Diffraction using a circular aperture

[Bolter]

Homework Statement

See image below

Relevant Equations

dsin(theta) = 1.22 lambda

Here is the following question I have been trying to answer

I have drawn a quick simple sketch of what I believe is happening in the set up.
Also because the angle that I am dealing with is very small, I made the assumption that sin(theta) = 12.9mm/X (where X in this case is the max distance I need from the observer)

Any help would be much appreciated! Thanks

 

[BvU]

Any help would be much appreciated!

Sure. What kind of help, except that we do your exercise ?

Do you think the answer is realistic ?

 

[Cutter Ketch]

This is pretty much right except that diagram offends. Labeling the Airy pattern with the object distances looks terrible, and you don’t show x. Why not draw a diagram with two objects a distance 12.9mm apart a distance x from your pupil with a diameter d = 4.0 mm and continue through the pupil to a screen on the other side of the pupil on which you draw your Airy patterns and label your angle. It will be clear from the construction that the angle to the Airy minimum is the same as the angle on the other side.

One other quibble. It doesn’t matter for such small angles, but technically the object side should be tangent. Very technically the half angle should be the tangent with the half separation. On the other hand, with such a small angle thinking in arc and calling it sine is fairly defensible, and it won’t change the answer. Nevertheless, given the construction, tangent is more correct.

 

[BvU]

Agree with Cutter. Unfortunately, nowadays a quick google permits a superficial look at the situation and devaluates the essence of a relevant sketch.
@Bolter: you did just fine, but do take cutter's advice !

Furthermore I am of the opinion that the Rayleigh criterion is to optimistic

 

[Bolter]

This is pretty much right except that diagram offends. Labeling the Airy pattern with the object distances looks terrible, and you don’t show x. Why not draw a diagram with two objects a distance 12.9mm apart a distance x from your pupil with a diameter d = 4.0 mm and continue through the pupil to a screen on the other side of the pupil on which you draw your Airy patterns and label your angle. It will be clear from the construction that the angle to the Airy minimum is the same as the angle on the other side.

One other quibble. It doesn’t matter for such small angles, but technically the object side should be tangent. Very technically the half angle should be the tangent with the half separation. On the other hand, with such a small angle thinking in arc and calling it sine is fairly defensible, and it won’t change the answer. Nevertheless, given the construction, tangent is more correct.

Thank you I can now see from what you have mentioned how flawed my sketch appears to be but it makes sense now :)