. Oct 23, 2012

### Erland

Why can't we see around corners? Or perhaps the right question should be: Why can we see anything at all other than blurred light where no shapes can be distinguished?

If we use a wave theory of light, these things seem mysterious (not so with a particle theory of light).

Let me explain by giving an example:

Assume that we are standing in complete darkness just behind a shining flashlight. Then we will see an illuminated conical region extending from the flashlight. This region is just a continuation of the shape of the flashlight. But the reason that we can see this illuminated region is that light is reflected by the air. In vacuo, we would not see anything at all, since we are not inside this cone. The light from the flashlight would then not reach outside this cone.

By why is it so? By the Huygens principle (considered as proved mathematically by Kirchhoff), every point to which the light reaches is the source of a new wave, extending spherically from this point. From every point inside the cone, sufficiently much in front of the flashlight, such a secondary spherical wave would reach our eyes. Thus, the light would leak around the edge of the flashlight, and by the same reasoning, we would be able to see around corners (although perhaps only blurred light, not any shapes).

Why is it not so? The only explanation I can think of would be that the contributions from all these secondary waves cancel each other out by interference at every point outside the cone. I know this is the case for points behind a wave front, which the wave has already passed, but I have never seen any proof (mathematical, based upon the wave equation) that this would happen outside such a cone. Does anybody know of such a proof?

As another example, consider light that just has passed the lense in an eye. Agian, light will spread by such secondary waves, and it is very hard to see how these waves can give rise to an image at the retina, instead of just being blurred.

So, the wave theory of light seems problematic. I don't doubt its truth, I just wonder how these things can be explaned, mathematically, from the wave equation. (So perhaps, this question belongs in the Mathematics section, but I post it here anyway.)

2. Oct 23, 2012

### HallsofIvy

You seem to be thinking that those "secondary waves" have a real existence. They don't. The waves referred to by the Huygens' principle are entirely imaginary, simply devices to help calculating wave fronts.

3. Oct 23, 2012

### CWatters

Under controlled conditions you can get defraction around an edge/corner..

http://www.mike-willis.com/Tutorial/PF7.htm

You can't see around the corner due to the small percentage of energy defracted and the fact that corners are not nice optically perfect sharp edges.

It's another matter for sound.

4. Oct 23, 2012

### Erland

Even so, I don't understand why the light waves from the flashlight must stay within the cone formed by the edge of the flashlight.

CWatters, I am aware of diffraction, but here we talk about objects who are large compared to the wavelengths of visible light.

5. Oct 23, 2012

### A.T.

Of course it is. Otherwise we wouldn't have the particle theory of light.

As for you question: I think it is a matter of wave length. Radio waves are not as easy to focus as visible light. Maybe with shorter waves there is more cancellation and the energy dissipates quicker.

6. Oct 23, 2012

### Born2bwire

That's the point. At high frequencies, the wave behavior because minimalized and the wave behaves very much like a particle theory. A flashlight has a tight beam because it has a very large effective aperture. If I have a radio antenna, the wave for a single dipole antenna is omnidirectional. However, if I construct an array of antennas then I can direct the beam. How tightly I direct the beam is generally proportional to the effective/physical aperture of the array. That is, the larger the electrical size of the array and the more elements that make it up, then the tighter the beam. This is because the large number of sources can create the necessary interferences to concentrate the power into a pencil beam, despite the fact that the individual sources are still omnidirectional.

7. Oct 23, 2012

### Erland

This might be the heart of the matter. But it is in no way obvious that it is as you say (at least not to me). Therefore, I would like a mathematical proof of this, based upon the Wave Equation. Can someone supply or give a reference to such a proof, preferably on the net?

8. Oct 23, 2012

### Born2bwire

At best, you may want to look at high frequency approximation methods. Something for example like geometric optics, physical optics, unified theory of diffraction, etc. The problem becomes that solving the wave equation at high frequencies becomes very difficult. Instead of diong the full wave equation, we make approximations based upon the electrical size and frequency of the incident radiation. But these approximations drop more and more of the wave nature so that it becomes increasingly like a particle theory.

9. Oct 23, 2012

### AJ Bentley

Performing this sort of calculation using the wave equations is difficult and complicated.
It's the sort of thing a Physicist would leave to the Engineering Department.

There's nothing (physically) interesting about it, nothing to learn, just a lot of numbers to crunch.

If you are having trouble convincing yourself that it's correct, try a simple experiment.
Find a big tray that you can fill with water. About half way down it, build a dam across with a gap in the middle then with the tip of your finger, make ripples at one end and see how they react when they hit the gap.

See how much of the wave spills around the corner. Depending on how wide the gap is, you'll find that it forms a cone, just like your flashlight.

10. Oct 23, 2012

### reasonableman

I believe this is because Huygens principle is incomplete. From wikipedia (http://en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle): [Broken]
"[Huygens] assumed that the secondary waves travelled only in the "forward" direction and it is not explained in the theory why this is the case"

The inability do define and explain 'forward' suggests there is a problem with this theory.

Last edited by a moderator: May 6, 2017
11. Oct 23, 2012

### Erland

Yoy mean someting like this?
One can clearly see a conical bounday in this image, but I don't know how it should be interpreted. Does the darker region indicate more or less intensity? In the latter case, this seems to contradict what you wrote about the experiment.

Unfortunately, I have no possibility to conduct such an experiment right know, but thanks for the suggestion.
However, I am a mathematician who wants a mathematical proof, no matter how complicated it is.

12. Oct 23, 2012

### Drakkith

Staff Emeritus
Read the first line in the very next paragraph. It explains that Fresnel took Huygens work and used it along with his theory of interference to explain exactly what you asking.

Last edited by a moderator: May 6, 2017
13. Oct 23, 2012

### Drakkith

Staff Emeritus
Erland, light DOES diffract around corners. However the number of photons that do so is very small with a large opening. Also, and perhaps most importantly are the following reasons we can't see around corners.

1. Light from everywhere else in your field of view completely drowns out the occasional photon that diffracts at a high enough angle around the corner to make it to your eye.

2. The light that does diffract around the corner is badly aberrated and mixed with light from every other point around the corner to such an extent that you cannot form a recognizable image even if the intensity was high enough for you to see. This is difficult for me to explain if you don't know anything about optics. See the following picture: http://en.wikipedia.org/w/index.php?title=File:Lens3.svg&page=1

For every infinitesimal point in your field of view, draw rays from it to the edge of your pupil in the manner like that image shows. If that "cone" of light is badly abberated or if other light is mixed in with it, much like driving into a sunrise/sunset with a very dirty windshield, then you cannot form a proper image of the object.

Last edited: Oct 23, 2012
14. Oct 23, 2012

### Erland

Yes, I understand that. But, as I understand it (correct me if I am wrong), in the limit when the wavelength tends to 0, there will no diffraction and no light (or radiation) will leak around corners.

What I am now struggling to understand, that is how one can go from a wavefront, with a width, to a ray, with zero width in the limit.

It is a little peculiar that there are two completely opposite idealizations of electromagnetic radiation: one is using planar waves, which extend through all of space, and have infinite width, and the other one is to use rays (geometric optics) with zero width. Interestingly enough, phenomena like reflection and refraction are very well explained for both of these. But what happens in between?

Anyway, I'll look up the links you all have provided, and other links I found myself. Thank you all!

15. Oct 23, 2012

### Drakkith

Staff Emeritus
Rays are not real. They are used to simplify optical calculations. Do not try to determine what light is doing by imagining it as a ray unless you just want a very general answer. Light IS an electromagnetic wave. End of story. (Until you get into quantum mechanics)

16. Oct 23, 2012

### Staff: Mentor

Did you check Wikipedia?
http://en.wikipedia.org/wiki/Fraunhofer_diffraction_(mathematics [Broken])

Last edited by a moderator: May 6, 2017