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.

 

[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?