# How can diffraction be explanined using the concept of wavelength.

1. Mar 28, 2013

### A Dhingra

hi..
I have been struggling, since a long time, to understand how is the diffraction pattern obtained by a slit of width of the order of the wavelength of light used is obtained, but found no answers!
As per the idea of wave theory (and Fermat's principle) it is the wave nature (specifically wavelength)of light that can explain how light "smells" the nearby paths selecting the one with extremum time taken. As per this, Wavelength is the distance between the successive crests of a periodic wave. How is this picture of wavelength anything to do with the diffraction pattern observed by a slit of specific width?
Can someone please explain this to me, physically, and using the Fermat's principle along with the pictorial definition of wavelength of light?

Looking forward to a satisfactory solution..

2. Mar 28, 2013

### Claude Bile

The diffraction pattern that arises is the Fourier transform of the aperture function of the slit. The Fourier conjugate variables w.r.t. space are the wavevector components in x,y,z (recipricol space) which scale with wavelength.

Claude.

3. Mar 28, 2013

### A Dhingra

Claude,
Can you explain how the definition of wavelength leads to diffraction pattern? I would rather want an answer that can be visualized in place of complicated looking maths...please.

4. Mar 28, 2013

### soothsayer

Imagine the crests and troughs of a typical sine wave. As a wave goes through a slit, it propagates through as though the slit were the source of the wave, and wavelength is preserved on both sides of the slit board. With multiple slits, you now have multiple sources of waves with identical wavelengths that interact. If you're familiar with the concept of constructive and destructive interference, you'd see that the separation of the slits and the wavelength of the light will determine where waves will interfere destructively or constructively. These patches of constructive interference correspond to high amplitude areas in a diffraction pattern, and vice-versa: the areas of destructive interference correspond to low amplitude areas in a diffraction pattern.

Hope that helps!

Soothsayer.

5. Mar 28, 2013

### Staff: Mentor

Do you understand what interference is? If not, then it is not possible to explain diffraction to you.

6. Mar 28, 2013

### technician

How do you explain interference without knowing what is meant by diffraction?
You seem to be initiating a circular ,not very helpful argument.

7. Mar 28, 2013

### BruceW

I don't know if this will help. But it is how I have come to accept this concept. OK, so we suppose that electromagnetic phenomena obey certain equations (Maxwell's equations). And we have some kind of impenetrable object. And we have a slit in the middle, which is essentially free space. So now, our equations are wave equations, with boundary conditions (that the fields must go to zero in the impenetrable object).

Further, we assume some incoming wave which has a certain wavelength. So when we solve those equations, we find that if the wavelength is of the same order as the size of the slit, to impose the boundary conditions, we find that diffraction must happen. But if the wavelength is much smaller, we find that we can still satisfy the boundary conditions by only making a small amount of diffraction. And if we say that the wavelength is much smaller than the slit, we can say that the diffraction is negligible. (Of course, there is still diffraction at the edges of the slit, but in the middle of the slit, not much diffraction will occur).

8. Mar 28, 2013

### BruceW

Also, I can understand that this is confusing. (since electromagnetism is a 'local' phenomena). So it seems strange that the fields in the middle of the slit can 'feel' the sides of the slit. It makes sense to me like this: when the light is first switched on, there will be waves reflected off the sides of the slit (and the fields will be a bit more complicated). But then, after a while, it all settles down, and we end up with a situation where the field in the middle of the slit depends on the width of the slit.

9. Mar 28, 2013

### Andy Resnick

Part of your confusion could be due to mixing up the wave and ray models of light. Fermat's principle and extremum paths are geometrical (ray) concepts and are not part of the wave model. Huyghens' principle is the basis for wave propagation.

10. Mar 29, 2013

### A Dhingra

Well, that is one confusing part.. though the answer you gave isn't very satisfactory. Can you please elaborate a bit, if possible?

11. Mar 29, 2013

### A Dhingra

I think, I am trying to club these two ideas together.
this is what is written in Feynman lecture, "Does it(light) smell the nearby paths, and check them against each other? the answer is , yes, it does , in a way. That is the feature which is, of course, not known in geometrical optics, and which is involved in the idea of Wavelength: the wavelength tells us approximately how far away light must "smell" the path in order to check it." To demonstrate this fact he considered a situation of diffraction of light, and measuring light at a point in the geometrical shadow region, which he explains as,"Initially with wide slit they check several paths nearby and say"those correspond to different times". On the other hand, if we prevent the radiation from checking the paths by closing the slit down to a very narrow crack, then there is but one path available and the radiation takes it! With a narrow slit, more radiation reaches the point than reaches it with a wide slit! "

I don't understand the highlighted portions. So can someone explain the dependence of wavelength and slit width, in this manner. Or explain those sections....
What i couldn't understand is that on reducing the slit width we are not affecting the path that ligth checks (because that's along the direction of propagation) as the slit width is reduced perpendicular to it, then how are these two affecting each other??

(i used an extract from the book lectures by feynman vol1 ch-26)

12. Mar 29, 2013

### soothsayer

What's happening here is that you're basically changing boundary conditions on the wavefunction of a photon passing through the slit. The wavefunction must be zero at the slit wall, so narrowing the slit will narrow the wavefunction, and limit the spread of possible photon positions.

Imagine trying to shoot bullets through a hole in a wall. If the hole is really big, the bullets can go through it in a number of different ways, and hit the wall behind it in all sorts of locations. If we limit the size of the hole to approximately the size of the bullet, there is only one way that bullet is fitting through that hole, and only one location that it will end up on the backboard. You can also simply consider the difference between a rifle and a shotgun: The bullets in a rifle are more funneled, and thus, don't spread out as much and hit targets much more accurately at long ranges. Shotguns are shorter, and allow for spreading between many small bbs resulting in a large, circular bullet pattern.

Since a photon's size is related to its wavefunction, which has a wavelength and areas of high and low probability amplitude, and since unobserved quantum objects must be in a superposition of all possible states, given multiple paths, a single photon wavefunction will effectively go through all those paths at one, with varying probabilities. It will then interfere with itself, either adding low and high amplitudes, creating areas of very low probability in destructive interference, or adding two high or two low amplitudes, causing a spike in probability--constructive interference--or something in between. This causes the diffraction pattern. It's very dependent on how many paths are available to the photon to take, which is dependent on slit size (just like in the above analogy).

At a certain point, you limit the slit size so much that the photon is localized to such a degree that the Uncertainty Principle begins to come into play. By restricting position, we unleash a greater uncertainty in momentum, which can cause photons to scatter in many more directions than before. I've actually seen a demonstration where the slit size in the apparatus could be modulated, and as it decreased, the diffraction became more and more concentrated, but then hit a point where the pattern began to quickly spread out again, due to momentum uncertainty caused by a highly constrained wavefunction.

13. Mar 29, 2013

### Andy Resnick

Ok, but this is also why you are confused:

In any case, if you must 'club them together', you could define the ray to be an extended object and assign a 'width' or 'radius' about equal to the wavelength.

14. Mar 30, 2013

### A Dhingra

If the answer lies in QM.....I doubt I will be able to appreciate it.(because I am yet to study it in my course).Yet I got the idea that Uncertainty Principle has a major role to play here.
After I will study QM I will come back to you if in case I would have any doubt regarding this...
Thanks a lot....
Thanks everyone..

15. Mar 30, 2013

### BruceW

I did a quick google search and read the chapter in Feynman's lectures. He is talking about rays. And he is talking about the 'principle of least time'. He is using an interesting way to explain diffraction. But as he says, in later chapters he will introduce the concepts of wavelength and that the 'ray' description is an approximation (definitely not the full picture). Also, as he says, classical physics is enough to describe diffraction. And the 'ray' description is not enough to give a proper explanation of diffraction.

Feynman very briefly introduces the concept of a path which extremises total time. This essentially means that if you make slight perturbations to the path, then to first order, the total time does not change. Also, the 'ray description' of light is an approximation that works when the length scales of objects are much larger than the wavelength of the light.

I'll try to re-word Feynman's explanation, hopefully that might help you. (also it will help me, to try to get the gist of what he was saying). OK, so he says that we place a detector head-on to the slit. If the slit is wide enough, the ray will be able to check several possible paths through the slit, and it will find that each path has the same total time (to first order). So it decides that the path to the detector extremises total time, and it takes the path. Also, if the slit is made smaller, then the ray still takes the path. Now, if we move the detector to the side, and keep the slit wide (compared to the wavelength of the light), then the light will be able to check several possible paths through the slit and will find that the total time does change (to first order), so it will decide that the path to the detector does not extremise total time, so it will not take that path. Now, if we make the slit small, then the ray can only take one path through the slit, so it cannot decide whether the path extremises total time, so it does take the path to the off-centred detector (i.e. diffraction happens).

This explanation is uh... Maybe more like an introduction to a full explanation which involves quantum mechanics. But rest assured that classical physics can explain diffraction as well. (It is just that you really need more than the ray description). Sorry if I repeated myself about that.

16. Mar 30, 2013

### soothsayer

Yes, technically the answer lies in QM, if you're studying the experiment using photons or electrons, but knowledge of simple classical wave mechanics work perfectly fine. In fact, there is no difference in the diffraction pattern between surface waves in a medium (like water) or photons, or electrons, or any other particle. Also, if you send sound through a single slit, it will diffract the same way. No quantum mechanics needed (though you do need to enforce the condition that the sound amplitude it zero at the boundary, otherwise sound will reverberate through the walls and tamper with the diffraction pattern.)

17. Mar 31, 2013

### sophiecentaur

This is a totally unreasonable request. The phenomenon of simple, idealised interference (leading to diffraction, in general) can only be described in terms of the way waves of different phases add together to produce a resultant. That has to involve some fairly sophisticated concepts in Maths. You would have to specify where your appreciation of Maths cuts off. If you can't deal with vectors and trigonometry then there is no satisfactory answer for you.
The "definition of wavelength" doesn't "lead to" anything.

If you can't understand what Feynman says in his lectures then you should go back along the line to a more elementary and geometric (text book) treatment. Feynman, despite being very entertaining, is definitely not the place to start on any Physics topic. What he says and what you hear will almost certainly be two different things if you are not 'listening' at the appropriate level. I have walked out of many lectures on advanced topics, given by excellent lecturers, convinced that I had it all sussed but it has been an illusion - as I have found out later, when trying to solve problems using only what I got from the lecture.
If you don't want to 'do the maths' for this problem then I would doubt that you are really ready to cope with the interference phenomenon. As with most Physics, there is not a viable 'seat of the pants' way into any serious understanding.

18. Apr 1, 2013

### A Dhingra

I really understand how important math is to understand physics(and I do like mathematics). I have done the maths in the classes but could not understand how was diffraction actually happening, i know the Huygens-Fresnel approach, but what i read in Feynman lectures stayed all the while that this was not a satisfactory solution to understanding diffraction as Feynman explained. That's why i asked only for stuff to visualize that can help me appreciate these works more (at my level).....
And you are right, there was a gap as to what he was saying and what I was listening..

19. Apr 1, 2013

### sophiecentaur

What do you mean by "actually happening"? What is happening is that a lot of waves are adding together to produce a resultant. You can't say what the result of that process will be without recourse to some 'maths'.
What is it about the Maths that makes it difficult to "visualise"? The geometry of the situation tells you the path lengths of the light for the various routes through the system. The vectors all add together to produce a resultant (demonstrated by the Cornu Spiral and other diagrams). Without those two bits of 'maths' there is no hope of getting a grasp.
As I wrote before, if Feynman is too hard (not surprisingly) then go back to the very basic two source interference theory and 'visualise' that, along with the familiar diagrams. Work on from that. Use the idea of Phasors to see how the path differences on the two possible paths will produce different resultant Phasors as the relative Phases of the two waves change across the projection screen. Then look at the effect of multiple sources (diffraction gratings etc). Finally, consider what happens when you have an array of elementary sources, spaced by a smaller and smaller gap - you then jump into the Integration approach. But this is only an extension of what you get with two, or more discrete sources.