I was wondering something about diffraction

[Byrgg]

I'm pretty sure the equation for finding the minima in a diffraction pattern is:

$$\sin\theta_m = m\frac{\lambda}{a}$$

Where m is an integer which is 1 for the first minima, 2 for second, etc, lambda is the wavelength of the wave and l is the width of the opening.

And I'm pretty sure the equation for finding maxima is:

$$\sin\theta_m = (m + 0.5)\frac{\lambda}{a}$$

Where m is an integer which is 1 for the first maxima, 2 for second, etc, lambda is the wavelength of the wave and l is the width of the opening.

What I'm wondering is why you multiply by n for the minima, and why you multiply by m + 1/2 for the maxima.

Please, someone respond soon, thanks in advance.

 

[Doc Al]

If you are talking about single slit diffraction (Fraunhofer limit), then the condition for the minima is:
$$\sin\theta_m = m\frac{\lambda}{a}$$
where a is the width of the aperture.

The condition for the maxima is more complicated; it's not simply a matter of replacing m by m + 1/2, though that's a rough approximation.

To find out why you use m, you'll have to learn how the formula describing the condition for minima is derived.

Have you bothered to study the simpler two-slit diffraction pattern yet?

 

[Byrgg]

Err, no, I haven't studied two-slit diffraction before...

Anyway, yeah, I'm wondering why you use m, basically, as you said, how the formula is derived.

 

[Doc Al]

My recommendation, as before, is to start with the easier stuff and then move on from there. Two-slit diffraction will give you the key concepts that will be reused many times over in exploring diffraction.

I think we've discussed this quite a bit before (https://www.physicsforums.com/showthread.php?t=123367); you refer to this equation (with m = 1) in your first post in that thread. Follow the links I gave you then.

 

[Byrgg]

I read through those links a bit again, but I'm still unsure as to why m is used in the formulae which are used to find the minima and maxima, would you be able to help me out a bit with this?

Also, what value of m, if any, do you use for the central maxima?

 

[Doc Al]

Here's the page that I would have you refer to to understand how the formula for the minima condition is derived: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.html#c1

Look at the wavefront at the aperture. Rays 3 and 4 meet at the minima; their difference in path length (and thus phase) must equal $\lambda/2$ if they are to be out of phase and thus cancel. Note that 3 starts at the top of the aperture while 4 starts at the middle. If those rays are out of phase, then the points just below the top and just below the middle will also be out of phase.

Follow this line of reasoning and you'll see that all rays (from all points along the wavefront at the aperture) leading to that point on the screen will cancel, making the point a minima. Express this mathematically (http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html#c1) and you'll have the formula for the minima.

As I've said several times, the formula is for minima. The condition for maxima is much harder to specify mathematically; similar logic applies except that you want to maximize constructive interference. Of course the central maxima is trivial: $\theta = 0$.

 

[Byrgg]

Why can't a ray traveling to the center be canceled out by another one? Since the central area is a maxima, I'm assuming that that means the rays can't arrive at this point out of phase, why is that exactly? And why are the angles maxima approximated to be $$\sin\theta_m = (m + 0.5)\frac{\lambda}{a}$$?

Hmm, I'm also wondering how the angles $$\theta$$ and $$\theta'$$ are approximated to be equal.

Also, if the condition for a minima is $$a sin\theta = m\lambda$$, then what does m signify? I still don't see how/why it's used. I know that each minima is a point where every ray is apparently canceled out by another one, but does the use of m mean the third minima is exactly 3 times as far from the centre line as the first minima? Why does it end up like this?

 

[Hootenanny]

Why can't a ray traveling to the center be canceled out by another one? Since the central area is a maxima, I'm assuming that that means the rays can't arrive at this point out of phase, why is that exactly?

Because the originate from a coherent source.

Hmm, I'm also wondering how the angles $$\theta$$ and $$\theta'$$ are approximated to be equal.

Because both angles are small when the distance from the slit to the screen is much greater than the slit width (D>>a in the diagram Doc provided), this is known as the small angle aproximation.

Also, if the condition for a minima is $$a sin\theta = m\lambda$$, then what does m signify? I still don't see how/why it's used. I know that each minima is a point where every ray is apparently canceled out by another one, but does the use of m mean the third minima is exactly 3 times as far from the centre line as the first minima? Why does it end up like this?

'm' is simply the order of diffraction, this identifies the order of the beam. For example the first order minima occurs when the there is a path difference of $frac{1}{2}\lambda$ between the two beams at this point. The second order minimum occurs when there is a path difference of $frac{1}{2}\lambda$ between the two beams.

Further Reading

• http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.html#c1

 

[Doc Al]

Why can't a ray traveling to the center be canceled out by another one? Since the central area is a maxima, I'm assuming that that means the rays can't arrive at this point out of phase, why is that exactly?

Think of the rays heading for any point on the screen as being in parallel (to a close approximation). The only reason for the rays to destructively interfere is if some are out of phase with others. But all the rays going towards $\theta = 0$ are in phase--they travel the same distance.

(Those rays don't look like they are in parallel in the diagrams because the diagrams are not drawn to scale. The screen is much farther away than show in the diagram. Of course, if drawn to scale, the angles would be too small to see.)

At any other angle, some of the rays (near the top of the slit in the diagram) have less distance to travel than other rays (near the bottom of the slit). Since they start out in phase, by the time they meet up at the screen they are not necessarily still in phase. At certain angles (the minima) the rays cancel out almost entirely; but at other angles, they don't.

And why are the angles maxima approximated to be $$\sin\theta_m = (m + 0.5)\frac{\lambda}{a}$$?

It should kind of make sense that in between the minima are the maxima; that's not an unreasonable approximation.

Hmm, I'm also wondering how the angles $$\theta$$ and $$\theta'$$ are approximated to be equal.

It's just geometry--they form similar triangles.

Also, if the condition for a minima is $$a sin\theta = m\lambda$$, then what does m signify? I still don't see how/why it's used. I know that each minima is a point where every ray is apparently canceled out by another one, but does the use of m mean the third minima is exactly 3 times as far from the centre line as the first minima? Why does it end up like this?

"m" is just an integer. You can think of it as signifying the phase difference between the top and bottom rays coming out of the slit--in terms of the number of wavelengths of phase difference. If m = 1; that means that the top and bottom are exactly one wavelength out of phase--which means they are in phase. But it also means the the top half of the slit is one half wavelength out of phase with the bottom half: it cancels out, that's why we get a minima. As you increase the angle, the waves come back into phase a bit (the cancelation is not complete, we get another maxima--not nearly as big as the central maximum, but a relative maximum nonetheless; the exact angle at which this relative maximum occurs is complicated to specify mathematically). At the angle where m = 2, the top and bottom rays are 2 wavelengths apart. That means the top 1/4 of the rays cancel the next 1/4 of the rays: so the top half of rays from the slit cancel out by themselves; same for the bottom half. Again we get a pretty good minimum. And so on...

And yes, in the small angle approximation that we are talking about here, the angle of the minima are proportional to m, so the 2nd minimum is twice as far from the center as the first minimum.

 

[Byrgg]

Hmm, I'm wondering something here, here is a crude diagram I'm going to use to help explain what I'm thinking:

Assume each of the numbered dots below is a point source.

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

When they ray coming from point 1 is exactly 1 wavelength out of phase with the ray at point 6, is it true that 6 and 3 are out of phase by 1 half, 5and 2 are out of phase by one half, and 4 and 1 are out phase by 1 half? Wouldn't the half-way point be right in the middle at 3.5? Also, why don't 6 and 1 constructively interfere, given the fact that they are in phase?

 

[Doc Al]

When they ray coming from point 1 is exactly 1 wavelength out of phase with the ray at point 6, is it true that 6 and 3 are out of phase by 1 half, 5and 2 are out of phase by one half, and 4 and 1 are out phase by 1 half? Wouldn't the half-way point be right in the middle at 3.5?

Sure, but that's a flaw of your model, not the physics. Draw 7 rays instead of 6. (In reality there are a zillion rays, not 6 or 7.)

Also, why don't 6 and 1 constructively interfere, given the fact that they are in phase?

Again, don't get too hung up on your model. A more accurate picture would model the light going through the slit as being split among 6 million rays, not 6. The fact that ray 1 and ray 6,000,000 might be in phase is countered by rays 2,999,999 and 3,000,000 also being in phase (very closely) with each other but out of phase with rays 1 and 6,000,000.

More importantly, realize that each "ray" represents a miniscule portion of the incident light. So if your model has an extra ray, forget about it.

 

[Byrgg]

Sure, but that's a flaw of your model, not the physics. Draw 7 rays instead of 6. (In reality there are a zillion rays, not 6 or 7.)

I think I read somewhere that each side of the gap produces it's own diffraction pattern, so if this is occurring on each half of the gap, the sides must have an equal number of rays passing through, and so there must be an even number of rays passing through, right? Or did I miss something?

Again, don't get too hung up on your model. A more accurate picture would model the light going through the slit as being split among 6 million rays, not 6. The fact that ray 1 and ray 6,000,000 might be in phase is countered by rays 2,999,999 and 3,000,000 also being in phase (very closely) with each other but out of phase with rays 1 and 6,000,000.

So if ray 1 and 6 000 000 are in phase, the rays 2 999 999 and 300 000 are also in phase, but out of phase with 1 and 6 000 000, so they destructively interfere? Ok, a few more things about this, instead of saying rays 3 000 000 and 2 999 999, wouldn't it be better to use rays 3 000 000 and 3 000 001? This way ray 3 000 000 and 6 000 000 would have a difference of 3 000 000, and so would rays 1 and 3 000 001, or do you have to do it the other way? Also, in this example, you mentioned the constructive interference of 2 pairs of rays, then how those rays desconstructively interfere, right? I take it if you apply the destructive interference first, then there is nothing left to constructively interfere, right? So regardless of how you approach it, the resulting effect is complete destructive interference, right? Also, is this really complete destructive interference, or just really close to it?

Think of the rays heading for any point on the screen as being in parallel (to a close approximation). The only reason for the rays to destructively interfere is if some are out of phase with others. But all the rays going towards $\theta = 0$ are in phase--they travel the same distance.

When you analyze the angle at which these rays destructively interfere, do you analyze the rays to be parallel? I was reading on another site, and got kind of confused, when it was describing the condition for the minima, the rays were all parallel in the diagram, it said because the rays were so close together, that they interfere. Is this right? I mean, when you take a point on the screen, are all the rays traveling to that point parallel? You said they are parallel to a close approximation, but now I'm getting slightly mixed messages. Could you clarify this for me?

 

[Doc Al]

I think I read somewhere that each side of the gap produces it's own diffraction pattern, so if this is occurring on each half of the gap, the sides must have an equal number of rays passing through, and so there must be an even number of rays passing through, right? Or did I miss something?

What's messing you up is thinking of the "rays" as being real chunks of light. No, they are just a mental construct to help visualize what's going on. So whether you represent a slit as having 10 or 10 million "rays" is arbitrary.

So if ray 1 and 6 000 000 are in phase, the rays 2 999 999 and 300 000 are also in phase, but out of phase with 1 and 6 000 000, so they destructively interfere? Ok, a few more things about this, instead of saying rays 3 000 000 and 2 999 999, wouldn't it be better to use rays 3 000 000 and 3 000 001? This way ray 3 000 000 and 6 000 000 would have a difference of 3 000 000, and so would rays 1 and 3 000 001, or do you have to do it the other way? Also, in this example, you mentioned the constructive interference of 2 pairs of rays, then how those rays desconstructively interfere, right? I take it if you apply the destructive interference first, then there is nothing left to constructively interfere, right? So regardless of how you approach it, the resulting effect is complete destructive interference, right? Also, is this really complete destructive interference, or just really close to it?

Sounds like you are getting the idea. You can add the interference of the various "rays" any way you like, you'll get the same answer. As far as whether you have complete destructive interference, that depends on the accuracy of your model. For example, we are ignoring the slight variation in intensity of light from the bottom of the slit compared to the top. (Since the bottom is further away from the screen, the intensity will be a little less, so the destructive interference cannot be "complete".) Within the model we are using, the interference is complete. (As opposed to the maxima where the constructive interference is far from complete--that's why the brightness of the higher order maxima decrease rapidly compared to the central maximum.)

When you analyze the angle at which these rays destructively interfere, do you analyze the rays to be parallel? I was reading on another site, and got kind of confused, when it was describing the condition for the minima, the rays were all parallel in the diagram, it said because the rays were so close together, that they interfere. Is this right? I mean, when you take a point on the screen, are all the rays traveling to that point parallel? You said they are parallel to a close approximation, but now I'm getting slightly mixed messages. Could you clarify this for me?

The way I visualize it is simple: For light "rays" from any two points in the slit to converge at the same point on the screen, they cannot be exactly parallel. But for the purposes of calculating their phase difference, you can certainly treat them as parallel. Realize that this model assumes that the screen is far away from the source.

 

[Byrgg]

What's messing you up is thinking of the "rays" as being real chunks of light. No, they are just a mental construct to help visualize what's going on. So whether you represent a slit as having 10 or 10 million "rays" is arbitrary.

Ok, so you're saying that the rays aren't really there? Also, do the 2 halves experience complete destructive interference with each other, or are there some parts that don't cancel out?

Sounds like you are getting the idea. You can add the interference of the various "rays" any way you like, you'll get the same answer. As far as whether you have complete destructive interference, that depends on the accuracy of your model. For example, we are ignoring the slight variation in intensity of light from the bottom of the slit compared to the top. (Since the bottom is further away from the screen, the intensity will be a little less, so the destructive interference cannot be "complete".) Within the model we are using, the interference is complete. (As opposed to the maxima where the constructive interference is far from complete--that's why the brightness of the higher order maxima decrease rapidly compared to the central maximum.)

The bottom is further away from the screen? Are you talking about just one point on the screen, or the screen as a whole?

The way I visualize it is simple: For light "rays" from any two points in the slit to converge at the same point on the screen, they cannot be exactly parallel. But for the purposes of calculating their phase difference, you can certainly treat them as parallel. Realize that this model assumes that the screen is far away from the source.

So then was the site's explanation wrong? It assumes that because the slit is so small, that all the rays in the gap are essentially on top of each other, and thus always interfere, but what about cases when the slit is much wider? Your explanation seems to hold true all the time, while the one I read on another site seems to hold true only if size of the opening is small, which isn't always the case, right? Is the explanation on that site a reasonable one, when the gap is small?

Also, you have to assume that the rays aren't parallel, so that they can meet, right? But you can still treat the rays to be parallel, for the sake of analyzing phase differences, and still achieve roughly the same answer, right? How does this relate to he fact that the screen is far away from the source? Oh, and the source you are assuming here is the gap, right?

 

[Byrgg]

Because both angles are small when the distance from the slit to the screen is much greater than the slit width (D>>a in the diagram Doc provided), this is known as the small angle aproximation.

Doc Al said that the angles are the same because the triangles are similar, why is this true? Also what is the difference between these 2 explanations, or are they related somehow?

 

[Hootenanny]

Doc Al said that the angles are the same because the triangles are similar, why is this true? Also what is the difference between these 2 explanations, or are they related somehow?

I myself, cannot see how the triangles are similar. If there were similar then $\theta'$ would be exactly equal to $\theta$.

Ok, so you're saying that the rays aren't really there? Also, do the 2 halves experience complete destructive interference with each other, or are there some parts that don't cancel out?

No, the rays do not really exist, they are simply a construct of physics which allows us to visualise the path of light. This is exactly the same as for the constructive interference, your model here assumes complete destruction.

The bottom is further away from the screen? Are you talking about just one point on the screen, or the screen as a whole?

Take a look at the diagram shown on http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html#c1". Note the length of the line drawn from the top of the slit compared to the length of the line drawn from the bottom of the slit. The line drawn from the bottom of the slit is longer, thus the light is further away from the screen.

How does this relate to he fact that the screen is far away from the source? Oh, and the source you are assuming here is the gap, right?

Yes, the slit is acting as the source. Basically, the further away the screen is from the slit, the more accurate the approximation that the rays are parallel.

 

[Doc Al]

Ok, so you're saying that the rays aren't really there?

Right, they are just a tool to picture things.

Also, do the 2 halves experience complete destructive interference with each other, or are there some parts that don't cancel out?

You tell me. At the angle where a minima occurs, add up all the contributions from each segment of the slit. Do they entirely cancel out?

The bottom is further away from the screen? Are you talking about just one point on the screen, or the screen as a whole?

I'm talking about a particular point on the screen, not the screen as a whole.

So then was the site's explanation wrong? It assumes that because the slit is so small, that all the rays in the gap are essentially on top of each other, and thus always interfere, but what about cases when the slit is much wider? Your explanation seems to hold true all the time, while the one I read on another site seems to hold true only if size of the opening is small, which isn't always the case, right? Is the explanation on that site a reasonable one, when the gap is small?

I don't know what site you are looking at, but I don't know what a statement such as "because the slit is so small, that all the rays in the gap are essentially on top of each other, and thus always interfere" is supposed to mean. Better to realize that light from every part of the slit is capable of reaching every part of the screen; to find the pattern, study the phase difference of light from different parts of the slit.

Also, you have to assume that the rays aren't parallel, so that they can meet, right?

Technically, sure. (Parallel rays won't intersect, right?) But the screen is so far away that the rays are as close to parallel as you want.

But you can still treat the rays to be parallel, for the sake of analyzing phase differences, and still achieve roughly the same answer, right?

Yes.

How does this relate to he fact that the screen is far away from the source?

Imagine light from different parts of the slit hitting a screen that was very close. No way to treat those rays as parallel. But as the screen gets further and further away, the rays become closer to parallel. (That's the small angle approximation.)

Oh, and the source you are assuming here is the gap, right?

Yes.

 

[Byrgg]

No, the rays do not really exist, they are simply a construct of physics which allows us to visualise the path of light. This is exactly the same as for the constructive interference, your model here assumes complete destruction.

Ok, so this model assumes complete destrcuctive interference, but in a real situation, does this still hold true? Also, where exactly is the point for being out of phase? I'm thinking it's not exactly at the midpoint, otherwise, there would be one ray that does not cancel.

Yes, the slit is acting as the source. Basically, the further away the screen is from the slit, the more accurate the approximation that the rays are parallel.

But why is the approximation more accurate when the screen is further away from the slit?

Also, for the central maxima, when I asked why the rays couldn't cancel out, it was said, "Because the originate from a coherent source."

What does that mean exactly? Why does that mean they can't cancel out?

 

[Doc Al]

Doc Al said that the angles are the same because the triangles are similar, why is this true?

It's certainly true that two right triangles with the same angles must be similar, but that might not have been very helpful. Instead, think of this: Imagine a T-shaped object, with a vertical segment and a horizontal segment. If I now tilt the horizontal segment so that it is some angle above the horizontal, then the (formerly) vertical section will be an equal angle away from the vertical. Examine the diagram to see how this relates: The horizontal segment corresponds to the ray from the center of the slit.

Also what is the difference between these 2 explanations, or are they related somehow?

You need the small angle approximation to make use of those two similar triangles: without the small angle approximation, you can't treat all rays as being parallel.

 

[Byrgg]

It's certainly true that two right triangles with the same angles must be similar, but that might not have been very helpful. Instead, think of this: Imagine a T-shaped object, with a vertical segment and a horizontal segment. If I now tilt the horizontal segment so that it is some angle above the horizontal, then the (formerly) vertical section will be an equal angle away from the vertical. Examine the diagram to see how this relates: The horizontal segment corresponds to the ray from the center of the slit.

Ok, I think I see it a bit more clearly now, but then shouldn't the angles be exactly the same, as opposed to approximately the same?

You tell me. At the angle where a minima occurs, add up all the contributions from each segment of the slit. Do they entirely cancel out?

What do you mean by adding up all the contributions of each segment of the slit?

 

[Doc Al]

Ok, I think I see it a bit more clearly now, but then shouldn't the angles be exactly the same, as opposed to approximately the same?

Where did I say they were approximately equal?

What do you mean by adding up all the contributions of each segment of the slit?

That's what we've been doing all along. (Think of each segment of light at the slit creating its own "ray".)

 

[Byrgg]

Where did I say they were approximately equal?

Well, in the example you gave me, with tht T shaped object, the angles are equal, but in on of the links you gave me, it is said that $$\theta$$ and $$\theta'$$ are approximately equal.

That's what we've been doing all along. (Think of each segment of light at the slit creating its own "ray".)

Could you show me how to model this correctly then? When I made a model of it earlier, you said it was flawed, in that I should draw 7 rays instead of 6, in which case not all of the rays cancel out, however, Hootenanny stated that the model assumed complete destructive interference.

 

[Doc Al]

Well, in the example you gave me, with tht T shaped object, the angles are equal, but in on of the links you gave me, it is said that $$\theta$$ and $$\theta'$$ are approximately equal.

It depends on how you define the angles, I suppose. (I only know how I would define the angles.) Once you get the point, you should be able to relate the angles to each other.

Could you show me how to model this correctly then? When I made a model of it earlier, you said it was flawed, in that I should draw 7 rays instead of 6, in which case not all of the rays cancel out, however, Hootenanny stated that the model assumed complete destructive interference.

What's with the 7 rays? Don't like 7? use 7 million!

Do this: Divide the top half of the slit into a million segments; do the same for the bottom half. Realize that the topmost segment of the top half is exactly out of phase with the topmost segment of the bottom half. Repeat for all two million segments.

Just to be clear: When I say that they are out of phase I mean that when the light from each of those two segments reaches the point on the screen where a minimum occurs they are out of phase. (They are in phase at the slit.)

 

[Byrgg]

It depends on how you define the angles, I suppose. (I only know how I would define the angles.) Once you get the point, you should be able to relate the angles to each other.

How you define the angles? What do you mean?

What's with the 7 rays? Don't like 7? use 7 million!

Do this: Divide the top half of the slit into a million segments; do the same for the bottom half. Realize that the topmost segment of the top half is exactly out of phase with the topmost segment of the bottom half. Repeat for all two million segments.

Just to be clear: When I say that they are out of phase I mean that when the light from each of those two segments reaches the point on the screen where a minimum occurs they are out of phase. (They are in phase at the slit.)

So then I should use an even number of rays in the model? This way everything cancels out, right? Then why did you say to use an odd number before?

Also, the ray at the top of the slit cancels out with the one just below the midpoint of the slit, and the ray at the bottom of the slit cancels out with the one just above the midpoint, right? If this is case, and the rays at the bottom and top of the slit are in phase, then it must mean that the rays just above and below the midpoint are also in phase, since they are each out of phase with the top and bottom rays, right? But then this means that 2 adjacent segments must be perfectly in phase with each other, whic can't be right... can it? I mean, if all adjacent segments were in phase, then there would be no phase differences.

Also, at the angle at which a minima occurs, the rays all start in phase then, as you said, right? I guess that means that the phase differences in the diagram in one of the links you gave were the ones which occur at the point on the screen, right?

 

[Doc Al]

How you define the angles? What do you mean?

Draw a line from a point on the screen to the middle of the slit. The angle that line makes with the horizontal is $\theta$. That line is also the "horizontal" piece of the T-shape I mentioned earlier. Now draw a line perpendicular to that line--the "vertical" piece of the T; the angle that that line makes with the vertical also equals $\theta$. That is always true, by my definition. (Of course, these angles are only useful for calculating phase differences when the angles are small.)

So then I should use an even number of rays in the model? This way everything cancels out, right? Then why did you say to use an odd number before?

As long as you are hung up on thinking that rays are real things, you will miss the point. (It's like asking how many points are on the number line between 1 and 2. For some purposes you can model it as having just a few points, but really it has an infinite number of points.) The purpose of drawing in rays is to paint a physical picture of what's going on--don't get hung up on the exact number of rays to use. Ask yourself: If I used a million times more rays, would I have the same problem? (Realize that a "ray" represents only a tiny portion of light.)

Also, the ray at the top of the slit cancels out with the one just below the midpoint of the slit, and the ray at the bottom of the slit cancels out with the one just above the midpoint, right? If this is case, and the rays at the bottom and top of the slit are in phase, then it must mean that the rays just above and below the midpoint are also in phase, since they are each out of phase with the top and bottom rays, right? But then this means that 2 adjacent segments must be perfectly in phase with each other, whic can't be right... can it? I mean, if all adjacent segments were in phase, then there would be no phase differences.

Right. Finite "rays" that are adjacent in your model cannot have exactly the same phase when they hit the screen. Find one way to add the phase differences that makes sense and stick to it: For the first minima, light from the top and bottom of the slit will exactly cancel each other. Keep thinking about this until it clicks.

Also, at the angle at which a minima occurs, the rays all start in phase then, as you said, right? I guess that means that the phase differences in the diagram in one of the links you gave were the ones which occur at the point on the screen, right?

All rays starting from the slit start out in phase regardless of where they end up on the screen. It's only because light from different parts of the slit take a longer (or shorter) path to the screen that we have phase differences leading to the diffraction pattern.

 

[Byrgg]

Draw a line from a point on the screen to the middle of the slit. The angle that line makes with the horizontal is $\theta$. That line is also the "horizontal" piece of the T-shape I mentioned earlier. Now draw a line perpendicular to that line--the "vertical" piece of the T; the angle that that line makes with the vertical also equals $\theta$. That is always true, by my definition. (Of course, these angles are only useful for calculating phase differences when the angles are small.)

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?

As long as you are hung up on thinking that rays are real things, you will miss the point. (It's like asking how many points are on the number line between 1 and 2. For some purposes you can model it as having just a few points, but really it has an infinite number of points.) The purpose of drawing in rays is to paint a physical picture of what's going on--don't get hung up on the exact number of rays to use. Ask yourself: If I used a million times more rays, would I have the same problem? (Realize that a "ray" represents only a tiny portion of light.)

I understand that there really aren't any rays, but what confused me, is that even though there are no rays, you told me to use 7 rays in my model instead of 6, if this is the case, then you can't cancel out all the rays. This makes it impossible for the model to show complete destructive interference, while Hootenanny said that the model assumes complete destructive interference. You also mentioned that there were an infinite number of points, doesn't that mean that the rays are real, just that there's an infinite number of them? Or are the points not even real, just a mental tool?

Right. Finite "rays" that are adjacent in your model cannot have exactly the same phase when they hit the screen. Find one way to add the phase differences that makes sense and stick to it: For the first minima, light from the top and bottom of the slit will exactly cancel each other. Keep thinking about this until it clicks.

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]

Perhaps the attached diagram will help to clear things up. The slit is at the left, and there are a bunch of parallel rays coming out at an angle. Imagine a wave with wavelength $\lambda$ traveling along each ray. We assume that all the waves are "in step" with each other, and we "freeze" their motion at a point in time when they are all at the beginning of their cycle at the slit (phase = 0 degrees). Then, all the points on each of the rays that are a multiple of $\lambda$ away from the slit, along the ray, also have phase = 0. I've marked these as blue dots along the rays.

I've drawn a red line perpendicular to the rays. Along that line, I've labeled the phase of each wave at the point where it crosses the line. I set up the diagram so those phases turn out to be 0, 20, 40, 60, ... 360 degrees. I usually think in terms of sine waves, so the "height" of a wave is zero at phase 0, 180 and 360 degrees, is a (positive) maximum at 90 degrees, and a (negative) minimum at 270 degrees.

Each pair of waves that is 180 degrees different in phase along the red line, will cancel at the screen. The waves that have phase 0 and 180 degrees along the red line will cancel, the waves that have phase 20 and 200 degrees will cancel, etc. [You may want to calculate sin (20 degrees) and sin (200 degrees), for example, to verify this.] You can also fill in the "in-between" waves if you like: the waves that have phase 47 and 227 degrees along the red line will cancel, the waves that have phase 103.1415926 and 283.1415926 degrees will cancel, etc.

You may be wondering, why did I draw the red line at an angle (so that it's perpendicular to the rays) instead of vertically? This has to do with the approximation that we're using here.

If the rays really were exactly parallel, they'd never meet, and therefore wouldn't interfere with each other! The rays must actually converge so that they meet at a single point on the screen. Nevertheless, if that point is very far away, well beyond the right edge of the diagram, the portion of the rays that are visible in the diagram appears to be very nearly parallel. The distance from the red line to the convergence point on the screen is very nearly equal for each ray, so the phase differences at the convergence point are very nearly equal to the phase differences along the red line.

 

[Byrgg]

I don't know why, but your diagram isn't showing up.

 

[jtbell]

I posted it just a little while ago. Attachments have to be approved by one of the moderators. Notice that above the attachment it says "pending approval".

 

[Byrgg]

In that diagram, there's one ray which doesn't cancel. This is why I'm confused, since Hootenanny mentioned earlier that the model assumes complete destructive interference.