# Why light diffract?

1. Mar 1, 2005

### scilover89

Why waves, such as light, will diffract when they pass through a small hole or slid?

2. Mar 1, 2005

### Staff: Mentor

It's because of interference between the light that goes through different parts of the slit or hole.

Remember two-slit interference? You add two waves (with different phases) together and they interfere. Now imagine dividing a single slit into lots of tiny "sub-slits," and add up the waves coming from each of those sub-slits. Each of those waves has a slightly different phase. Take the limit as the number of sub-slits becomes infinite, and the width of each slit becomes infinitesimally tiny. You end up with an integral that gives you the overall intensity.

Note that when we do two-slit interference, we usually start by assuming that the slits are very narrow so that their width doesn't matter, only the spacing between them. When we do take the slit width into account, we get a combination of the single-slit and double-slit interference patterns: the maxima in the double-slit pattern have different intensities, in a pattern that matches the single-slit pattern.

3. Mar 1, 2005

### ZapperZ

Staff Emeritus
Actually, this doesn't answer the question. If what you said is true, then how come you don't see such effects when the slit is WIDER? After all, why can't you still divide the slit into many, many more smaller slits? Why would the diffraction effects go away?

Note that a diffraction effect is really the "broadening" of the beam path. Within such broadening, you can get diffraction patterns. I've explained why this really is a manifestation of the uncertainty principle! Rather than repeat everything that I've written, I will simply refer to my journal entry below:

[11-15-2004 09:26 AM] - Misconception of the Heisenberg Uncertainty Principle

So diffraction is caused by the HUP. What causes the HUP? The 1st Quantization rule of QM that deals with non-commuting operators. What causes the 1st Quantization rule of QM? That, we don't quite know, and part of QM's postulate.

Zz.

4. Mar 1, 2005

### dextercioby

Well Zz,how come the classical theory of LIGHT DIFFRACTION (you know,the one made by Fresnel,Fraunhofer,Kirchhoff and Sommerfeld) is both correct & elegant,while the one proposed by Feynman (with those photons taking a bunch of paths) is horrible and noninteligeable...?

I'll say diffraction is a wave-like phenomenon and applying to it the famous duality,though correct,would complicate things beyond one's power of comprehension.It would be interesting to view water waves' diffraction through the eyes of the HUP...

The diffraction is so nice with Fresnel integrals and Bessel functions,why bring in operators and propagators...?

Daniel.

5. Mar 1, 2005

### ZapperZ

Staff Emeritus
No, no. You read way too much into what I just said. I said it can be EXPLAINED by the HUP. I didn't say anyone in their right mind would USE it. It is the same with the 2-slit expt. I can show you how to do that using photons, but why do that when the wave picture is so much more convenient for most purposes?

This is why we continue to treat EM radiation as waves in many cases. But the diffraction from a single slit is the CLEAREST example of the HUP, something that many people simply do not realize. Such effects are not just restricted to microscopic, atomic scale.

Zz.

6. Mar 1, 2005

### HallsofIvy

Staff Emeritus
A wave (light, sound, water, etc.) can always be thought of as constantly "regenerating" itself from every point.

If at some instant, the wave front is a plane, then Drawing an sphere out from each point with a fixed radius (representing the distance light would travel from that point) gives an "envelope" which is another plane parallel to the original one- the wave front at that later time.

If the wave is constrained to go through a pin pointopening (or a very small opening compared to the wave-length), the wave front will become a sphere centerd on that point.

If a plane wave meets a barrier with two pinpoint openings, the wave front on the other side of the barrier will be two spherical waves centered on the two points which can interfere with one another.

I remember doing that with water waves in highschool physics.

7. Mar 1, 2005

### dextercioby

Yes,Halls,that picture of Huygens' principle is really nice.Though i'd still vote for Kirchhoff's integral...

Daniel.

8. Mar 1, 2005

### Staff: Mentor

You still can!

As the slit becomes wider and wider, the "diffraction effects" become smaller and smaller relative to the overall intensity at each point on the screen. But they're still there, in principle. Try putting a point source behind a large aperture, and a screen close to the aperture. Look closely and you'll see diffraction effects at the edge of the shadow of the aperture's edge. They still exist in the middle of the "beam", but they're imperceptibly tiny.

You have to be careful when doing the integral in such a situation, though. When we derive the standard single-slit diffraction formula, we use the Fraunhofer approximation, which assumes that the rays from various points in the slit to the image point on the screen are practically parallel, and that all points in the slit are equidistant from a point source. If the slit is wide enough, this approximation breaks down.

We can do better with the Fresnel approximation. Look in an optics book or on the Web, and you can probably find some nifty diffraction patterns generated that way. Try searching for "Cornu spiral", which is a graphical device to aid in generating such patterns. Once upon a time, I did some that recognizably approach the "no-diffaction" regime.

And finally for the ultimate accuracy and generality, do the integral numerically, customized for the specific source, slit and screen configuration.

9. Mar 1, 2005

### dextercioby

For large apertures it's almost the same as "edge diffraction" (semiinfinite plane,check chapterin Born & Wolf).

Daniel.

10. Mar 1, 2005

### ZapperZ

Staff Emeritus
But I think that's the point I was trying to make. We can't just leave the explanation as "many point sources in the slit". This by itself cannot explain why the diffraction pattern gets wider as we make the slit smaller. Unfortunately, once we try to go beyond that, then we run into the things you mentioned and the explanation gets highly mathematical. I don't see a simple and intuitive way of doing that.

Zz.

11. Mar 1, 2005

### Staff: Mentor

At a fundamental level, at least in terms of the Huygens-Fresnel diffraction theory, that's really all it is. More precisely, if the slit is illuminated by a point source, the basic idea is "many paths from the source to the point P on the screen," each path connecting the source and one sub-slit with a straight line, and the sub-slit and point P with another straight line. Actually, we should use the term "sub-aperture" rather than sub-slit, because this formulation applies also to two-dimensional apertures.

For example, see Hecht's Optics, which starts its discussion of the rectangular aperture on p. 497 with the contribution to the field at P, from a single sub-aperture:

$$dE_P = \frac {K(\theta) E_0} {\rho r \lambda} \cos [k(\rho + r) - \omega t] dS$$

where E_0/\lambda is a constant related to the intensity of the source, \rho is the distance from the source to the sub-aperture, r is the distance from the sub-aperture to P, and K(\theta) is Kirchhoff's obliquity factor

$$K(\theta) = (1+\cos \theta) / 2$$

where \theta is the angle between the two lines described above, which prevents the aperture from producing a wave that goes backwards towards the source. The \rho and r in the denominator produce an inverse-square falloff of intensity with distance, going from the source to the sub-aperture, and from the sub-aperture to P, as we should expect.

This is equivalent to having the sub-apertures as our "original" sources, producing waves with different amplitudes and phases, from different locations within the aperture.

Starting from here, the only difference between different situations (single slit, double slit, circular aperture, rectangular aperture, whatever) is the geometry of integration and the approximations that one uses to make solving the integral less complicated.

At this level, the only 'ad-hoc' thing in the expression above is the obliquity factor. But it emerges naturally when you do a rigorous derivation starting from the differential wave equation.

12. Mar 1, 2005

### ZapperZ

Staff Emeritus
Unfortunately, I don't think I'm getting through, here.

You'll notice that, as I've said in the previous posting, that as soon as you try to explain this, you cannot avoid but invoke a complete mathematical explanation - which is fine and which I understand completely (I did, after all, used Hecht and Zajak for undergrad optics course eons ago). However, it is difficult to explain conceptually why the diverging beam that goes way beyond the size of the slit as the slit gets smaller and smaller, and yet, it is about the same size as the slit when it gets larger and larger (for plane waves impinging on the slit).

I could easily explain this mathematically by saying "Look, take the Fourier transform of the slit (i.e. assume it is a square pulse, the width of which equals to the width of the slit), and you'll see the fourier transform of this (in k space) corresponds to the actual diffraction pattern that you get. This is as accurate as any. But it says nothing conceptually.

Let's put it this way... if you had given your original answer, and included the additional explanation as you did in your 2nd response, I think I wouldn't have said anything, since that clearly indicated that to know what happens as the slit width gets larger, the superpostion of many, many sub-slits will conspire to produce a nasty mathematics that eventually will give us what we observe. I don't believe that there is (at least, I can't do it) a conceptually easier or hand-waving way to explain this.

Zz.

13. Mar 1, 2005

### Claude Bile

Kirchhoff's Integral is inexact. It is fine for far-field optics, but for near-field optics, it cannot be applied. Maxwell's equations are the only equations that can precisely predict any diffracted field, given an incident field and an aperture.

Claude.

14. Mar 1, 2005

### dextercioby

Kirchhoff's Integral is inexact,but not for the reason you mentioned(BTW,Fresnel's theory is based on this integral,too,check Jackson or,better,Born & Wolf).It's because it's SCALAR THEORY OF DIFFRACTION.The correct theory is the vector one...

Daniel.

15. Mar 1, 2005

### Claude Bile

Sorry, I meant near-field, not in relation to Fresnel's theory, but in relation to sub-wavelength apertures and the field within one aperture length of the aperture (If that makes sense). Near-field is just one of those terms that can mean different things in a given context, I guess I got caught out (again!).

I seem to recall that there was another reason why Kirchhoff's Integral fails for sub-wavelength apertures in the near-field (not just because it is a scalar equation), something to do with the way it assumes the form of the field inside the aperture and so forth.

Claude.

16. Mar 2, 2005

### T.Roc

all

Why doesn't this happen with a laser (1 coherent freq.) through the slit?

TRoc

17. Mar 2, 2005

### dextercioby

Happen what...?

Diffraction...?

Daniel.

18. Mar 2, 2005

### ZapperZ

Staff Emeritus
What made you say that this doesn't happen with a laser? The ONLY way to see clear diffraction pattern IS to use a source with plane, monochromatic light.

Zz.

19. Mar 2, 2005

### dextercioby

Probably never walking into a lab...??

Though i have a hunch he's not exactly a theorist...

Daniel.

20. Mar 2, 2005

### T.Roc

dex, zz

I simply mean that I don't get the same results through the same slits when I use a laser, and when I use "white" light. What does this mean?

TRoc