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Fresnel zone plate diffraction

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data

    A zone plate is to be set up on an optical bench, where it will be used as an enlarging lens. Its innermost zone is to be 0.2250 mm in diameter and monochromatic blue-green light of wavelength 480 nm from a cadmium arc is to be used. If the magnitude of the enlargements is to be eightfold in diameter, find:

    (a) the focal length of the zone plate and
    (b) the object distance and
    (c) the image distance.

    2. Relevant equations

    qIKvT.png

    where n is an integer, r is the radius, and r_N and delta r_N are the outermost zone radius and its width.

    3. The attempt at a solution

    LPzkt.png

    this is my diagram which shows my current understanding of whats going on. i dont think it was taught very well in lectures and either im not searching for the right thing on the net, or theres not information on this problem, or its so basic no one has bothered to put up any information on it. at the moment i dont really know where to go from here. we're given r_1 or r_0 (not sure if its 1 or 0), the wavelength and the enlargement magnitude. it looks like im gonna have to use some tricky geometry or something but im honestly not sure what to do.
     
    Last edited: Oct 30, 2011
  2. jcsd
  3. Oct 30, 2011 #2

    Ibix

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    You should be able to tell from the first equation whether the quoted radius is [itex]r_0[/itex] or [itex]r_1[/itex].

    Given that, it looks like you've got (a) from the first equation.

    For monochromatic light you can basically treat a Fresnel lens as a "regular" thin lens (they have brutal chromatic aberration, though, and their off-axis performance isn't great, if memory serves). So parts (b) and (c) come from the usual lens formulae relating focal length and magnification to object and image distances.
     
  4. Oct 30, 2011 #3
    ahh i see, first equation gives 10.5cm for f. that was simple, why didnt i see that before?? now i was thinking we just use the thin lens equation in the first place since all it was doing is the same thing a thin lens would do, but i wasnt sure because i dont fully understand whats going on in the zone plate. alright so thats answered everything basically, thanks! one last thing, is my diagram correct?
     
  5. Oct 31, 2011 #4

    Ibix

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    No. The recipe for geometric optics is to draw a point off-axis at the object distance. Draw a line from this point through the center of the lens. Draw another line from the point parallel to the axis until it strikes the lens, then draw a line from this intersection through the focal point on the other side of the lens. Where this line crosses the first line is where the image will be formed. You can then add arbitrary rays from the point to the lens and from the lens to the image of the point.

    A zone plate is basically a diffraction grating designed to give a diffraction maximum at a distance f from the plate. The Wikipedia article isn't bad.
     
  6. Oct 31, 2011 #5
    cheers
     
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