Resolution of a circular aperture

[usfz28]

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!

 

[Doc Al]

Raleigh Criterion

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

To find the minimum separation distance that can be resolved, use some trig: \tan(\theta_{min}/2) = (d/2)/L, 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 \theta_{min}/2. (The two right triangles your professor spoke of would be A-M-X and B-M-X.)

 

[usfz28]

Hey thanks for clearing it up! Big help!