# Resolution of points by human eye

1. Jul 8, 2012

### Cruikshank

I'm trying to teach myself optics from Frances Sears' book Optics from 1949. I'm attempting every problem, and there are answers to the odd ones in the back. I've gotten a lot of wrong answers and don't know why, had a few I just couldn't even see how to start, and at this point, I'm seeing multiple things to try giving different answers and none of them are right. It's getting very frustrating, all the more so because some of these sound just like the problems I solve for freshmen all the time.

Here's the first problem for Chapter 10, on limit of resolution:

The headlights of a distant automobile may be considered as point sources. The distance between the headlights is 1.5m and the automobile is 6000m away. (a) What is the distance between the centers of the images of the sources on the retina? (b) What is the radius of the central diffraction disk of each image? (c) What is the maximum distance at which the headlights could be resolved? Assume a wavelength of 550nm and a pupillary radius of 1mm.

The book claims the answers are 4.7x10E-6m, 6.3x10E-6m, and 4500m.

My attempts:
If the retina distance is 2cm, then -di/do = m =.02m/6000m = hi/ho = hi/1.5m, so that gives hi = 5x10E-6m. I'm guessing that is wrong because of the different indices of refraction meaning I can't use the thin lens formula?

But if I try n/s + n'/s' = (n'-n)/R, I get gibberish.

For part b, I thought I could just use θ = 1.22λ/D for resolution limit angle, and then the retina distance 2cm, but that gives me 6.71x10E-6m, not 6.3x10E-6m.

In another attempt, I used a text formula sin α' = 0.61λ/n'R, giving 2.51x10E-4 rad, but that gives 5.02x10E-6m.

For part c, the text said the separation of cones was .01mm, so I set that as height of image and solved for do and got 3000m, not 4500m.

I don't understand what I'm doing wrong. Help?

2. Jul 9, 2012

### haruspex

In (a) and (b), looks to me like you're using a slightly too large value for the lens/retina distance. A quick net search shows me 1.7cm. Maybe 2cm is from the eye surface?

In (c), not sure that cones would have much to do with it. Should be rods, no? Judging from the wording of the question, it's not to do with the spacing of the rods. Rather, it's to do with overlap of the diffraction discs and loss of focus due to pupil size.