# Determine the distance between barely resolvable objects

1. Nov 3, 2013

### NasuSama

1. The problem statement, all variables and given/known data

The closely packed cones in the fovea of the eye have a diameter of about 2 μm. For the eye to discern two images on the fovea as distinct, assume that the images must be separated by at least one cone that is not excited.

If these images are of two point-like objects at the eye's 25-cm near point, how far apart are these barely resolvable objects? Assume the diameter of the eye (cornea-to-fovea distance) is 2.0 cm.

2. Relevant equations

$\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}$

3. The attempt at a solution

I attempt to determine $d_i$ by setting $f$ to be the distance between the cornea and the tip of fovea (difference between the diameter of fovea and the diameter of eye) and then solving for $d_i$, BUT this doesn't work.

I'm stuck in this problem.

2. Nov 3, 2013

### cepheid

Staff Emeritus
I'm assuming that the "near point" of the eye is the closest distance to which it can focus, and that's where the two sources are located. So, it sounds like do is 25 cm. The image distance, di, is the distance between the lens and the image. So, that certainly sounds like the cornea-to-fovea distance to me, since, if the eye is focused on the objects, then their images should form on the retina. So, it sounds like di is 2.0 cm. That allows you to solve for f. So, the mistake you made was mistaking di for f. Keep in mind that f is not necessarily equal to the physical distance between the lens and the image. That's only true when focused to infinity (parallel rays), in which case 1/do becomes 0, and f = di. More generally, f is just a number that depends on the curvature of the lens. In the case of the eye, the lens curvature (and hence f) can change depending on whether you're trying to focus on nearby or more distant objects.

However, for this problem, I don't think you even need to solve for f. You just need to answer this question: how far apart must the two objects be in order to have the required image spacing on the retina? The problem indirectly tells you what the image spacing on the retina must be, because it tells you that the images must fall on two cones that are separated from each other by at least one unexcited cone in the middle. So that tells you how far apart the images must be in units of cone radii.

To answer this question about the spacing: suppose the two sources have separation y. Rays from each object that pass through the *centre* of the lens are undeviated. So, given the object-to-image distance, how far apart will be the points on the retina where these two rays land?

Last edited: Nov 3, 2013
3. Nov 3, 2013

### NasuSama

What does that suppose to mean? Then, $f$ is a solution for the first paragraph? But I need to determine the amount of spacing of the images, right?

4. Nov 3, 2013

### haruspex

The spacing of the images is determined by the information about the width of a cone and the need for a full cone's separation between stimulated cones. I would assume that each image is to be in the centre of a cone. If so, how far apart are the images?
As cepheid notes, you don't need need to find f. This is not a question about lenses at all. You can treat the lens as a pinhole here. If the object is 25cm from the lens on one side and the image is 2cm from the lens the other side, simple geometry will give you the relationship between object separation and image separation.

5. Nov 3, 2013

### NasuSama

I think I got it; I see the relationship between the distances and the width of the object (and fovea? I think that takes in account of the situation).

I got 25 μm by setting the proportion of two lengths and widths. That is:

$\dfrac{2\,\text{cm}}{25\,\text{cm}} = \dfrac{2\,\text{\mum}}{x}$

where $x$ is what we want to solve. Let me know if I'm on the right track.

6. Nov 3, 2013

### NasuSama

Nevermind. The answer is wrong. It's not 25.

7. Nov 3, 2013

### cepheid

Staff Emeritus
As I said in my previous post, you should pay attention to the hint the problem gave you about the image spacing. The image spacing is NOT 2 microns. There has to be a whole unexcited cone in between the two cones where the images land. So, there is a row of cones: the first cone has image 1, the middle cone is blank, and the third cone has image 2. So, how many cone radii are there between the first cone and the third cone? Draw a picture.