Resolution of the Eye-Light Rays and Diffraction

[alingy1]

Please look at the problem.
The problem is solved for me. However, I don't understand why the angle alpha is considered to stay constant when the rays go through the lens. The ray along the distance s DOES NOT get refracted, which I agree. But the ray along the top point of the circle DOES get refracted once it gets to the liquid inside the eye. Can someone clarify why the textbook does not take that into account?
http://imgur.com/0USADP7

 

[BvU]

I have the impression the circle center is on the optical axis in the drawing; it looks as if you interpret the bottom ray is ?

For any (ideal...) lens, rays passing through the center go straight through.

The picture is a little deceiving: the angle of incidence is closer to perpendicular than you seem to think (the curvature of the cornea is closer to that of the lens than to the curvature of the eyeball as a whole).

For the angular resolution determination, I should think any refraction doesn't really matter anyway: the entire diffraction pattern is influenced proportionally, so resolution isn't affected.

 

[alingy1]

I have the impression the circle center is on the optical axis in the drawing; it looks as if you interpret the bottom ray is ?

What? I don't understand what you are saying and why that is relevant.

For the angular resolution determination, I should think any refraction doesn't really matter anyway: the entire diffraction pattern is influenced proportionally, so resolution isn't affected.

The question states that the angles alpha is equal on both sides. That is the whole equality on which this problems lays.

What I am asking is: for a ray that just got out of the lens and that is going through the liquid of the eye, will there be refraction there? In case of refraction: then alpha will not be equal on both sides of the lens...

I really appreciate your help and your involvement for a troubled student like me (:P) but please give a bit more information because I got lost on your post.

 

[BvU]

"What? I don't understand what you are saying and why that is relevant" My comment was was triggered by your "The ray along the distance s DOES NOT get refracted, which I agree". I thought you were referring to a line in the drawing that isn't drawn; but perhaps you were not.

And yes, $\alpha$ is equal on both sides. If alone from symmetry. A ray passing through the center of a thin lens continues straight on.

Approaching from the other end: suppose there is non-zero refraction, would that change the diffraction pattern ? If so, would it change the resolution ? I am inclined to say no, as I tried to state in the post.

 

[alingy1]

Ok. I got to get something clear because any other discussion would be futile.

When a lens is located at the barrier between two liquids of different refractive indexes, (in this case air and the liquid of the eye), and a ray passes right through the center of the lens, will the ray continue a straight trajectory or not. If yes, why?

 

[alingy1]

Then, according to the formula, theta=1.22(lambda)/(diameter of the lens), the refraction pattern would not change. But why is the question? I think, as you said, every ray is influenced accordingly, so the resolution criteria still remains the same.

 

[Dadface]

Please look at the problem.
The problem is solved for me. However, I don't understand why the angle alpha is considered to stay constant when the rays go through the lens. The ray along the distance s DOES NOT get refracted, which I agree. But the ray along the top point of the circle DOES get refracted once it gets to the liquid inside the eye. Can someone clarify why the textbook does not take that into account?
http://imgur.com/0USADP7

The book has simplified the ray diagram and ignored refractions at other media boundaries. Since the ray in question makes a small angle to the axis, the ray path shown would be very close to the real path followed by the ray.

 

[alingy1]

Dadface, how can you prove me that the refraction would be negligible? There is still a pretty big difference between n=1 and n=1.33, even if the ray still makes a small angle.
1*sin(theta 1)=1.33sin(theta 2)
sin approximation for small angles:
theta 1= 1.33 theta 2 Seems pretty non negligible to me.

 

[BvU]

My compliments for your tenacity. Not taking anything that can't be explained properly for granted is a good quality as far as I'm concerned.
I'm sorry for trying to defend something quite established without actually being able to completely understand it myself, let alone being able to explain it exlpicitly and clearly. I'm not in the business, just a curious physicist.

So: as far as I can make out, refraction is "responsible" for the image formation on the retina. The refractive index of the eye liquid has an influence, e.g. on the magnification.

However, the subject under study is diffraction. Having to do with wavelength (hence the n_eye) and aperture shape and size. A single point is distorted to an Airy disc and that's where the acuity formula pops up.

Some math is needed t answer your yes/no question in post #5. Working it out probably requires some assumptions ("thin lens, lens maker's formula"etc.) that are not very realistic for an eye. I don't blame the writers of the book for avoiding to dig so deep.

My money is on yes, but not well-founded enough to bet the farm. Anyone else listening in ? (Hihi or should I submit a new post ?)

 

[Dadface]

Dadface, how can you prove me that the refraction would be negligible? There is still a pretty big difference between n=1 and n=1.33, even if the ray still makes a small angle.
1*sin(theta 1)=1.33sin(theta 2)
sin approximation for small angles:
theta 1= 1.33 theta 2 Seems pretty non negligible to me.

Consider the ray in question. As you suggested it will not follow a straight line path. It will deviate from linearity at two places at least, on entering the lens and on emerging from the lens. But since the ray makes only a small angle to the principle axis the deviation wil be small also. A simple way to look at this is to look at your ray diagrams which show that as a ray gets closer to the principle axis the angle of deviation gets smaller. You should also look up "lateral displacement" and you will find that for a parallel block the diplacement decreases from a maximum equal to the block thickness for grazing incidence and approaches zero as the incident angle approaches zero. The situation will be slightly different for the non parallel faces of the lens.

Possibly the best way for you to understand this is to draw some detailed ray diagrams of rays passing through the central portion of a lens at different angles. You should show refractions at both surfaces for surrounding media of different refractive indices.