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Diffraction at slits

  1. Feb 2, 2014 #1
    1. The problem statement, all variables and given/known data

    A system of 8 slits, each separated from its neighbour by 0.05 mm, is illuminated with light of
    wavelength 576 nm. Using phasor analysis, evaluate at what angle on a distant screen there is the first
    minimum in intensity.

    2. Relevant equations



    3. The attempt at a solution

    Ok firstly, I'm not at all comfortable with phasor diagrams as I don't really understand them so I've tried to go about this a different way. My thinking is that dsin(theta) = (2n-1)/2 x wavelength . I got this by thinking that if the extra distance travelled is a multiple of half a wavelength there will be destructive interference. Now by using n =1 and d= 0.05 I do not get the correct answer for theta(I have taken into account unit conversions), I know this is straightforward.
    Help would be much appreciated!
     
  2. jcsd
  3. Feb 2, 2014 #2

    SammyS

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    Hello Scientist94. Welcome to PF !
    What is the correct answer?

    What is the ratio of the correct answer to your answer?

    What is the ration of the sines of those angles?
     
  4. Feb 2, 2014 #3
    Hi

    The correct answer is 1.44 x 10^-3 radians, which is exactly 4 times the value of my answer, this is the same for the sines. However I do not understand why exactly it is the correct answer. I know the fact that there is 8 slits makes a difference, whereas my method does not include that into the equation.
     
  5. Feb 2, 2014 #4

    SammyS

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    You must have figured the angle required for the path difference for light from say the first slit to be 1/2 wavelength farther than light from the second slit.

    If you compare light from the first & fifth slits, for example, the path difference will be 4 times as much for a given (small) angle. So, to get this difference to be 1/4 as much, the angle will be 1/4 of what you got. Why do you suppose that might be the correct thing to look at ?
     
  6. Feb 2, 2014 #5
    Is it something to do with the intensity of the light? i.e the intensity of the maxima directly opposite the 5th slit would be the highest? I'm afraid I don't understand this topic very well.
     
  7. Feb 2, 2014 #6

    rude man

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    No, intensities from all 8 slits are presumed to be equal. What is going on here is total destructive interference of 8 equally intensive beams.
     
  8. Feb 2, 2014 #7

    SammyS

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    No. That's not it.

    Another way to look at this ...

    You must have found the angle necessary for the light from neighboring slits to destructively interfere, that is the light from slit #1 interferes destructively with the light from slit #2, etc.

    At what angle will the light from slit#1 interfere (destructively) with the light from slit #3? How can you pair other slits to have destructive interference at the same angle? How does this angle compare with you initial angle? ... or comapre with the correct answer?
     
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