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I was wondering something about diffraction

  1. Jul 13, 2006 #1
    I'm pretty sure the equation for finding the minima in a diffraction pattern is:

    [tex]\sin\theta_m = m\frac{\lambda}{a}[/tex]

    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:

    [tex]\sin\theta_m = (m + 0.5)\frac{\lambda}{a}[/tex]

    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.
     
    Last edited: Jul 13, 2006
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  3. Jul 13, 2006 #2

    Doc Al

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    If you are talking about single slit diffraction (Fraunhofer limit), then the condition for the minima is:
    [tex]\sin\theta_m = m\frac{\lambda}{a}[/tex]
    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?
     
  4. Jul 13, 2006 #3
    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.
     
    Last edited: Jul 13, 2006
  5. Jul 13, 2006 #4

    Doc Al

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    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.
     
  6. Jul 14, 2006 #5
    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?
     
  7. Jul 14, 2006 #6

    Doc Al

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    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 [itex]\lambda/2[/itex] 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: [itex]\theta = 0[/itex].
     
  8. Jul 14, 2006 #7
    Why can't a ray travelling to the center be cancelled 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 [tex]\sin\theta_m = (m + 0.5)\frac{\lambda}{a}[/tex]?

    Hmm, I'm also wondering how the angles [tex]\theta[/tex] and [tex]\theta'[/tex] are approximated to be equal.

    Also, if the condition for a minima is [tex]a sin\theta = m\lambda[/tex], 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 cancelled 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?
     
  9. Jul 14, 2006 #8

    Hootenanny

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    Because the originate from a coherent source.
    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.
    '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 [itex]frac{1}{2}\lambda[/itex] between the two beams at this point. The second order minimum occurs when there is a path difference of [itex]frac{1}{2}\lambda[/itex] between the two beams.

    Further Reading
     
  10. Jul 14, 2006 #9

    Doc Al

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    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 [itex]\theta = 0[/itex] 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.

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

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

    "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.
     
    Last edited: Jul 14, 2006
  11. Jul 15, 2006 #10
    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?
     
  12. Jul 15, 2006 #11

    Doc Al

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

    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.
     
  13. Jul 15, 2006 #12
    I think I read somewhere that each side of the gap produces it's own diffraction pattern, so if this is occuring 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?

    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?

    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 travelling 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?
     
    Last edited: Jul 15, 2006
  14. Jul 16, 2006 #13

    Doc Al

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    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.



    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 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.
     
  15. Jul 16, 2006 #14
    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?

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

    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?
     
    Last edited: Jul 16, 2006
  16. Jul 16, 2006 #15
    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?
     
  17. Jul 16, 2006 #16

    Hootenanny

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    I myself, cannot see how the triangles are similar. If there were similar then [itex]\theta'[/itex] would be exactly equal to [itex]\theta[/itex].
    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.
    Take a look at the diagram shown on this page. 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.
    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.
     
  18. Jul 16, 2006 #17

    Doc Al

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    Right, they are just a tool to picture things.
    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?



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


    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.


    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.
    Yes.
    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.)

    Yes.
     
  19. Jul 16, 2006 #18
    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.

    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?
     
  20. Jul 16, 2006 #19

    Doc Al

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    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.


    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.
     
  21. Jul 16, 2006 #20
    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?

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