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Single slit diffraction interference

  1. Nov 7, 2009 #1
    erm anyone knows why for destructive interference to occur for a fraunhofer's diffraction pattern, the equation is a/2sin(theta) = m(lamda)/2 , ==> sin(theta) = m(lamda)/a ?

    where a = slit width
    lamda = wavelength
    theta = angle the ray makes with the central axis

    the problem i have here is with the value "m" takes. the textbook i am using says that "m" ranges from 1,2,3,4, etc but no 0 . but if m takes on the value 2,4,6 etc, wouldn't the path difference of a/2sin(theta) be equals to 1,2,3 etc lamda ? which is a complete wavelength, hence the light would meet in phase and be constructive?

    so why is "m" be able to take on the values 1,2,3,4,5 etc...

    i know it doesn't take on 0 because the central fringe is always bright.
     
  2. jcsd
  3. Nov 7, 2009 #2

    Doc Al

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    You're trying to find the angles for which the light from the slit "cancels out" and thus produces a dark fringe. For m = 2, 4, 6 etc, it's certainly true that the light from the middle of the slit will be in phase with the light from the edge of the slit. But that's not enough to draw any conclusions from. You have to look at all the light.

    Look at it this way. For any integer value of m, imagine the light at the slit divided into m sections. For each of those sections, you can show that each point of light from the first half is exactly canceled by a corresponding point of light from the second half. Thus each section contributes zero light--and as long as you have an integral number of sections you'll have a dark spot. If m is not an integer, you'll have some light left over so it wouldn't be a dark spot.

    Make sense?
     
  4. Nov 7, 2009 #3
    okay.... it makes sense...
    but i am a bit confuse also...
    so what exactly is "m"?

    from what i understand from your message,
    it would mean that when we take a/2sin(theta) for the path difference, we are actually taking "m" as 2? so if we were to take m as 1, we would be considering the entire light source? top edge to bottom edge gives "a/1"

    so now if we were to consider the slit to be made up of many many "m" , so we would have all the rays from the top and bottom of each "m" canceling each other out at a point P?

    so the formula is now (a/m) sin(theta) = (lamda)/2 for destructive fringes?

    so the "m" here is not the same as the double slit experiment where dsin(theta) = m(lamda) ? or are they the same?

    so what exactly is m? what does it mean when we say m =5? does it mean like the double slit, 5 means the fifth maxima because the fifth maxima occurs at the angle where the path difference is 5(lamda)? or...something else? thanks!
     
    Last edited: Nov 7, 2009
  5. Nov 7, 2009 #4

    Doc Al

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    For single slit diffraction, m is the order of the minima--the dark fringes. m = 1 is the first minimum, etc.

    Not exactly. When you take (a/2)sinθ as the path difference, you are comparing light from the center of the slit to light from the edge of the slit. To find a minimum, you want that path difference to equal λ/2, so m = 1.

    When you divide the slit into "m" regions, you want the light from each region to cancel itself out. Thus you are comparing the path difference of light from the top and center of each region (not top and bottom). So you want (a/2m)sinθ = λ/2.
     
  6. Nov 7, 2009 #5
    oh i see...

    so now if i were to refer to the example where (a)sin(theta) = 2(lamda) where m =2 in this case,

    the center a/2 would have a path difference from the top by lamda.

    so similarly, every point just below the top ray and every point just below the center would arrive at point P with a path difference of also lamda. so wouldn't all the rays be constructive at m =2,4,6 even when we consider the whole light ray of (a)?
     
  7. Nov 7, 2009 #6

    Doc Al

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    True, but not useful.

    Nope. All you've demonstrated is that for each point in the top half of the slit, there will be a point in the bottom half with the same phase. That's very different from saying that all the light will be in phase and thus constructively interfere. (An analogy: Imagine a roomful of people. Just because for every person on one side of the room there's a person on the other side wearing the same color shirt, doesn't mean that everyone in the room is wearing the same color shirt.)

    Since m = 2, look at the slit as having two independent regions. In each each region, the top half exactly cancels the bottom half. That's a much stronger--and more useful--statement than the one you made above.
     
  8. Nov 7, 2009 #7
    oh isee...

    so it means that the top ray and the center will arrive in phase with each other for m=2 since path difference is lamda, but not necessarily in phase with the corresponding adjacent points just below them?

    but that would mean that all these individual pairs of points will produce constructive interference on point P and there would be a certain amount of bright fringe form?

    or is it so happen that all these points cancel each other off exactly?

    so it is the top and center points PAIR that cancel themselves off, and NOT the pair cancelling another pair just below the first pair that causes destructive interference at point P?

    oh 1 more thing , why is the approximation to bright fringes given by asin(theta) = (m+1/2)(lamda) now? why do we add a 1/2(lamda) for bright fringes?

    thanks a lot!
     
  9. Nov 8, 2009 #8

    Doc Al

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    Those rays will be in phase with each other, but out of phase with the rays from the 1/4 and 3/4 positions along the slit. (I don't know what you mean by "the corresponding adjacent points just below them".)

    No.

    Right.

    :confused: For m = 2, the light from the top 1/4 of the slit is exactly out of phase with--and thus cancels--the light from the second 1/4. The same thing is true for the bottom half of the slit.

    That just says that the bright fringes are (approximately) in the middle between the dark fringes.
     
  10. Nov 8, 2009 #9
    erm ok,

    |
    |
    |
    |

    the 4 lines represent the 4 sections that we divide the slit into, for m=2.
    where the middle is the center of the slit.

    so the top half (comprising 2 sections ) cancel each other out(1st/4 - 2nd/4) due to the light being exactly out of phase with each other.

    so within the section itself,

    | (1st/4),

    the light Point from the top of this section and the top point of the 2nd section,

    | (2nd/4),

    exactly cancels out each other due to their pathdifference being (lamda)/2. so they arrive at a point P with a (PI) phase difference.

    so am i right to say that it is this PAIR-CANCELING that causes dark fringe at P, so every point below the 1st/4 section will cancel out every other point below the 2nd/4 section.

    and NOT because the resultant phase of light that the top pair of the 1st/4 and 2nd/4 section ,produced at Point P, cancels out the resultant phase of light that the 2nd points(just below the top) of the 1st/4 and 2nd/4 section produces at P.

    so this would explain why even if the top point and the center of the m=2 have a lamda path difference, the light that this 2 points may be in phase with respect to each other at P, but not necessarily in phase with the subsequent points of light just below them respectively.

    is this correct?
     
  11. Nov 8, 2009 #10

    Doc Al

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    Sounds good to me!
     
  12. Nov 8, 2009 #11
    hmm thanks a lot!
     
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