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Young's fringes experiment, interference fringes

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data

    (this is high school level)

    The pic below shows interference fringes observe in Young's fringes experiment:

    attachment.php?attachmentid=199901&d=1361797885.jpg

    Explain why wider slits at the same spacing could produce fringes at the outer edges brighter than some of the fringes nearer the centre.


    Bear in mind there was a question (part a) just before which asked me to explain why fringes at the outer edges are dimmer than fringes nearer the centre. I answered by saying that because these interference fringes have to be contained within the 'envelope' intensity shape of the central diffraction fringe, it therefore follows that intensity decreases away from the centre.


    2. Relevant equations

    fringe spacing of interference fringes = (lamda * D) / slit separation
    width of central diffraction fringe = (2 * lamda * D) / aperture width

    D = slit-screen distance


    3. The attempt at a solution

    I know from the second formula above that wider slits will mean that the central diffraction fringe will be smaller in width. But thats where I'm stuck. I don't understand how its possible that fringes (just to make sure they mean interference fringes, right?) at the outer edges could be brighter than some of the fringes nearer to the center just because the slit is made wider. Surely these interference fringes would still follow the 'envelope' shape in intensity variation set out by the diffraction fringe in which they are contained within.

    and just to confirm something basic, is the picture I've attached above showing only the central diffraction fringe for the red and blue light?
     
  2. jcsd
  3. Feb 26, 2013 #2

    rude man

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    With slit width b < wavelength λ you get the familiar I ~ cos2(πdsinθ/λ) corresponding to two-slit diffraction, but for wider slits the diffraction pattern of a single slit becomes visible also so then you also get the effects of single wide-slit diffraction.

    So you need to combine the effects of both kinds of diffraction to get the two wide-slit picture. Hint: include Fraunhofer diffraction effects for the case of two slits of finite width, which results in a sinc2 term along with the aforementioned cos2 one for the total intensity for each fringe.

    I suggest you pick suitable values of d, b and λ, run the Fraunhofer equation on e.g. Excel, and show that this can result in deeper fringes having more intensity than close-in ones. I have never done this so 'bonne chance'!

    Your images show a central fringe in both images. The red has 3 fringes on either side for a total of 7 fringes. The blue has a total of 9 fringes.
     
  4. Feb 26, 2013 #3
    I haven't come across the formula 'I ~ cos2(πdsinθ/λ' nor 'Fraunhofer diffraction', I'm still in high school and this was a question from my textbook.

    I'm using only high school level knowledge here so I may have completely misunderstood you, can you tell me if I've got the right idea from your post:

    so basically if you increase the width of the slits whilst maintaining their separation, it almost becomes a single wide slit such that a single slit diffraction pattern emerges and it is this which causes outer fringes to be brighter than some fringes closer to the centre. My image at the top shows a central fringe for red and blue light, so there would be a single slit pattern within this central fringe, right?
     
  5. Feb 26, 2013 #4

    rude man

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    What you get is the superposition of two wide-slit diffraction patterns plus the interference pattern characteristic of a narrow-slit setup. The result is actually quite a complex pattern of "fine" and "coarse" fringes, with the fine representing narrow-slit interference and the coarse due to wide-slit diffraction by each slit separately.

    This subject is very difficult to discuss without the math. I would just answer that as the slit width exceeds the wavelength the fringe pattern becomes more complex and intensity variations such as you describe are made possible.

    It's also possible there is another effect present that I don't recollect or even knew! I could only think of wide-slit Fraunhofer diffraction.
     
  6. Feb 26, 2013 #5
    so what I said about the single slit pattern appearing within the central fringe is correct?

    isn't this like single slit pattern within a single slit pattern? because the central fringe comes from a single slit pattern

    and I didn't realise this before you mentioned it, so usually the slit width is less than the wavelength?
     
  7. Feb 26, 2013 #6

    rude man

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    You get a central maximum for both Young's interference pattern and for a single-slit diffraction pattern.

    If d = distance between slits and θ = angle a fringe makes with the central fringe middle,
    for Young interference, maxima (bright) fringes are given by dsin(θm) = mλ so there is a central bright fringe at m = 0 and another one for m = 1, m = 2, etc. all the way to θ = 90 deg.

    For a single-slit aperture, with w = slit width, the diffraction pattern is similar except that the position of dark fringes is given by
    w sin(θm) = mλ. From this you can see that you get lots of fringes if the slit width >> 1λ but if w = 1λ the entire screen is filled with the central maximum since then θ1 = 90 deg already.

    For dual-slit (Young's) experiments, the slits are usually << 1 wavelength to avoid the need to consider diffraction effects. Point sources (radiators) are assumed, which means very narrow slits.

    To see wide-slit diffraction effects the opposite is the case, as I have described above.

    What makes life complicated is if you have two wide slits so that diffraction effects come into play on top of the interference pattern ascribable to a narrow-slit setup. I find it difficult to describe the total pattern without the math.
     
  8. Feb 26, 2013 #7
    You said that if w = 1λ then the entire screen is filled with the central bright fringe, which I understand because from your formula, this means that the first order dark fringes would be at 90 degrees as you've pointed out. But if w = less than one wavelength, how would this look? is just like when w = 1λ, whereby the entire screen is filled with the central bright fringe?

    I tried finding the angle of the first order dark fringe by using your formula: w*sin(θm) = mλ, and then subbing w = (a figure less than 1)*λ , and m = 1, into it, the calculator came up with an error :confused:

    Let me check if this is right, below is a diagram for the intensity of fringes in a single and double slit diffraction pattern, I've circled in blue the central diffraction fringe of both:

    attachment.php?attachmentid=200224.jpg

    So is it true, that if the slit(s) width is equal to the wavelength, then all we would see in the diagram for both single and double slit is the bit I've circled?

    and if the slit width is greater than 1 wavelength, then we would see the other smaller diffraction fringes on either side on the central diffraction fringe?
     
    Last edited: Feb 26, 2013
  9. Feb 27, 2013 #8

    rude man

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    Yes.

    You just finished observing that for w < 1λ there are no dark fringes. So why did you try to find one anyway? It's pretty hard for your calculator to solve sin(x) for x when sin(x) > 1!
    Yes. It's the part of both images covering from - λ/D to +λ/D. BTW those are great images and should answer all your questions by themselves.

    Exactly as your images show. If the slit width is greater than 1 lambda you get diffraction minima (one or more, depending on how wide the slits are).

    I can make no better recommendation than that you study both your images at great length because they answer all your questions. In fact, I'm keeping them for myself since I've never seen the combined waveform pictured (the bottom one).
     
  10. Feb 27, 2013 #9
    Oh, because I wasn't totally sure about what I observed was true. Glad to know that though, it makes sense now.

    Really? I have another similar one in my textbook as well, I thought it was pretty common. Surprised you haven't come across a diagram like that, since you've probably been around physics a lot longer than I have.

    You see the thing is I got confused about when I saw that diagram in my textbook, is that it never mentioned the condition for it to exist (w > wavelength). So I made all sorts of false assumptions, and ended up mistaking that there would always be other diffraction fringes on the screen, rather than just one central diffraction fringe. It also took me a while to realise that the double slit diffraction fringes of the blue and red light shown in the image, was only showing the central diffraction fringe (so only the bit I circled in both of my diagrams in post #7)

    I'm glad the pictures are useful to you as well, but real credit goes to University of Colorado at Boulder, I found their diagram off google image which I then found on their website here: http://www.colorado.edu/physics/phys2020/phys2020_sum98/lab_manual/Lab5/lab5.html [Broken]

    Seems like there are other useful info as well on there. Anyway thanks a lot for the help you've provided.
     
    Last edited by a moderator: May 6, 2017
  11. Feb 27, 2013 #10

    rude man

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    But the x axis clearly shows the ratio of wavelength to slit width so I don't understand your confusion. It doesn't have to mention w > lambda, it's clearly shown on the x axis what the difraction pattern is as a function of the ratio of w to lambda.

    If w < 1 lambda you clearly get only that part of the graph for which w < 1 lambda.
    OK. I'm a (retired) EE so I haven't been around that kind of physics for a while ... had to double-check a thing or two myself!

    Good luck to you in pursuing your studies. You have one of the prime prerequisites - lots of curiosity and want to really understand. Most OP's on this board could use a little more of that.
     
    Last edited by a moderator: May 6, 2017
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