I was wondering something about diffraction

[jtbell]

Would someone be able to help me answer this part of my post as well?

I have to read it again carefully and think about it, and I've got other things on my plate the next day or two. My first impression is that it's similar to things that I've tried to address before in this thread.

Also, I was wondering how the calculation for phase differences(360 degrees * path difference/wavelength) is derived, could someone explain it?

The quantity (path difference / wavelength) is the number of wavelengths that fit into the path difference. One wavelength is one cycle of the wave. One cycle is 360 degrees (or 2 \pi radians).

 

[Doc Al]

Another point about the central maxima I'd want to make as well. Say you divided the gap into the 2 halves, couldn't you just say that the light from both halves travels the same distance, and since they start in phase, they will end up in the same phase, and thus constructively interfere?

Modeling the light passing through the slit by rays works as long as the rays represent light with the same phase. Of course, as we have calculated, all the light from the slit that hits the screen at the central maxima travels the same distance to a very close approximation. So, within that approximation, it doesn't matter how many rays you use to represent the light through the slit.

On the other hand, if you want to take into consideration the small phase difference between the various rays, using just two rays won't do at all.

I thought this might be the case, since when calculating the first minima, you can simply dividde the segments into 2 halves, which are 180 degress out of phase, or am I even able to make the same assumption for the maxima?

In calculating the minima it would not do to just use two rays. You have to use enough rays so each ray represents light with a single phase when it reaches the screen. (As explained before, the real deal is to use an infinite number of rays and calculus. But you don't need an infinite number of rays to see the pattern.) Calculating the maxima is more complicated.

On the other hand, however, you can divide the gap into 4 segments, and achieve an answer which shows a slight path difference among segments of the light coming from the gap. Also, since every segment on one half matches up with one segment on the other half, couldn't you say, no matter how many segments there are, that the 2 halves are in phase, and thus constructively interfere?

Your logic here is flawed. I believe you are saying: If rays A and B are in phase, and rays C and D are in phase, then the combination of A+B+C+D will be in phase and thus produce a maximum. Not true. The combination of A+B+C+D depends on the details of the phases. (For example, what if A and C were out of phase? Then A+B+C+D would completely cancel.)

On the other hand, the following reasoning is valid: If rays A and B are out of phase, and if rays C and D are out of phase, then the combination of A+B+C+D must add to zero. This is the reasoning we applied when figuring out where the minima would be.

 

[Byrgg]

Your logic here is flawed. I believe you are saying: If rays A and B are in phase, and rays C and D are in phase, then the combination of A+B+C+D will be in phase and thus produce a maximum. Not true. The combination of A+B+C+D depends on the details of the phases. (For example, what if A and C were out of phase? Then A+B+C+D would completely cancel.)

I don't think I was saying that A and B were in phase, while C and D were also in phase. It was more like this:

A
B
C
D

I meant since at the maxima, rays A and D are in phase, as are rays B and C, wouldn't you be able to say that the two halves are in phase? I think I was thinking more along the terms of if ray A has phase 40, and B has phase 20(these are just arbitrary values used to explain the problem, if these numbers are inaccurate, please tell me) at the maxima, then wouldn't ray C have phase 20, and ray D have phase 40? If this is the case, I thought that you could treat the two sides as being in phas.

On the other hand, the following reasoning is valid: If rays A and B are out of phase, and if rays C and D are out of phase, then the combination of A+B+C+D must add to zero. This is the reasoning we applied when figuring out where the minima would be.

But what if the phase difference between rays A + B is small, as is the phase difference between C+ D? Then this models the maxima, not minima, doesn't it?

Also, along with this, I have yet another question to ask. I was thinking about diffraction through much wider gaps, say sunlight through a window. I had some problems trying to understand what I observed. Basically, the light travels through the window, and then hits the floor or a wall, and the whole area beyond the window receives light. When I thought about the math behind this, I became very confused, a gap like a window is very large when compared to the wavelength of light, therefore, the angle of the first minima must be very small. Using this reasoning, I thought that what I should see, would be a very bright, narrow band along the center of the square shape made on the floor by the light passing through the window, and I wouldn't see much other light. However, this wasn't the case, there is a full square on the floor, that is bright because of the light coming through the window, this area appears to be of uniform brightness, which doesn't seem to make any sense, given the calculations of the maxima and minima. Sorry if this was a poor explanation, I can try to clarify if no one understood exactly what I was talking about, but I can only ask that someone try to help me understand the situation which I described.

 

[Doc Al]

I don't think I was saying that A and B were in phase, while C and D were also in phase. It was more like this:

A
B
C
D

I meant since at the maxima, rays A and D are in phase, as are rays B and C, wouldn't you be able to say that the two halves are in phase?

Not in any meaningful sense that I can see. A "ray" picture only makes sense when each ray represents a single phase. Certainly each half slit does not represent a single phase. You still seem to be reasoning like this: If A & D are in phase, and B & C are in phase, then A+B+C+D must be in phase, which is not the case.

I think I was thinking more along the terms of if ray A has phase 40, and B has phase 20(these are just arbitrary values used to explain the problem, if these numbers are inaccurate, please tell me) at the maxima, then wouldn't ray C have phase 20, and ray D have phase 40?

I'm not sure what you mean by a phase of 40 and 20. Wavelengths? Degrees? Compared to what?

Assuming you are talking about the central maximum, and are measuring phase difference from the center in degrees, then your phase differences are much greater than in any realistic example. We calculated the maximum phase difference (for the example I gave earlier) to be a fraction of a degree--nowhere near 40 degrees.

If this is the case, I thought that you could treat the two sides as being in phas.

Again, I don't really know what that means.

But what if the phase difference between rays A + B is small, as is the phase difference between C+ D? Then this models the maxima, not minima, doesn't it?

Doesn't seem to model anything to me. Again, to draw a conclusion you must know the phase between A and C.

 

[Byrgg]

I think I understand this much about the maxima a little better now. So, at the maxima, basically, it's incorrect to say that the two halves are in phase, simply because A is not in phase with C? Doesn't this also mean that in a real life application, constructive interference is nearly impossible to accomplish, given the fact that any two rays can be dvided into segments that travel different distances than the others? After this, I'd be greatful if someone would help me with the other question in my last post as well.

 

[jtbell]

I don't think I was saying that A and B were in phase, while C and D were also in phase. It was more like this:

A
B
C
D

I meant since at the maxima, rays A and D are in phase, as are rays B and C, wouldn't you be able to say that the two halves are in phase? I think I was thinking more along the terms of if ray A has phase 40, and B has phase 20(these are just arbitrary values used to explain the problem, if these numbers are inaccurate, please tell me) at the maxima, then wouldn't ray C have phase 20, and ray D have phase 40?

As you go across the bundle of rays, the phase increases linearly with transverse position. So if your rays are equally spaced, and A and B have phases 20 and 40 degrees respectively, then C and D must have phases 60 and 80 degrees. See the diagram that I attached to post #27 in this thread, for another example.

In the Fraunhofer approximation (parallel rays), this is exact. In the real world, where the rays actually converge to a point, it's not exact, but for a realistic situation it's very very close. The four rays might actually have phases of 20.00002, 40.00001, 60.00001, 80.00002 degrees.

 

[Byrgg]

As you go across the bundle of rays, the phase increases linearly with transverse position. So if your rays are equally spaced, and A and B have phases 20 and 40 degrees respectively, then C and D must have phases 60 and 80 degrees. See the diagram that I attached to post #27 in this thread, for another example.

I meant at the maxima, if A has phase 20 and B has phase 40 degrees, then the symmetry of the situation makes it so that C has phase 40 degrees, and D has phase 20 degrees, doesn't it?

 

[jtbell]

You're talking about the "height" of each wave, y = A \cos {\phi}, not the phase, which is \phi.

 

[Byrgg]

You're talking about the "height" of each wave, y = A \cos {\phi}, not the phase, which is \phi.

Won't the phase of A and D end up being the same at the central maxima though? They bothe travel the same distance, and since they start wwith the same phase, shouldn't they end up with the same phase?

Also, for the equation for calculating phase difference, I tried to figure out the reasoning behind the equation, and I came up with the following:

By dividing the path difference by the wavelength, you get how many wavelengths fit into the path difference. Multiplying by 360 degrees is simply for conversion into phase difference, for each wavelength that fits into the path difference, there will be a full cycle of 360 degrees. Does this sound right? Did I miss anything?

Howver, one thing that I had trouble figuring out with this equation, is the reasoning behind doing thi in the opposite way, first multiplying by 360 degrees, and then dividing by the wavelength, and I couldn't come up with anything reasonable, I couldn't figure out these questions: Why do you multpliy by 360 degrees right away? And what do you do with these strange units you get(path difference * 360 degrees gets you very strange units in the result that I didn't know how to work with, metre * degrees?)?

So if someone could help me figure out my questions about that equation, along with my other questions, it would be appreciated.

 

[Doc Al]

I think I understand this much about the maxima a little better now. So, at the maxima, basically, it's incorrect to say that the two halves are in phase, simply because A is not in phase with C?

While you might say two rays are in phase, I don't see the sense of saying two halves are in phase, since the halves represent many phases. If every ray comprising the two halves were in phase (which is a close approximation to what really happens) then you could say that all the light is in phase.

Doesn't this also mean that in a real life application, constructive interference is nearly impossible to accomplish, given the fact that any two rays can be dvided into segments that travel different distances than the others?

Not at all. As I thought we've shown, to a very good approximation you will easily have a central maximum.

 

[Doc Al]

Won't the phase of A and D end up being the same at the central maxima though? They bothe travel the same distance, and since they start wwith the same phase, shouldn't they end up with the same phase?

Sure.

Also, for the equation for calculating phase difference, I tried to figure out the reasoning behind the equation, and I came up with the following:

By dividing the path difference by the wavelength, you get how many wavelengths fit into the path difference. Multiplying by 360 degrees is simply for conversion into phase difference, for each wavelength that fits into the path difference, there will be a full cycle of 360 degrees. Does this sound right? Did I miss anything?

That's all there is to it.

Howver, one thing that I had trouble figuring out with this equation, is the reasoning behind doing thi in the opposite way, first multiplying by 360 degrees, and then dividing by the wavelength, and I couldn't come up with anything reasonable, I couldn't figure out these questions: Why do you multpliy by 360 degrees right away? And what do you do with these strange units you get(path difference * 360 degrees gets you very strange units in the result that I didn't know how to work with, metre * degrees?)?

I have no idea what you are asking here. Are you asking whether you can first multiply by 360 degrees and then divide by the wavelength? Sure, why not?

 

[Byrgg]

Not at all. As I thought we've shown, to a very good approximation you will easily have a central maximum.

I know, but true constuctive interference would only result when all segments are in phase, right? So isn't a real life example of constructive intereference almost impossible to create, seeing as all if the segments of the light have to travel the same distance?

 

[Byrgg]

I have no idea what you are asking here. Are you asking whether you can first multiply by 360 degrees and then divide by the wavelength? Sure, why not?

I understand that the numbers would work out to be the same, but how would you explain the reasoning(as I had done in my example) behind this? Of what importance is the value of the path difference multiplied by a single cycle(and what's witht he weird units you get, metre*degrees?)? And why would you choose to multiply by a single cycle right away? I understand the reasoning if you do the division first, but the reasoning behind doing the multiplication first is what I find confusing, as demonstrated by my questions about it.

 

[Doc Al]

I know, but true constuctive interference would only result when all segments are in phase, right? So isn't a real life example of constructive intereference almost impossible to create, seeing as all if the segments of the light have to travel the same distance?

If you are talking about PERFECT, 100% constructive interference from a single slit, then sure that's impossible. But so what? You must realize that constructive interference is a matter of degree, not an all or nothing affair. Perfectly ordinary slits give perfectly acceptable, very bright, central maxima due to a high degree of constructive interference.

 

[Doc Al]

I understand that the numbers would work out to be the same, but how would you explain the reasoning(as I had done in my example) behind this? Of what importance is the value of the path difference multiplied by a single cycle(and what's witht he weird units you get, metre*degrees?)? And why would you choose to multiply by a single cycle right away? I understand the reasoning if you do the division first, but the reasoning behind doing the multiplication first is what I find confusing, as demonstrated by my questions about it.

I have no idea why you would multiply path difference (in meters) by 360 degrees and retain that number. Why would you want to do this? Seems pretty meaningless to me except as a step in calculating phase difference.

 

[Byrgg]

If you are talking about PERFECT, 100% constructive interference from a single slit, then sure that's impossible. But so what? You must realize that constructive interference is a matter of degree, not an all or nothing affair. Perfectly ordinary slits give perfectly acceptable, very bright, central maxima due to a high degree of constructive interference.

If the segments are even slightly out of phase, isn't it considered destructive interference? I thought that constructive interference only occurred if the two segments were perfectly in phase.

 

[Byrgg]

I have no idea why you would multiply path difference (in meters) by 360 degrees and retain that number. Why would you want to do this? Seems pretty meaningless to me except as a step in calculating phase difference.

By the step in calculating phase difference, do you mean the step after dividing the path difference by the wavelength? If so, it seems I'm on the right track, because this is similar to what I was thinking, I could think of no real reasoning behind multplying by 360 degrees first, even though, mathematically, you can still do this.

 

[Doc Al]

If the segments are even slightly out of phase, isn't it considered destructive interference? I thought that constructive interference only occurred if the two segments were perfectly in phase.

If two segments are exactly in phase, you will have perfect constructive interference; if exactly out of phase (by 180 degrees), you will have perfect destructive interference. For other phase differences, you'll get something in between.

 

[Byrgg]

If two segments are exactly in phase, you will have perfect constructive interference; if exactly out of phase (by 180 degrees), you will have perfect destructive interference. For other phase differences, you'll get something in between.

I know, but what is 'something in between'? Isn't is a partial amount of destructive interference(reduces the overall amplitude, but not entirely)?

 

[Doc Al]

I know, but what is 'something in between'? Isn't is a partial amount of destructive interference(reduces the overall amplitude, but not entirely)?

Perhaps you think that the sum of two segments of light that are slightly out of phase has smaller amplitude than either of the two segments alone? Not true at all. The amplitude of the sum is not quite twice as big, but close.

 

[Byrgg]

Perhaps you think that the sum of two segments of light that are slightly out of phase has smaller amplitude than either of the two segments alone? Not true at all. The amplitude of the sum is not quite twice as big, but close.

So then at what point does this phase difference not cause in increase in amplitude, but rather a decrease? I know that at 180 degrees phase difference, they will completely cancel, but at what phase difference will they partially cancel?

 

[jtbell]

So then at what point does this phase difference not cause in increase in amplitude, but rather a decrease?

Use the formula in post #49 in this thread, and let A_2 = A_1 to make the two initial waves the same amplitude. Either experiment with various values for the phase difference \delta, or do a bit of algebra to find the value of \delta that makes A = A_1.

 

[Byrgg]

Use the formula in post #49 in this thread, and let A_2 = A_1 to make the two initial waves the same amplitude. Either experiment with various values for the phase difference \delta, or do a bit of algebra to find the value of \delta that makes A = A_1.

Ok, thanks for that info, but what about my other questions, could someone help me those please?

 

[Byrgg]

I have no idea why you would multiply path difference (in meters) by 360 degrees and retain that number. Why would you want to do this? Seems pretty meaningless to me except as a step in calculating phase difference.

I was thinking about this a little more, and I thought, sure the path difference multiplied by 360 degrees doesn't seem to hold any significance, but couldn't the same be said for the path difference divided by the wavelength? I'm just trying to sort this out, I have a hard time trying to understand how one way makes sense, and the other doesn't.

 

[Doc Al]

... the path difference multiplied by 360 degrees doesn't seem to hold any significance, but couldn't the same be said for the path difference divided by the wavelength?

Path difference is just a distance. What matters (in this context) is the phase difference, which depends on the wavelength. Dividing path difference by wavelength is meaningful, as it gives you the phase difference in terms of the number of wavelengths. Multiplying by 360 degrees just converts the number of wavelengths to degrees.

That last step is optional: A phase difference can be equally well expressed as a number of wavelengths or a number of degrees.

 

[Byrgg]

Path difference is just a distance. What matters (in this context) is the phase difference, which depends on the wavelength. Dividing path difference by wavelength is meaningful, as it gives you the phase difference in terms of the number of wavelengths. Multiplying by 360 degrees just converts the number of wavelengths to degrees.

That last step is optional: A phase difference can be equally well expressed as a number of wavelengths or a number of degrees.

That last part really helped, thanks. I think I see now, because the 360 degrees is just for conversion, right? And because you can express it as a number of wavelengths, you are technically already complete, right? And, even though it IS possible, mathematically, to mutliply by 360 first, there's really no reasoning in doing so, because the resulting product is a fairly meaningless number, right?

 

[Doc Al]

Right. Sounds good to me.

 

[Byrgg]

And also, you can't express the phase difference in terms of the path difference because it's dependent on the wavelength, like you said, right? It's just a distance, as you said earlier, and because it does not reflect the wavelength in any way, it alone, cannot be used to describe the phase difference, right?

 

[Doc Al]

You are correct.

 

[Byrgg]

This keeps bugging me, I don't know why, for some reason, I just can't get over the fact that there's no real point in multiplying the path difference by 360 degrees. It just seems so weird that you can mathematically do it, but there's no real explanation as to why.

 

[Byrgg]

Someone please help, I think that another example, similar to this one(preferably a simple one, I've only completed gr.11 physics), would be helpful.

 

[Byrgg]

Could someone please help?

 

[Byrgg]

Someone please help me with this? I can try clarifying what it is I'm wondering exactly, if it's necessary.

 

[Doc Al]

I can try clarifying what it is I'm wondering exactly, if it's necessary.

I think it's necessary. I, for one, have no idea what's troubling you.

 

[Byrgg]

Alright, I'll explain what I'm trying to figure out here.

Basically, I'm confused about the reasoning behind doing the calculation for phase difference in ways that don't seem to make sense, but they do, in fact, mathematically work out to be correct.

The equation: phase difference = 360 degrees * path difference/wavelength, comes about from phase difference being able to be expressed as a number of wavelengths, or a certain number of degrees.

Earlier in this thread, something was said, similar to the following: The path difference divided by the wavelength gives the number of wavelengths that will fit into the path difference, and multiplying by 360 degrees is just a way to convert the number of wavelengths to to degrees(since 1 wavelength = 360 degrees).

As I said earlier in the post, what's confusing me, something that's been discussed a bit before in this thread, is the reasoning behind doing the calculation another way. I originally figured out that, mathematically, it's possible to first multiply the path difference by 360 degrees, and then divide by the wavelength, and achieve the same answer as you would if you did it the normal way, which is to divide the path difference by the wavelength, and then multiply by 360 degrees. More recently, I have also figured out that it would be possible to divide 360 degrees by the wavelength, and then multiply by the path difference, which, again, results in the same answer you would achieve from doing this the normal way.

I understand that these other two methods are mathematically possible, but I'm having trouble trying to figure out the reasoning behind these methods. As said earlier, in the normal way to do this, you start by getting the number of wavelengths which fit into the path difference, this is one way to express the phase difference, and is thus a meaningful number. However, the other methods you can take, involve steps which do not seem as logical, the result of the path difference, multiplied by 360 degrees, while possible to do mathematically, does not seem to bear any relevance when taken at face value, nor does 360 degrees divided by the wavelength, which is like the other value, mathematically possible to do.

I'm having trouble understanding how it is exactly, that these numbers are irrelevant, one of my early impressions was that, the units you get when you carry out these steps do not seem to make any sense, 360 degrees multplied by the path difference gets you a number which is units of degrees*metres(or any other valid unit that can be used to measure distance). 360 degrees divided by the wavelength, also gets you strange units(different from the previously mentioned strange units) of degrees/metres(or any other valid unit that can be used to measure distance). One thing that I thought, was that these strange units may be reason enough to claim the steps that get them to be nonsensical, though I'm not sure if it correct to say so, and so this is one source of my confusion. I also thought that there may still be other reasons for these steps to not seem relevant.

One other reason that I thought these steps did not seem to bear relevance, was that 360 degrees seems only to be the conversion factor to multiply a number of wavelengths by(similar to 100 centimetres being the value which you multiply metres by, in order to convert metres to centimetres), being nothing but a conversion factor to turn a number of wavelengths into degrees, I figured it was meaningless to multiply this number by the path difference, or divide it by the wavelength. However, I don't know if it is correct to assume that 360 degrees is only the conversion factor used to convert a number of wavelengths to a number of degrees, another source of confusion.

Despite my reasons for thinking these steps were meaningless, though mathematically valid, I thought that if someone had great enough knowledge about wavelengths, and how they relate to degrees and such, should be able to use their knowledge as reasoning for these steps. However, I thought that, despite someone having this knowledge, my other reasons may still stand, and so in order to say you can take these steps, you would have to have knowledge of the normal way to do the equation. This is my main source of confusion, would it be accurate to say that someone with enough knowledge on the subject could take these steps, which seem nonsensical, when taken at face value, or would they not be able to provide reasoning for these steps at all, without having knowledge of the normal way to do the equation?

I know there's a lot there, but if someone could help me figure out these questions, I'd be greatful.

 

[Doc Al]

I'm still not quite zeroing in on your confusion, but here are some ideas.

What matters is expressing the path difference in terms of wavelengths. To do that, you need to know the wavelength. In a sense, that first step can be viewed as a conversion--from meters to wavelengths. Here's an example:

Let 1 wavelength = 600 nm (= 600 x 10^-9 m)
Let the path difference = 400 nm

So:
path difference = 400 nm * ( 1 wavelength/ 600 nm) = 2/3 wavelengths

Now a number of wavelengths can be expressed as a phase difference, since 1 wavelength = 360 degrees. (That's all there is to that.) So, putting it all together we get:

Phase difference = 400 nm * ( 1 wavelength/ 600 nm) * (360 degrees/ 1 wavelength) = 240 degrees

That's how you find the phase difference in degrees. I think you understand that. I think your question is "Is there some secret significance to multiplying all the numerators together first, then dividing by the denominators?" Like this:

Phase difference = [400 nm * 1 wavelength * 360 degrees]/[600 nm * 1 wavelength] = 240 degrees

Does [400 nm * 1 wavelength * 360 degrees] have some special meaning? Not to me!

 

[Byrgg]

path difference = 400 nm * ( 1 wavelength/ 600 nm) = 2/3 wavelengths

Why did you multiply by (1 wavelength/600nm)? I thought you only divided by the wavelength, which, in this case, is 600nm. Does have something to do with unit conversions?

Does [400 nm * 1 wavelength * 360 degrees] have some special meaning? Not to me!

Why did you multiply the path differece(400 nm) by 360 degrees and one wavelength? Why do you multiply by one wavelength? If it's conversion related, wouldn't you multiply by (360 degrees/1 wavelength) instead of (360 degrees*1 wavelength)? Or am I missing something?

Does [400 nm * 1 wavelength * 360 degrees] have some special meaning? Not to me!

It seems logical to say that this value does not have any relevance or special meaning, but I find it strange how something that can be mathematically, does not have any real explanation or reasoning behind it. This is where my points of of confusion came up, I'll summarize them again here, and maybe someone could help me figure them out.

One source of confusion is the strange units that you'll get when you multiply 360 degrees by the path difference, assuming you measured the path difference in metres, here, you would get units of degrees*metres. The same goes for dividing 360 degrees by wavelength, which is possible to do mathematically, but, assuming you measured the path difference in metres, results in units degrees/metres.

In any equation I've learned, the units always seem to make sense in one way or another, though, I can't think of any reason why a number in the units of degrees*metres, or degrees/metres would be useful. Here, the confusion arises from not knowing whether or not these are useful units, and if they aren't useful units, is this reason enough to dismiss these steps as nonsensical?

Another source of confusion relates to the purpose of the 360 degrees. Is it correct to say that this number is only a conversion factor, which when multiplied by a number of wavelengths, gets you the number of wavelengths as a number in degrees? If this is only a conversion factor for a number of wavelengths, is this also reason enough to dismiss these steps as nonsensical?

My final source of confusion is, would a person with great enough knowledge on the subject be able to use their knowledge as reasoning to first take one of these seemingly nonsenical steps? Or, would you assume, that, in order for anyone-- no matter how great their knowledge-- to take these steps, they would have to have knowledge of the normal way to do the equation?

I hope that explained exactly what's confusing me a little better. I can try to explain it a bit more, if no one really understands what it is I'm having trouble with here.