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Homework Help: Resolution of a circular aperture

  1. Apr 12, 2005 #1
    Just want to say hi. Iam new to the forum but here is my question. Today we were talking about resolution of a circular aperture and the instructor told us to make sure we know this he gave us some equations, but i see no examples of it in the book. could someone help me out? The equatuons he gave me were. theta(min)in rad.=1.22(lamda)/D and tan (theta(min)/2)=(d/2)/L
    and theta(min)=2inverse(tan) (d/2)/(L). I didnt really understand what he was saying but he was talking about breaking it up into two right triangles. Thanks for any help you can give me or any examples!!
  2. jcsd
  3. Apr 12, 2005 #2

    Doc Al

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    Staff: Mentor

    Raleigh Criterion

    These equations represent the "Raleigh Criterion" for the minimum angular resolution possible due to diffraction from a circular aperture:
    [itex]\sin\theta_{min} = 1.22 \lambda/D[/itex] where D is the aperture diameter. (When the angle--in radians--is small, [itex]\sin\theta_{min} = \theta_{min}[/itex].)

    To find the minimum separation distance that can be resolved, use some trig: [itex]\tan(\theta_{min}/2) = (d/2)/L[/itex], where d is the separation distance and L is the distance to the aperture. To understand the geometry involved, imagine this. Call the two points you are trying to resolve A and B. (The distance between them is d.) Now draw a line from the aperture (at X) to the midpoint (at M) between A and B. The right triangle A-M-X is what we are talking about. A-M = d/2; M-X = L; the angle at corner X is [itex]\theta_{min}/2[/itex]. (The two right triangles your professor spoke of would be A-M-X and B-M-X.)
    Last edited: Apr 12, 2005
  4. Apr 12, 2005 #3
    Hey thanks for clearing it up! Big help!!
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