<p>Here's how I would do it.</p>
<blockquote>
<p>Estimate the angular resolution of the human eye at the centre of the visible wavelength range during daylight.</p>
</blockquote>
<p>you mentioned in your comment</p>
<blockquote>
<p>The first part is easy you just use the angular resolution formula.</p>
</blockquote>
<p>Likely you mean the Rayleigh criteria discussed in Wikipedia's <a href="https://en.wikipedia.org/wiki/Airy_disk" rel="nofollow noreferrer">Airy disk</a>.</p>
<p><span class="math-container">$$\theta = 1.22 \frac{\lambda}{d}$$</span></p>
<p>where <span class="math-container">$d$</span> is the daytime pupil diameter and <span class="math-container">$\lambda$</span> is the</p>
<blockquote>
<p>centre of the visible wavelength</p>
</blockquote>
<p>Make sure to keep your units straight, convert both of them to the same units (nm or um or mm or m) before division.</p>
<p>The question asks about sample density in mm^-2 (&quot;per mm2&quot;) at the back of the eyeball. so draw a triangle from the center of the pupil to the retina, and use the small angle approximation <span class="math-container">$\tan \theta \approx \sin \theta \approx \theta$</span> and multiply your <span class="math-container">$\theta$</span> by the diameter of the eyeball to get the resolution in mm at the retina.</p>
<p>Then google &quot;critical sampling&quot; and find Wikipedia's</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Nyquist_frequency" rel="nofollow noreferrer">Nyquist frequency</a></li>
<li><a href="https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem" rel="nofollow noreferrer">Nyquist–Shannon sampling theorem</a></li>
</ul>
<p>to calculate your sampling frequency (neurons per mm).</p>
<p>Then square that to get neurons per mm^2.</p>

CC BY-SA 4.0
2023-12-31T23:15:23.117Z
2023-12-31T23:15:23.117Z
uhoh
<p>Estimate the angular resolution of the human eye at the centre of the visible
wavelength range during daylight. Using your answer, estimate the surface den-
sity of rods (per mm2) on the retina, assuming that they critically sample the
image.</p>
<p>The first part is easy you just use the angular resolution formula.It's the second part I'm having trouble with namely the critically sampling bit.</p>

CC BY-SA 4.0
2023-12-30T16:18:26.570Z
This sounds like a homework assignment, which we don’t usually answer on AstronomySE.
Pierre Paquette
CC BY-SA 4.0
2023-12-30T17:44:56.390Z
It’s also a really bad homework question, because you have no rods in the very center of your vision, which is the only place you actually have high resolution.
Topcode
CC BY-SA 4.0
2023-12-30T23:19:50.850Z
@PierrePaquette from [Community Policy Repository](https://astronomy.meta.stackexchange.com/q/560/7982) in meta, find [Are homework questions allowed?](https://astronomy.meta.stackexchange.com/a/88/7982) See also [Stop the off-hand "this sounds like a homework question" comments that offer no guidance and raise false flags?](https://astronomy.meta.stackexchange.com/q/762/7982) I think we can offer some help improving the question.
uhoh
CC BY-SA 4.0
2023-12-30T23:23:23.487Z
Hi @Rob, this question, and your other question [Hi,Just wondering if anybody could help me with this exam question on radio telescopes, shown below?](https://astronomy.stackexchange.com/q/55640/7982) show no evidence of effort. The first comment someone should have left is "What did you try?" For this type of "homework-like" question where you state a detailed problem to work and just ask for the answer rather than insight, it's necessary to explain exactly where the problem is - how far you got and where you are getting stuck. Until then, down votes for lack of research will continue.
uhoh
CC BY-SA 4.0
2023-12-31T10:55:07.653Z
I’m voting to close this question because this is rather a biology or physiology question. I fail to see the relation to astronomy. That said, please follow uhoh's advice on how to ask meaningful questions: dive into them yourself, see how far you get, with whatever assumptions you can make and their justifications, share your results and thought process and ask for clarification and guidance for exactly the point you seem to get stuck.
planetmaker
CC BY-SA 4.0
2023-12-31T17:05:06.867Z
It's on an observational astronomy past paper question so yes it is an astronomy question.I did try to do it.Wouldnt be asking on here if I knew how to do it.The first part is easy you just use the angular resolution formula.It's the second part I'm having trouble with namely the critically sampling bit.
Rob1
CC BY-SA 4.0
2023-12-31T23:17:37.903Z
@planetmaker no it's not at all about physiology, it's about the [Nyquist frequency](https://en.wikipedia.org/wiki/Nyquist_frequency) and you can be sure that anybody choosing a CCD pixel size for their optical system starting from the modulation transfer function of their optics will explain what they are doing to their friends in terms of pupils and retinas. **voting to leave open**
uhoh
CC BY-SA 4.0
2024-01-01T23:26:34.770Z
The eye alone was a critical tool for astronomy until about 1600 AD.  Some of us still use it for astronomy, sometimes in combination with a telescope or binoculars.
eshaya
<observational-astronomy><angular-resolution>
Just wondering if anybody knows how to do this question about angular resolution of the human eye?
Rob1
# Just wondering if anybody knows how to do this question about angular resolution of the human eye?

(Asked by: username_0 on 2023-12-30)

Estimate the angular resolution of the human eye at the centre of the visible
wavelength range during daylight. Using your answer, estimate the surface den-
sity of rods (per mm2) on the retina, assuming that they critically sample the
image.

The first part is easy you just use the angular resolution formula.It's the second part I'm having trouble with namely the critically sampling bit.

**Tags:** observational-astronomy, angular-resolution
> **Comments:**
> **username_1**: This sounds like a homework assignment, which we don’t usually answer on AstronomySE.  
> **username_2**: It’s also a really bad homework question, because you have no rods in the very center of your vision, which is the only place you actually have high resolution.  
> **username_3**: @PierrePaquette from [Community Policy Repository](https://astronomy.meta.stackexchange.com/q/560/7982) in meta, find [Are homework questions allowed?](https://astronomy.meta.stackexchange.com/a/88/7982) See also [Stop the off-hand "this sounds like a homework question" comments that offer no guidance and raise false flags?](https://astronomy.meta.stackexchange.com/q/762/7982) I think we can offer some help improving the question.  
> **username_3**: Hi @Rob, this question, and your other question [Hi,Just wondering if anybody could help me with this exam question on radio telescopes, shown below?](https://astronomy.stackexchange.com/q/55640/7982) show no evidence of effort. The first comment someone should have left is "What did you try?" For this type of "homework-like" question where you state a detailed problem to work and just ask for the answer rather than insight, it's necessary to explain exactly where the problem is - how far you got and where you are getting stuck. Until then, down votes for lack of research will continue.  
> **username_4**: I’m voting to close this question because this is rather a biology or physiology question. I fail to see the relation to astronomy. That said, please follow uhoh's advice on how to ask meaningful questions: dive into them yourself, see how far you get, with whatever assumptions you can make and their justifications, share your results and thought process and ask for clarification and guidance for exactly the point you seem to get stuck.  
> **username_0**: It's on an observational astronomy past paper question so yes it is an astronomy question.I did try to do it.Wouldnt be asking on here if I knew how to do it.The first part is easy you just use the angular resolution formula.It's the second part I'm having trouble with namely the critically sampling bit.  
> **username_3**: @planetmaker no it's not at all about physiology, it's about the [Nyquist frequency](https://en.wikipedia.org/wiki/Nyquist_frequency) and you can be sure that anybody choosing a CCD pixel size for their optical system starting from the modulation transfer function of their optics will explain what they are doing to their friends in terms of pupils and retinas. **voting to leave open**  
> **username_5**: The eye alone was a critical tool for astronomy until about 1600 AD.  Some of us still use it for astronomy, sometimes in combination with a telescope or binoculars.  

---

## 1 answers

### Answer by username_3 on 2023-12-31. Score: 1

Here's how I would do it.

> Estimate the angular resolution of the human eye at the centre of the visible wavelength range during daylight.

you mentioned in your comment

> The first part is easy you just use the angular resolution formula.

Likely you mean the Rayleigh criteria discussed in Wikipedia's [Airy disk](https://en.wikipedia.org/wiki/Airy_disk).

$$\theta = 1.22 \frac{\lambda}{d}$$

where $d$ is the daytime pupil diameter and $\lambda$ is the

> centre of the visible wavelength

Make sure to keep your units straight, convert both of them to the same units (nm or um or mm or m) before division.

The question asks about sample density in mm^-2 ("per mm2") at the back of the eyeball. so draw a triangle from the center of the pupil to the retina, and use the small angle approximation $\tan \theta \approx \sin \theta \approx \theta$ and multiply your $\theta$ by the diameter of the eyeball to get the resolution in mm at the retina.

Then google "critical sampling" and find Wikipedia's

* [Nyquist frequency](https://en.wikipedia.org/wiki/Nyquist_frequency)
* [Nyquist–Shannon sampling theorem](https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem)

to calculate your sampling frequency (neurons per mm).

Then square that to get neurons per mm^2.


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