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Anonymous Poster

# Diffraction and wavelength question

01/14/2007 5:51 PM

Hi

When waves are diffracted through an opening the wavelength plays a role in determining the extent of the diffraction.

If the size of the gap is much larger than the wavelength there is little diffraction. If the wavelength and the gap size are similar the diffraction is larger.

Why is this? The way I see it is that the wavelength is perpendicular to the length of the gap and therefore something like amplitude would be a more relevant factor.

Thanks in advance.

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Anonymous Poster
#1

### Re: Diffraction and wavelength question

01/15/2007 11:42 PM

The degree of diffraction is a function of wavelength to hole size. I would only adjust one variable at a time when thinking about this; i.e. I would hold the wavelength constant and let the hole size change or I would change the wavelength and hold the hole size constant. But as the wavelength gets smaller and a hole size is constant the diffraction angle gets greater and greater. I can put up some equations if you want but I assume you already have those.

One way to disprove your reasoning that amplitude doesn't have anything to do with diffraction is to assume you have a standing wave going through a hole. It can go through a really little hole and not get blocked. But I think how things are normally derived is that you have a plane wave and it goes through a hole so the standing wave thing doesn't really make sense.

Anonymous Poster
#3
In reply to #1

### Re: Diffraction and wavelength question

01/16/2007 12:39 PM

Hi,

Thanks for the replies, but I'm still a little way from understanding.

What I'm really after is why, "as the wavelength gets smaller and a hole size is constant the diffraction angle gets greater and greater".

Why should the wavelength dictate the extent of diffraction. How are the two related?

Thanks again.

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#4
In reply to #3

### Re: Diffraction and wavelength question

01/17/2007 9:57 AM

The basis for this is where the contribution from different parts of the wave reinforce or cancel each other. The contributions reinforce each other when paths from a wavefront to the point of interest are the same as each other or whole multiples of a wavelength, and tend to cancel when the distance is (n+1/2) wavelengths (n any integer). The smaller the size of the hole, the larger an angle you can move from the initial direction of propagation for any given difference between the path lengths. Therefore, you can move further away from the original propagation direction before the wave becomes self-cancelling. Long wavelengths conversely mean you need a greater path difference. This is why (for a given hole size) it is the ratio of hole size to wavelength that determines the angular pattern of the diffraction.

BTW, the power diffracted into a particular region varies directly with the input power, just as you would expect.

Angular distribution will depend on the detailed shape of the hole, but typically the first zero of the output will occur in the region where the angle from the centre of the beam (in radians) is slightly greater than the wavelength divided by the diameter of the hole. For chapter and verse as to how to calculate values for specific shapes of hole, you'd probably do best to consult Wikipedia or a book on wave optics.

(Neglecting polarisation effects (and for off-axis beams diffraction near the hole) and assuming a uniform beam, the relative amplitude density comes out as
Integral(h.exp(2.pi.j.w/lambda.theta)dw/area
where theta is the angle of interest, w is the displacement across the wavefront in the direction of interest for the diffraction, and h is the hole dimension at position w in the direction along the wavefront but perpendicular to w)

Fyz

Anonymous Poster
#5
In reply to #4

### Re: Diffraction and wavelength question

01/17/2007 12:19 PM

Hi

Thanks for your reply but I'm still a little confused. Mainly concerning these two lines...

"The smaller the size of the hole, the larger an angle you can move from the initial direction of propagation for any given difference between the path lengths. Therefore, you can move further away from the original propagation direction before the wave becomes self-cancelling."

Would it be possible for you (or anyone else) to expand on this little. Sorry to keep asking but I've been doing a Chemistry degree for four years and every time someone mentions it in a lecture it annoys the hell out of me but I always forget to ask someone.

Thanks again.

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#6
In reply to #5

### Re: Diffraction and wavelength question

01/17/2007 3:41 PM

First, you need to accept that the only thing that controls where a wave will propagate is interference. For a wave starting from a point source, there is no interference, so the wave goes in all directions equally*

Now it becomes a job for a drawing. We can demonstrate the principle in two dimensions - on a sheet of photocopier paper. Place it in landscape mode. Draw a horizontal 1" long straight line at centre of the bottom of a sheet of paper. Measure (or calculate using trig) the difference between the distances from each end of that line to various points along the top of the sheet. Now try the same thing with a line that is 2" long. You will find that the differences are about twice as great for the longer lines. Now imagine that identical waves start out from each of the ends of the lines. Assume first a wavelength of 0.1". There is no interference, these travel omnidirectionaly. Where the path lengths are the same, the waves will add (constructive interference). Where the paths differ by 0.05" the movement (or field, or strain, or height) of the two waves will be in opposite directions, so they will cancel. Now you can draw an approximate graph of amplitude versus position across the top for each of the two cases. Next draw the graphs for a wavelength of 0.2".

You will see that the distance (hence the angle) between the zeros decreases as the length of the lines increase. But you will also see a sequence of peaks and troughs of roughly equal height**. This is exactly what you see in the classic Young's slit interference experiment, so we seem to be on the correct track.
**Not if you include the inverse square law - but that's just an extra unnecessary complication for now.

[There is a calculator and diagrams for multiple slit experiments at http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html]

Now consider the output from 5 points equally spaced along the 2" line (you've already measured four of these). You will see that fully constructive interference (the differences between all distances being being integer multiples of a wavelength) becomes less frequent. But it does occur (approximately) at the centre of the top of the paper. It should by now be becoming apparent that when you look at the waves coming from everywhere along the line that the only place where constructive interference occurs is at the centre. We can make this more rigorous by pretending that the end of the paper is at infinity (this is called the "far field approximation" or Fraunhofer diffraction), and consider the angle. Integration of the relative field due to waves starting all the way along the originating line now becomes relatively straightforward:

Integral(over -w/2, w/2) of cos(2.pi.sin(theta)*x/lambda) dw = constant.sin(pi.sin(theta).w/lambda)/sin(theta).lambda/w

This is the classic "sin(x)/x" distribution that you may have heard about. The first zero is when sin(theta)=lambda/w, which again shows the width reducing with wavelength and increasing with w.

Hope this helps

Fyz

*Before someone objects, this is not exactly true for transverse waves, but there's no point in trying to run before we can walk

PS I think you already know this, but never worry about looking silly when you ask a question - provided you've done your damnedest first.

Anonymous Poster
#7
In reply to #6

### Re: Diffraction and wavelength question

01/17/2007 6:14 PM

Thanks, this is very helpful.

I was unaware of what you stated in your first line, I guess thats the kind of thing you miss out on if you don't do a physics degree.

So, am I right in thinking that there are waves in all directions, but constructive interference only occurs at the centre and destructive interference occurs at the sides, or larger angles? Would I also be right in saying that when a wave goes through a slit, every position in the slit becomes a point source with waves traveling omni directionally from them all?

I think I get it now, just need some time for it to sink in, waves always confuse the hell out of me. If a double-slit experiment is set up with a slit gap twice the length of the wavelength, won't there be constructive interference at angles of 90 degrees? When considering wave-particle duality of something like electrons, what does the wave 'look' like, or what does it refer to? Is there a small, localised area of 'wave' moving around in space, or does it mean the distribution of the probability within the area?

Thanks.

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#9
In reply to #7

### Re: Diffraction and wavelength question

01/18/2007 5:07 AM

You are right about the sideways output of a double slit - though in practice the baffle will get in the way (and, for future reference, with light only one one of the polarisations is well-behaved in this sense - the other starts out with a figure-of-eight distribution even without interference).

BTW, you really should have covered the double-slit experiment in high school; this suggests that you will be finding unexpected holes in your background that will make life harder than it might otherwise have been. But you are showing the character to overcome this.

Best not to think too hard about matter as waves until you are fully comfortable with travelling waves and Fabry-Perot etalons.

{The electron is treated as a wave packet. The wave function is not strictly a probability distribution, but this is the closest you are likely to come intuitively. The quantum mechanical solution method for interaction of charged particles appears to many workers to be somewhat arbitrary, and remains one of the controversial areas in physics; the justification is that it works, and the assumption is that a more complete picture will eventually emerge (hopefully, a picture that provides justification)}

Fyz

Anonymous Poster
#10
In reply to #9

### Re: Diffraction and wavelength question

01/18/2007 11:04 AM

What is 'the baffle'?

I'm from England and did not study physics at an advanced stage in High School. I'm not sure if they even cover it then. They certainly do in Chemistry degrees although not with any particular depth and only with reference to wave-particle duality and not actually the wave behaviour itself. I actually read an excellent description/explanation of the experiment written by Feynman quite recently.

I get the feeling we're approaching this from very different perspectives, I've never heard of 'Fabry-Perot etalons'.

I found this place recently when looking for stuff on Einstein's thought experiments, seems pretty good, I might have to register.

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#11
In reply to #10

### Re: Diffraction and wavelength question

01/18/2007 11:36 AM

I'd heard from my UK academic friends that British school education had become inadequate to prepare students for university, and they were having to use the first year to fill in the gaps. Maybe your establishment should consider providing catch-up classes if it doesn't already?

A baffle is (in general) simply something that gets in the way. In this case it is the (sheet of?) material that is used to block the waves (and in which the slits are cut)

Regarding Fabry-Perot - loads of references on the web, it's basically about waves between a pair of mirrors, and the main topic of interest is the interference between plane waves travelling in opposite directions. The relevance is that this is the simplest case of a standing wave (where the location of the energy is confined) that you can find.

N.B. My personal prejudice would be for intending physicists to study mathematics for their first degree, and for intending chemists to study physics first - but then I'm an old fuddy-duddy (I think that is the standard euphemism)

Fyz

Anonymous Poster
#12
In reply to #11

### Re: Diffraction and wavelength question

01/18/2007 12:15 PM

Hah!

Indeed, the education system is becoming increasingly less adequate. I've heard complaints from many academics in my department that almost half of the intake last year did not study one of maths or physics at A-Level, which is our highest, pre-university, qualification. It certainly seems to draw the emphasis of chemistry away from the physical area to the organic area. They do indeed provide catch-up classes (compulsory ones in the above case) but usually have to dedicate the time to calculus rather than what would be considered a specialist physics topic.

Surprisingly enough I took optional first year courses in both maths and physics and am now working in theoretical chemistry. Sometimes I find quite significant gaps in my knowledge... quite worrying!

Thanks again for all the help, this probably won't be the last you hear from me.

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#13
In reply to #12

### Re: Diffraction and wavelength question

01/18/2007 1:14 PM

Hi Guest, if you are genuinely trying to learn and to me you are, then you will find most people here at CR4 will gladly help you. I know I will try an help whenever I can so please don't hesitate to ask. Don't be afraid of asking a stupid question either because the only stupid question is the one that doesn't get asked.. There are a lot of very experienced people here and as you are now learning there is absolutely no substitute for experience. I wish there was something like this when I was studying engineering but that was some time ago.

Anyway I suggest that you register that way we know with whom we are talking.

I still havn't been able to post that diagram of how the constructive addition works and how the wavelength effects the angle but I will try again in a few hours so keep your fingers crossed.

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#14
In reply to #12

### Re: Diffraction and wavelength question

01/18/2007 4:59 PM

I would like to second everything that Masu says in posting #13. Regarding registering, it is not always realistic to declare everything about yourself on a public website; in my case, I already get more than enough personalised junk mail, and that is why I simply call myself Fyz.

From your viewpoint, it would be worth registering just so that we can recognise you as a non-time-waster, even though there can be many reasons why you mightn't want your identity known.

I also recognise what you say about the trend to organic chemistry. The truth there is that, although there is a great deal of "system" to learn, in the end you will require all the tools of physical chemistry as well if you are to make real progress.

Good luck, and keep on trying to understand (it's a life-long struggle).

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#15
In reply to #14

### Re: Diffraction and wavelength question

01/19/2007 12:14 AM

Hi Guest, Fyz & Bernie,

The CR4 editor is accepting images again so here we go. I believe that the age old saying that a picture is worth a thousand words is an understatement so whenever possible I like to include a diagram. This drawing is to scale so all the measurements I give will be relative to each other and can be used in the equations that Fyz has posted earlier. What we have is an arrangement of 2 slits that are small enough to allow to allow the waves to reradiate and are 4 units apart. The red wavelength is 3.14 or π units, the green wavelength is √5 units and the Blue wavelength is 1.75 units.

Looking at the red waves we have concentric circles that have a radius of multiples of π centered on each of the slits. Where these circles intersect you will get constructive addition taking place. If you now connect the points of intersection you can clearly see that they form three straight lines, one perpendicular to the baffle and centered between the two slits and two further lines that are at an equal angle to the first on either side.

If we look at the green example we have the same thing taking place except this time the radius of the circles is a multiple of √5 and the angle of the two diffracted lines is smaller. In other words shorter wavelength results in a smaller angle of diffraction.

Finally if we look at the blue example we have a similar thing happening except there are now 5 lines in total. This is because the wavelength of 1.75 is less than half the distance between the slits but we still have the same thing as before with a smaller wavelength producing a smaller diffraction, its just that it has started to repeat.

From this you can clearly see that it is the distance between the concentric circles or wavelength that governs the angle of the diffracted lines and since the colour of a light source is dependant on the wavelength we end up with a separation of colours.

As I said earlier these diagrams are to scale so you may wish to plug the numbers into the equations and see how accurate my drawings are.

Sorry it took so long but we got there in the end and I hope the diagram has helped you visualize what is going on and how the wavelength governs the angle of diffraction.

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#16
In reply to #15

### Re: Diffraction and wavelength question

01/19/2007 1:16 AM

Masu's latest posting reminded me of the famous "dual slit interference experiment", in which a chamber containg an intermediate wall with two slits produces interference fringes on the back wall. The interference results from reinforcement and cancellation of the diffracted light from each of the two slits.

Of interest is the fact that the interference pattern is yet formed when the light flux is very low, only one photon at a time. Thus diffraction from only one slit occurs at any given time. Being primarily an electronic circuit designer, this phenomenon reaches to the boundary of my knowledge, but is said to result from Maxwellian wave-particle duality. Does anyone have any comments?

Bernie Katz

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#17
In reply to #16

### Re: Diffraction and wavelength question

01/19/2007 8:05 AM

Hi Bernie

This goes way beyond the basic scope of the thread - but never mind...

I'm not familiar with Maxwellian wave-particle duality. All the versions I'm familiar with arise from later workers. The two-slit results could be explained in terms of the wave being a probability field, but this fails when it is known that exactly a single-photon was transmitted - because exactly a single photon is then detected. To explain this we do indeed require full duality. So far, so uncontroversial.

However, conceptual problems arise as soon as we try to explain the combination of a wave with finite spatial extent and non-duplication of the particle. The dominant interpretation is that the wave pertaining to each particle collapses instantaneously - i.e. the occurrence of detection has global impact without restriction of the speed of light. There is an alternative explanation - that the evolution of the detection arrangement is adequately determined (by effects propagating at the speed of light) before the wave/particle is detected, so the decision is effectively made before the particle is detected. So far as I know there is currently no experimental basis for discriminating between these explanations. However, the first is easier to support under the (admittedly incomplete) models of Quantum Electro-Dynamics; for myself, I feel compelled to accept the field-collapse model (at least until something else emerges, but with emotional reservations), as I am not working directly in the area and so not in a position to do anything about it.

Plenty to chew on, I hope, but I fear not really an answer?

Regards

Fyz

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#18
In reply to #15

### Re: Diffraction and wavelength question

01/19/2007 10:35 AM

I agree, diagrams are undoubtedly the most helpful tool in understanding such problems.

Thanks for all the help on this topic, hopefully I'll have many more questions with which to 'exploit' the knowledge of people here.

As an aside, what is the general knowledge base of people here? I see it is advertised as "The Engineer's Place for News and Discussion", but the scope seems far greater.

AD (formerly the 'guest' in this thread).

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#19
In reply to #18

### Re: Diffraction and wavelength question

01/20/2007 2:52 AM

You will also find that threads have a tendency to wander completely off track and by the end it not uncommon for them to have switched to a completely different topic. It's pretty close to what happens when you put a number of engineers together, ultimately the conversation ends up being about their favorite subjects.

Anyway welcome aboard AD, I am glad we could be of help.

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#2

### Re: Diffraction and wavelength question

01/16/2007 1:39 AM

In very elementary terms, diffraction occurs because the slit or other opening, when compared with wavelength, is very small. Thus, the wave front on the other side of the slit or opening acts as a point source (a slit acts as a line source), which spreads the energy at a wide angle.

You can conduct an experiment in your bathtub by placing a thin barrier with a vertical slit in it maybe 1/4 of the way from one end, and dropping a small object into the 1/4 length end. You will see the amplitude waves go through the slit, and spread out on the other side. Light waves are actually made up of electric and magnetic field components at right angles to each other, but the effect is the same.

Optically the same effect can be seen in a pinhole camera, with a very small pinhole with a well defined source of light facing it. The film or ground glass will show a series of concentric rings around the central image.

I'd like to give a better explanation, but the hour is late here.

Bernie Katz

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#8

### Re: Diffraction and wavelength question

01/18/2007 2:22 AM

Hi Guest, Fyz & Bernie,

I did a pretty coloured picture that shows what's going on with a diffraction grating but for some reason the CR4 editor is not accepting images at the moment. This has happened before so when it starts working again I will post it here.

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