I was wondering something about diffraction

[Doc Al]

So you're saying that the angles are equal? Then why are they only approximated to be equal in the diagram in one of the links you gave me?

Because that page defined the angle a bit differently than I did. They defined \theta' by constructing a line that starts at the top of the slit and is perpendicular to the "ray" drawn to the bottom of the slit. But in the "far screen" approximation, all the rays are treated as parallel, thus that line is perpendicular to all of them. (Just like in my definition.)

For reference, here's that page again: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/fraungeo.html#c1

This is the problem, I can't find a way to add the phase differences that makes sense. When you said light from the top and bottom of the slit will exactly exactly cancel each other, you meant the light from the top and bottom halves of the slit, right?

In order for the light in the top half to cancel with the light in the bottom half, they must be exactly half a wavelength out of phase, but how is this achieved exactly?

jtbell has kindly drawn a very useful diagram explaining the whole thing. Study it. (Maybe his explanation will click better than mine.)

In the meantime, let me try once more to describe what I'm talking about. Divide the slit into small segments, from top to bottom. How many? As many as we need to understand what's going on! Let's label the segments for reference, from top to bottom. I'm going to label the segments in the top half of the slit like this: T-1, T-2, T-3, etc. Similarly, I'll label the segments in the bottom half of the slit like this: B-1, B-2, B-3, etc. I hope this is clear.

Now the light from each segment goes out everywhere, hitting all points of the screen. But at any particular point on the screen, since it's at an angle to the slit, the light from each segment ends up just a little bit further out of phase than the segment above it, since the light has further to travel.

Now let's discuss a special point on the screen, the location of the first minima. That occurs when the angle is such that light from the first segment of the bottom half of the slit (B-1) travels exactly one half wavelength farther than the light from the top segment (T-1) by the time the light hits the screen. Since the light from B-1 and T-1 are exactly out of phase, they cancel out.

If T-1 and B-1 are out of phase, then it follows that T-2 and B-2 are also out of phase by the same amount. (Got that?) That means the light from segments T-2 and B-2 cancel out. Using this same reasoning, you should agree that every segment from the top half will exactly cancel out every segment from the bottom half. Thus, at the special angle where a minimum occurs, all the light from the slit cancels and the spot is dark.

Let me know if this is any clearer.

 

[jtbell]

Let me guess (although you really should have told me which ray you're thinking of so I don't have to guess )... you're either thinking:

(a) If rays 0 and 180 cancel, what does ray 360 cancel?

or

(b) If rays 180 and 360 cancel, what does ray 0 cancel?

This problem arises because the light heading from the slit towards the convergence point on the screen isn't a finite number of discrete rays. In terms of our diagram, what we really have is a continuous band of light, filling the spaces between the rays.

Let's sharpen up our procedure by associating each ray with a narrow band of light, such that these narrow bands completely fill up the broad band coming from the slit. For example, let the ray with phase 20 correspond to the band beween 10 and 30; let the ray with phase 40 correspond to the band between 30 and 50; etc. This gives us bands with width 20. But the two rays at the end are different! The ray with phase 0 can correspond only to a band between 0 and 10, and the ray with phase 360 can correspond only to a band between 350 and 360. Both of these "edge bands" have width 10.

The "edge bands" at 0 and 360 each carry only half as much light as one of the "internal bands," so neither of them alone can cancel the band at 180. But they're really the same phase, so they add constructively, and therefore, together they do cancel the band at 180! :!)

 

[Byrgg]

Let me guess (although you really should have told me which ray you're thinking of so I don't have to guess )... you're either thinking:

(a) If rays 0 and 180 cancel, what does ray 360 cancel?

or

(b) If rays 180 and 360 cancel, what does ray 0 cancel?

This problem arises because the light heading from the slit towards the convergence point on the screen isn't a finite number of discrete rays. In terms of our diagram, what we really have is a continuous band of light, filling the spaces between the rays.

Let's sharpen up our procedure by associating each ray with a narrow band of light, such that these narrow bands completely fill up the broad band coming from the slit. For example, let the ray with phase 20 correspond to the band beween 10 and 30; let the ray with phase 40 correspond to the band between 30 and 50; etc. This gives us bands with width 20. But the two rays at the end are different! The ray with phase 0 can correspond only to a band between 0 and 10, and the ray with phase 360 can correspond only to a band between 350 and 360. Both of these "edge bands" have width 10.

The "edge bands" at 0 and 360 each carry only half as much light as one of the "internal bands," so neither of them alone can cancel the band at 180. But they're really the same phase, so they add constructively, and therefore, together they do cancel the band at 180! :!)

That explanation was very helpful, but how would it work out if you drew an even number of rays?

 

[jtbell]

If the rays in my diagram are at phases 10, 30, 50, ..., 350, then each of the associated bands has width 20 (0 to 20, 20 to 40, etc.), and you simply pair them up so each pair has a phase difference of 180 degrees. In that case there's no "middle ray" at 180. The rays at 10 and 190 cancel, and so do the ones at 170 and 350.

If you're thinking of an even number of rays, with two of them coming from the edges of the slit (phases 0 and 360), I suspect that in that case it's not possible to have rays that are both both equally spaced and can be put in pairs that are 180 degrees apart.

For example, I came up with a set of 16 equally-spaced rays starting at 0 and ending at 360 (0, 24, 48, ..., 312, 336, 360) but no two of them are 180 degrees apart.

 

[Byrgg]

So then it's impossible to make that diagram with an even number of rays if they(0 and 360) start at the edges of the slit?

One of the major problems I'm having here, is that, when using an even number of rays, I can't understand how the ray at the top of the slit can be in phase with the ray at the bottom of the slit, while those rays are still canceled out by rays in the center, it doesn't make any sense to me seeing as the the 2 rays in the center would need to be in phase, and yet adjacent rays can't be in phase. I also remember reading somewhere that each corner produces it's own part of the overall diffraction pattern, thus I would assume the model should use an even numbre of rays. Is there anything else someone could say to help clarify this whole thing a bit more?

 

[Doc Al]

You'd be better off thinking of the "rays" in your model as representing a segment of the light passing through the slit. You can represent that light with any number of segments or "rays" that you wish. Of course, to draw reasonable conclusions about interference effects you need to use segments that are: (1) small enough to be described as having a single phase, and (2) the same size, thus representing the same amount of light.

You can model it anyway you want, but if you don't follow those rules you'll have a hard time drawing a clear conclusion. In addition, if you wisely divide the slit into an even number of segments you will be able to reason as I described in my last post. You'll be able to compare corresponding segments (rays) in the top and bottom halves of the slit and see that they produce perfect destructive interference.

Please point out the exact statements in this post and my preceding post that you don't understand.

 

[jtbell]

The real way to do this is to use calculus: set up an integral that simultaneously adds the waves from all parts of the slit (each wave with a different phase) that arrive at a point on the screen specified by the angle \theta. This gives you the amplitude of the light wave as a function of \theta, and squaring it gives you the intensity ("brightness") of the light. Then you identify the values of \theta that give you an intensity (or amplitude) of zero. This is covered in intermediate-level optics books.

The procedure that we've been struggling with in this thread is a shortcut method that requires the rays to be set up in certain symmetrical ways in order to work. Even then you have to fuss with certain specific rays in order to get all the details just right, as I did a few posts ago. It's not a general procedure, and in my opinion you shouldn't waste too much time on it.

 

[Byrgg]

You'd be better off thinking of the "rays" in your model as representing a segment of the light passing through the slit. You can represent that light with any number of segments or "rays" that you wish. Of course, to draw reasonable conclusions about interference effects you need to use segments that are: (1) small enough to be described as having a single phase, and (2) the same size, thus representing the same amount of light.

You can model it anyway you want, but if you don't follow those rules you'll have a hard time drawing a clear conclusion. In addition, if you wisely divide the slit into an even number of segments you will be able to reason as I described in my last post. You'll be able to compare corresponding segments (rays) in the top and bottom halves of the slit and see that they produce perfect destructive interference.

Please point out the exact statements in this post and my preceding post that you don't understand.

You said that if I divide the slit into an even number of segments, I will be
able to reason as you described in your last post. Here's what I don't understand, you used 6 rays in your explanation. In order for your explanation to hold true, T-1 has to half a wavelength out of phase with B-1, and this continues with the pairs T-2 and B-2, as well as T-3 and B-3. But as far as I know, T-1 and B-3 are at the edges of the slit, and so they must be in phase, since the condition for a minima is asin\theta = m\lambda. If they are in phase, and they both cancel with rays in the center, then those 2 rays in the center must be in phase, and it is my understanding that adjacent segments can't be in phase. This is what I don't understand.

 

[jtbell]

But as far as I know, T-1 and B-3 are at the edges of the slit, and so they must be in phase,

No, they're not. T-1, B-3, etc. are strips or bands of light rays that include a range of phase angles. If there are six of them, and \theta is such that we have the first minimum in the total intensity, then T-1 spans a range of phase angles from 0 to 60 degrees, and B-3 spans a range of phase angles from 300 to 360 degrees.

When setting up the cancellation of rays, we need to choose the most representative ray from each band, that is, the one with the most representative phase angle. In a range of 0 to 60 degrees, which phase angle is the most representative one? Not 0, the one on the very edge of the range, but rather, 30, the one in the middle!

Working through the six bands, you should be able to see that the most representative phase angle for each band is as follows:

T-1: 30 degrees (not 0!)
T-2: 90 degrees
T-3: 150 degrees
B-1: 210 degrees
B-2: 270 degrees
B-3: 330 degrees (not 360!)

The representative rays for T-1 and B-1 are 180 degrees apart, as are the ones for T-2 and B-2, and the ones for T-3 and B-3.

You may object that there's something fishy about using a ray with a phase of 30 degrees to represent a ray that actually has a phase of 0, or 60, or 45, or any other number between 0 and 60 that isn't actually 30. And indeed, in general this sort of thing gives only an approximation to the actual result. In this particular case, it works OK because of the symmetries of the situation, but in general you can't count on it.

To get a better approximation, we use more bands, which are narrower: 10 bands that each span 36 degrees in phase, 36 bands that each span 10 degrees in phase, 360 bands that each span 1 degree in phase, 3600 bands that each span 0.1 degree in phase, etc. In this particular case, we still get the same kind of cancellation as we use more and more bands, which indicates that our choice of representative rays for each bands is valid. In general, though, we have to add up the contributions from each band and find the limiting value of the sum as the number of bands tends toward infinity, and the width of each band tends toward zero. This is precisely what integral calculus does.

 

[Byrgg]

Alright, I know it's been a while, but I came up with a few more questions about this, some may have already been explained earlier, at least partly, but I'm still going to ask about the things I'm not sure about.

First, when using the diagram, do you really treat the rays as parallel? If you are assuming the rays have phase differences, then they can't be parallel, right? I mean, In jtbell's diagram, the rays were parallel, but at the end, there was an angled screen, which allowed you to analyze the rays as not being parallel, since they all hit the screen at different phases.

Second, how is it that the central maxima is where total consstructive interference occurs? It was said earlier that all rays must travel the same distance to the center, but I don't believe this to be true, here's a rough diagram to explain my thinking:

|
.1
.2
.3
.4
.5
.6
.7
.8
|

The central maxima would occur at a point on the screen directly across from the point between rays 4 and 5, right? If this is the case, then sure, rays 4 and 5 travel the same distance to this point, as do rays 3 and 6, 2 and 7, and 1 and 8. But the thing is, the pair of say, rays 4 and 5 has a shotrer pathlength then the pair of rays, 3 and 6. Therefore, not all rays travel the same distance. If someone could explaine how this causes the central maxima, I'd be greatful.

 

[Doc Al]

First, when using the diagram, do you really treat the rays as parallel? If you are assuming the rays have phase differences, then they can't be parallel, right? I mean, In jtbell's diagram, the rays were parallel, but at the end, there was an angled screen, which allowed you to analyze the rays as not being parallel, since they all hit the screen at different phases.

Yes, you treat the rays as parallel. Why do you think that rays with phase differences cannot be parallel? Two parallel rays can certainly travel different distances and thus reach the screen with different phases.

Are the rays exactly parallel? Of course not, since parallel rays would not reach the same point on the screen. But it's an excellent approximation. To see how good an approximation it is, figure out the angle between two rays from opposite ends of the slit. (Realize that for visible light, a typical slit size would be a fraction of a millimeter while the distance to the slit would be about a meter.)

Second, how is it that the central maxima is where total consstructive interference occurs? It was said earlier that all rays must travel the same distance to the center, but I don't believe this to be true, here's a rough diagram to explain my thinking:

|
.1
.2
.3
.4
.5
.6
.7
.8
|

The central maxima would occur at a point on the screen directly across from the point between rays 4 and 5, right? If this is the case, then sure, rays 4 and 5 travel the same distance to this point, as do rays 3 and 6, 2 and 7, and 1 and 8. But the thing is, the pair of say, rays 4 and 5 has a shotrer pathlength then the pair of rays, 3 and 6. Therefore, not all rays travel the same distance. If someone could explaine how this causes the central maxima, I'd be greatful.

Again, to an excellent approximation, the rays from the slit that produce the central maximum are parallel and thus travel the same distance and arrive at the screen in phase. Just for fun, why not calculate the difference in distance traveled for rays from the edge of the slit and from the middle of the slit. Use these values: slit width = 0.1 mm, distance to screen = 1.0 m. Then figure out the resulting phase difference, assuming light of wavelength = 600 nm.

 

[Byrgg]

Yes, you treat the rays as parallel. Why do you think that rays with phase differences cannot be parallel? Two parallel rays can certainly travel different distances and thus reach the screen with different phases.

Rays with phase differences couldn't be parallel if they meet at the same point on the screen, if they all travel at the same angle towards the screen, then they will be parallel, but will not meet, if some parts of teh screen were further away, then they could have different phases when reaching the screen, or am I wrong? This is what I'm getting at, you assume the rays aren't parallel, since they do meet at a point. If they were parallel, the only thing that could cause phase differences would be for them to travel different distances, right? This could only happen if the surface of the screen was not perpendicular to the gap, right?

Again, to an excellent approximation, the rays from the slit that produce the central maximum are parallel and thus travel the same distance and arrive at the screen in phase. Just for fun, why not calculate the difference in distance traveled for rays from the edge of the slit and from the middle of the slit. Use these values: slit width = 0.1 mm, distance to screen = 1.0 m. Then figure out the resulting phase difference, assuming light of wavelength = 600 nm.

They are parallel to an excellent approximation? Is this because the screen is far away from the gap? What if the screen wasn't so far away? Since these rays are parallel to an excellent approximation, then would be accurate to say that some, though a very little amount destructive interference occurs at teh central maxima?

 

[jtbell]

See the attached diagram for my version of Doc Al's exercise. It uses slightly different numbers because I didn't think to look at his numbers carefully until after I had drawn the diagram and scanned it, and I'm too lazy to do it all over again.

It shows two different versions of a slit and screen setup. The top one is the "exact" version, with two non-parallel rays coming from the top and bottom edges of the slit, with lengths R_1 and R_2. Part (a) of your assignment is to calculate the path difference R_2 - R_1, and the corresponding phase difference, (360 degrees) * (path difference) / (wavelength).

The bottom version is the "Fraunhofer approximation" version, showing two parallel rays coming from the top and bottom of the slit, and hitting an angled screen squarely. These rays have lengths R_1' and R_2'. Part (b) of your assignment is to calculate the path difference R_2' - R_1' and the corresponding phase difference.

Note that in the bottom version it doesn't actually matter what R_1' is, because moving the screen backwards or forwards along the rays doesn't change the path difference. That's why I didn't specify the distance from the slit to the screen. You can pick any distance you like, if you need one.

Part (c) of your assignment is to compare the two phase differences, and judge how good an approximation part (b) is to part (a).

 

[Byrgg]

For, jtbell's little assignment, I'm having trouble trying to sort thimgs out. Could someone lend me a hand?

For the first diagram, R_1 would be equal to the square root of 200^2 + 0.995^2, right?

And R_2 would be equal to the square root of 200^2 + 1.005^2, right?

Someone please let me know if I'm doing this right, I don't want to mess this up somehow. I followed through with my work and ended up with some huge number for the path difference, I think I messed up my converisons somewhere, but could someone just let me know if I'm on the right track?

 

[jtbell]

For the first diagram, R_1 would be equal to the square root of 200^2 + 0.995^2, right?

And R_2 would be equal to the square root of 200^2 + 1.005^2, right?

If you're using cm for distance, those are correct. You're on the right track. Just make sure you consistently use cm for everything else, including wavelength. And be careful to avoid roundoff errors when you're doing the arithmetic. Keep intermediate results in your calculator's memory if possible, and don't round them off.

 

[Byrgg]

In response to Doc Al's post, I think I calculated the phase difference(using the method in jtbell's post) between the paths taken by an outer ray, and an inner ray, and I got 0.6, could someone verify if that's correct please?

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? 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? 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? There's a lot of back in forth in my thinking about this, which is probably why I cna't figure it out, so if someone could help clarify this, it would be greatly appreciated.

 

[jtbell]

In response to Doc Al's post, I think I calculated the phase difference(using the method in jtbell's post) between the paths taken by an outer ray, and an inner ray, and I got 0.6, could someone verify if that's correct please?

You didn't do it for the two outer rays as in my diagram? By "inner ray", do you mean the one starting at the center of the slit?

For that diagram, the phase difference should be 360 degrees between the two outer rays, and 180 degrees between an outer ray (either one) and the center ray.

Show us your work and I (or someone else) can probably spot where you went wrong.

 

[Doc Al]

In response to Doc Al's post, I think I calculated the phase difference(using the method in jtbell's post) between the paths taken by an outer ray, and an inner ray, and I got 0.6, could someone verify if that's correct please?

That's a phase difference of 0.6 degrees, I presume. Assuming you are giving your solution to the exercise I proposed at the end of post #41, using my numbers, then your answer is close. Here's how I would do it:

R^2 = D^2 + (d/2)^2
where R is the path length of the "outer" ray, D (= 1 m) is the path length of the central ray (and equals the distance between screen and slit), and d (= 0.1 mm) is the size of the slit.

Thus:
R = \sqrt{D^2 + (d/2)^2} = D \sqrt{1 + (d/2D)^2}

Since the second term under the square root sign is small, you can approximate the answer like this:
R = D (1 + (1/2)(d/2D)^2) = D + (1/8)(d^2/D)

That last term is our path length difference; it equals 1.25 nm. Using my wavelength of 600 nm gives a phase difference of (1.25/600)*(360) = 0.75 degrees. That's not much.

 

[jtbell]

]You didn't do it for the two outer rays as in my diagram?

Oops, I now realize that you [Byrgg] were talking about Doc Al's diagram.

For his setup, the rays arrive at the middle of the central maximum, so ideally we should get a phase difference of zero. In fact, we don't quite get zero, as you've calculated. Nevertheless a phase difference of 0.75 degree (I assume Doc Al's figure is more accurate) is negligible in this situation. Two waves with amplitude A_1 and A_2 and a phase difference \delta combine to give a new wave with amplitude

A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \delta}

Letting A_2 = A_1 (same amplitude for both waves) and \delta = 0.75 degree gives A = 1.999957 A_1. If there were perfect constructive interference, we would have A = 2A_1 exactly. The two situations are indistinguishable for all practical purposes.

I constructed my example so the rays meet at the first minimum, so the phase difference between rays from the two edges of the slit is ideally 360 degrees. Between a ray from one edge and a ray from the middle, the phase difference is ideally 180 degrees. To the resolution of my calculator, there's no difference between the parallel-ray version and the non-parallel ray version of the calculation. Either way, I get 360 degrees or 180 degrees depending on which two rays I use.

 

[Byrgg]

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? 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? 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? There's a lot of back in forth in my thinking about this, which is probably why I cna't figure it out, so if someone could help clarify this, it would be greatly appreciated.

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

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

 

[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.