# Diffraction and Wavelength Dependency

1. Nov 3, 2015

### Jimmy87

I have found lots of forum threads on this but I still can't see the connection between wavelength and diffraction. Here is a picture I found:

If you increase the wavelength then all is this will do is just increase the time between the crests that arrive and I don't see how that can have anything to do with the diffraction effect. Another example is when you go inside a faraday cage you can block radiowave signals coming in/out but you can still see the person. This is explained by the fact that the radiowave has a wavelength similar to the spacing of the gaps in the Faraday cage whereas the light wavelength is much smaller but I don't see how increasing the distance between crests can do anything to the diffraction as the length of the wavefront (as seen in the diagram above) doesn't change. Could somebody please help explain.

Thanks

2. Nov 3, 2015

### Khashishi

Divide the space across the hole into a small grid and consider each point along the hole as a secondary starting point for waves.
Try to map out how the waves interfere with each other.

3. Nov 3, 2015

### Jimmy87

I get that each of these points in the grid will act as their own point sources and this will causes them to cross over and interfere but I don't see the significance of the wavelength to this effect?

4. Nov 3, 2015

### Merlin3189

If the point sources are close, they will overlap almost in step.
If they are further apart, when they overlap they may be out of step in some directions.
The distance they need to be apart (when viewed from a particular direction) in order to cancel out, is half a wavelength. Then the peaks from one point will land in the troughs from the other point.
Notice, the distance is not 3mm nor 0.000000007m nor any other fixed measurement: it is half a wavelength. (And for partial cancellation, some other fraction of wavelength.)
So in a situation where light of one wavelength goes through a gap and, at a particular viewing angle, light from points in one half of the gap exactly cancels with light from the points in the other half of the gap (because the path differences are half a wavelength apart), then no light is seen at that angle. But if we change the wavelength, then the path differences will no longer be exactly half a wavelength, so some light will go in that direction.

For all interference effects, what matters is whether different waves (or perhaps better, different parts of a wave) reach a particular point in step (when they add), out of step (when they cancel out), or something in between (where they partially add or partially cancel). So the only significance of all distances, is how many wavelengths (including fractions) they are. (For waves, a distance of half a wave is indistinguishable from a distance of one and a half waves, or any whole number of waves and a half wave. And similarly for any other fraction.)

5. Nov 4, 2015

### Jimmy87

Thanks for the reply. I have read your reply carefully and done some background reading and totally get what you mean. However, I still don't see how the wavelength of the approaching wave-front bears any significance on the spreading out (diffraction) of a wave as it goes through the slit?

6. Nov 4, 2015

### sophiecentaur

It doesn't surprise me that you are finding a problem with this because you are jumping in, half way through the problem.
You either need to accept the Maths behind the way Diffraction works or start with the most elementary forms of interference (the two slit experiment) and work towards the effect across a wide slit. If you are prepared to start with the basics then you have a chance of understanding this.

7. Nov 5, 2015

### Merlin3189

This is a "conversation" reply, but I can't get upload to work in conversation, so I'm posting here.

Firstly you need to understand "interference". When two waves combine, the sum may be a larger wave, a smaller wave or even no wave at all.
In these picture we add two equal waves (shown red & blue) and show the result in green

So when two waves are in phase (no phase difference), they produce a wave of double the amplitude.
When they are in antiphase (phase difference of half a wave), they cancel each other and produce zero resultant wave.
As the phase difference increases from 0 to 1/2 wave they support each other less and less and eventually cancel each other completely.
As the phase difference continues to increase from 1/2 to 3/4 wave, the cancellation becomes less complete and from 3/4 wave to 1 full wave they reinforce each other more and more until they combine to give double amplitude again.

When the phase difference is 1 whole wave, that is indistinguishable from zero phase difference. Then the whole cycle repeats again.
After any whole number of waves difference, the waves are in phase again. So only the fraction of a wave phase difference is important.

Why do waves combine out of phase? In the next diagram identical waves are sent out from A & B and we pick two points on the screen P & Q and show how the two waves can arrive out of phase. (I will talk as if the waves are light, but to make it more human sized, I am using lengths which are much too big - by about 10^7 )

The distances AP and BP are equal at the centre of the screen, so the two waves arrive exactly in phase (top left diagram.) They combine to give a larger wave at P. P will be a bright point on the screen.

For the point Q away from the centre, AP is shorter than BP. So when the two waves arrive at Q, they may no longer be in phase.
The difference in distance is about 1.44m. But what is the difference in phase? That depends on the wavelength.

Say we used waves with 3m wavelength, then 1.44m is about 1.44/3 = 0.48 of a complete wavelength. So the two waves are out of phase by nearly half a wave (the bottom left diagram ) and almost cancel each other. The point Q on the screen will be dark.

But if the wavelength were instead 2m, then 1.44m is about 1.44/2 = 0.72 of a complete wave. So the waves are out of phase by nearly 3/4 of a wave (the bottom right diagram) and partially support each other. Now the point Q will be bright, though not as bright as the centre point P where the waves fully reinforce each other.
(BTW notice that saying, wave A is 3/4 wave ahead of wave B, is the same as saying, wave B is 1/4 wave behind wave A. There is a symmetry to phase differences. Which is why the two right hand diagrams are similar.)

If you change the wavelength again to 4m, the phase difference at Q becomes about 1.44/4 = 0.36 of a wave and the interference is partially destructive(*). So point Q is slightly dimmer than if it had been illuminated by only one of the waves.

(*) Interference is constructive from 0 to 0.25, destructive from 0.25 to 0.75, with complete cancellation at 0.5, and constructive from 0.75 to 1 (and on to 1.25, which of course is 0 to 0.25 of the next wave.) Some examples are shown here

Notice the similarity between 40% and 60%. Similarly 70% will be like 30%, 80% like 20%, etc.

8. Nov 5, 2015

### Jimmy87

Thanks for such a detailed reply. After reading through this a few times I get the physics you are trying to explain. My point is why do you get more spreading of the wave when the wavelength is similar to the size of the gap? (as shown in my picture). I have tried to apply my question to your last post. In your diagram, if you used waves with wavelength of 100m (which is one I picked as it is much larger than the 1m gap) then at point Q there would be no phase difference because you are dividing 1.41m by 100. This means all the waves would be in phase. Does this mean there would be less spreading of the wave somehow (which is what I am trying to get at)? So basically if you have a small gap then the path length difference to arbitrary points is small and if you have a big wavelength then all the waves stay in phase and don't diffract/spread? Is that what you are saying?

9. Nov 5, 2015

### sophiecentaur

See my post above. Take it one step at a time, starting with the most basic situation and you have a chance of understanding the process. It really is not a problem that's suited to hand waving.

10. Nov 6, 2015

### Merlin3189

Apologies for the hand waving. You have me bang to rights! I nearly added a comment in my post to ask real physicists to excuse me.
My excuse is simply that the question of why the wavelength matters, seemed to imply a lack of understanding of basic interference and I wanted to explain that idea. Also the questions are qualitative, so I don't think we are looking for a detailed mathematical analysis. My numbers are just "illustrative" as the financial ad's say. I would have sent it in a conversation, so that the real scientists would not be offended, but I wanted to add pictures.

11. Nov 6, 2015

### sophiecentaur

I know I can get a bit tetchy when people ask for non-maths explanations to things but there are some things that can't really be done without it. Try discussing the odds in horse racing or the gear ratios on a racing bike. Just how could you decide whether one particular combination was higher of lower geared, without doing the calculation? Interference depends upon the way the various paths change as the directions change. Those pictures of sinusoids adding and cancelling are only half the story. What really shows the way you get maxes and mins is to look at the way lengths from the sources to the 'screen' change as you move across the screen. The Hyperphysics link earlier, shows it simply and in a way that hand waving just can't approach. I think you would find the effort of getting into the basics of the maths very rewarding.

Last edited: Nov 6, 2015
12. Nov 6, 2015

### Mister T

If you increase the wavelength then you increase the time between the arrival of the wave fronts, giving the elastic medium more time to respond to the disturbances.

13. Nov 7, 2015

### sophiecentaur

This is nonsense. There is no elastic medium in EM and no 'response time'. Why not just read some proper Physics?

Last edited: Nov 7, 2015
14. Nov 7, 2015

### Jimmy87

Ok thanks I will give that a try. Could someone just confirm whether or not my qualitative explanation in post 8 in response to 7 is right or wrong?

15. Nov 7, 2015

### sophiecentaur

The important picture is the one with the slits and the screen. When the two paths differ by a whole number of wavelengths you will get constructive interference (max) and when they are differ by an odd number of half wavelengths, they will cancel (mins). The details of the variation of intensity between the maxes and mins doesn't matter much and involves 'harder sums', which are not necessary for this thread.
The simplified formula to show the spacing (y) of the fringes is that y = mλD/d for the maxima in the picture in the hyperphysics link. As λ increases, so, accordingly, does y.

16. Nov 7, 2015

### Mister T

Doesn't make it nonsense. It's the standard explanation for how waves propagate through a medium. The medium takes time to respond because of both its elasticity and its density in the same way as a mechanical oscillator. It's the standard treatment of the topic found in textbooks. In the case of electromagnetic waves there's an analogous argument involving the finite speed of the field propagation, but that's usually not covered in introductory physics textbooks.

Last edited: Nov 7, 2015
17. Nov 7, 2015

### Jimmy87

Thanks. I do see the point you are trying to make about the quantitative approach which I understand now. I can see how interference is wavelength dependence because it defines the phase difference. I still don't see how this explains the SPREADING out of a wave through a slit/gap though. For example we were taught that you can hear someone around a door but not see them because the gap in the door is comparable size to the wavelength of sound whereas the wavelength of light is much smaller. Also another example we were given is where people who live in between hilly areas can receive long length radio-waves but not short length ones (the longer ones diffract and spread out around the hill so they can reach the house). Why is the SPREADING of a wave dependent on the wavelength and gap size? Or can you argue with the same physics you and Merlin3189 were discussing in terms of path length difference and phase differences to explain these two examples?

18. Nov 7, 2015

### Mister T

For me to answer this question for myself, I had to first ask myself why a wave spreads out when it goes through a barrier. I think about a water wave doing this, like in the drawing shown your first post. Once I thought about that, then I could think about why the effect is wavelength-dependent.

19. Nov 7, 2015

### sophiecentaur

EM waves do not involve any 'elasticity'. The wavelength of mechanical waves depend upon the modulus but that just determines the wavelength and has nothing to do with the diffraction effects. Interference and diffraction are dependent on the wavelength and not what you were stating. The reason is the geometry of the situation - as already described.

20. Nov 7, 2015

### sophiecentaur

To progress from the two slits interference to the effect of diffraction through a finite gap requires more than an arm waving explanation.
This hyperphysics link give a formula but not aformal derivation of how it works. You will either need to accept it or get into the maths. I cannot think of any other way than to take on board the necessary language for dealing with these things. The derivation is based on Integration of the contributions (phase) of all the infinitessimal sections of the gap. Again, you 'just' add the contributions of each section, vectorially, and that gives you the resulting amplitude in any direction. The Fresnel Integral for calculating the intensity in a direction as a wave hits a slit can be arrived at using the Cornu Spiral. See this other link.

21. Nov 7, 2015

### Mister T

I agree. Never said otherwise.

If diffraction depends on the wavelength, and the wavelength depends on the modulus, then diffraction depends on the modulus.

22. Nov 8, 2015

### sophiecentaur

There is no direct link at all. Take two liquids, one has a much higher modulus than the other. Put them, one at a time, in a test rig, which has a sound source and an aperture between two ideal, rigid walls. You can adjust the frequencies used in the diffraction experiment to produce the same wavelengths in each liquid. The diffraction patterns will be precisely the same in both cases. The diffraction pattern for EM or sound-in-air waves would also be the same, in the same rig. The 'time' factor that you are introducing, has no relevance to the diffraction pattern because it is just the wavelength that governs path lengths and, hence, the diffraction pattern.

23. Nov 8, 2015

### Mister T

24. Nov 8, 2015

### Mister T

I'm not "introducing" any time factor. You can think of it as the period if you want. Yes, if you have a different medium you will have a different wave speed, which means you can adjust the period to produce the same wavelength. So you now have the same diffraction pattern even though you have a different period.

Here's an equivalent way of looking at it. Imagine that the waves shown in the OP are water waves passing through an opening in a barrier. Replace the water with a different liquid that responds more slowly to a disturbance. Waves travel at a lower speed on its surface, so that you have to decrease the frequency to produce the same wavelength. The wave still spreads in the same way as it passes through the barrier, but that's because you've made it oscillate up and down (at the barrier) at a lower rate to compensate for the fact that the medium responds more slowly.

25. Nov 8, 2015

### sophiecentaur

Your earlier post is specifically introducing a time factor. That is what I was objecting to. Are you changing your ideas now?
Here, you are just pointing out the wavelength and frequency relationship with wave speed. This is basic theory but has nothing to do directly with diffraction. In what way can you claim that it is? Perhaps a reference would help here, to show it's more than just your personal attempt at an interpretation. I could refer you to any text book that you can get hold of, for a justification of what I am saying.