Resolution of points by human eye

In summary, the conversation is about a person trying to teach themselves optics from a book but having trouble with the problems and getting incorrect answers. They are specifically working on a problem involving the limit of resolution for distant point sources and are struggling with finding the distance between images on the retina, the radius of the diffraction discs, and the maximum distance at which the headlights can be resolved. Despite trying various methods, they are still unable to arrive at the correct answers and are seeking help.
  • #1
Cruikshank
82
4
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?
 
Physics news on Phys.org
  • #2
Cruikshank said:
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?
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.
 

Related to Resolution of points by human eye

What is the resolution of the human eye?

The resolution of the human eye refers to the ability of the eye to distinguish between two points that are close together. It is measured in pixels per degree (PPD) and varies depending on factors such as age, genetics, and lighting conditions.

How does the resolution of the human eye compare to that of a camera?

The resolution of the human eye is much higher than that of a camera. The average human eye has a PPD of 60, while the highest resolution camera has a PPD of around 120. This is due to the complex structure and function of the eye, which allows for greater detail and clarity.

Can the resolution of the human eye be improved?

No, the resolution of the human eye cannot be improved. It is determined by the physical limitations of the eye, including the number of photoreceptors and their distribution on the retina. However, certain corrective surgeries, such as LASIK, can improve visual acuity and overall visual quality.

Does screen resolution affect the resolution of the human eye?

No, screen resolution does not affect the resolution of the human eye. The resolution of a screen is a measure of how many pixels it can display, while the resolution of the human eye is a measure of its ability to distinguish between two points. However, a higher screen resolution can provide a sharper and more detailed image, which can be perceived as better resolution by the human eye.

How does age affect the resolution of the human eye?

As we age, the resolution of the human eye decreases. This is due to changes in the structure and function of the eye, such as a decrease in the number of photoreceptors and changes in the lens and retina. This can result in a decrease in visual acuity and an increased need for corrective lenses.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
903
Replies
21
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
5K
Back
Top