# Resolution of the Eye-Light Rays and Diffraction

1. Aug 17, 2014

### alingy1

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?

Last edited: Aug 17, 2014
2. Aug 17, 2014

### 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.

3. Aug 17, 2014

### alingy1

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

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...

4. Aug 18, 2014

### 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.

5. Aug 18, 2014

### alingy1

Ok. I gotta 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?

6. Aug 18, 2014

### 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.

7. Aug 18, 2014

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.

8. Aug 18, 2014

### 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.

9. Aug 19, 2014

### 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 neye) 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 ?)

10. Aug 19, 2014