<blockquote>
<p>For a project, I'm trying to calculate the photon flux from the zenith galactic background that a sensor would see. From <a href="https://www.sciencedirect.com/science/article/abs/pii/S0168900201012645" rel="nofollow noreferrer">ScienceDirect</a> I get a value of <span class="math-container">$2.6 \times 10^{12}$</span> photons sr<span class="math-container">$^{−1}$</span> s <span class="math-container">$^{−1}$</span> m <span class="math-container">$^{−2}$</span> for 300nm &lt; λ &lt; 650nm. I want to make sure that I'm interpreting this properly: to get photons/second in the cited band, I multiply this by the FOV of the telescope in steradians and the telescope aperture in square meters (e.g. .785 m<span class="math-container">$^2$</span> for a 1 meter diameter telescope)?</p>
</blockquote>
<p>I'll venture a &quot;yes&quot; answer.</p>
<p>So for example if you have a 0.785 m<sup>2</sup> aperture and a 1° diameter FOV (π/4 square degrees or (π/180)<sup>2</sup> × π/4 sr, then you have a flux of <span class="math-container">$4.88 \times 10^{8}$</span> photons per second.</p>
<p>You may also want to do it per pixel.</p>
<p>In that case the solid angle is the pixel area divided by the focal length squared. For a 30 um pixel and a 10 meter focal length, you'd get <span class="math-container">$2.6 \times 10^{12} \times (3 \times 10^{-5} / 10)^2$</span> or about 26 photons per second.</p>
<p>The corresponding photoelectron production rate would be somewhat lower and the corresponding charge units collected and sent to the amplifier would be somewhat lower than that.</p>

CC BY-SA 4.0
2022-11-21T08:46:16.323Z
2022-11-21T08:46:16.323Z
uhoh
<p>For a project, I'm trying to calculate the photon flux from the zenith galactic background that a sensor would see. From <a href="https://www.sciencedirect.com/science/article/abs/pii/S0168900201012645" rel="noreferrer">ScienceDirect</a> I get a value of <span class="math-container">$2.6 \times 10^{12}$</span> photons sr<span class="math-container">$^{−1}$</span> s <span class="math-container">$^{−1}$</span> m <span class="math-container">$^{−2}$</span> for 300nm &lt; λ &lt; 650nm. I want to make sure that I'm interpreting this properly: to get photons/second in the cited band, I multiply this by the FOV of the telescope in steradians and the telescope aperture in square meters (e.g. .785 m<span class="math-container">$^2$</span> for a 1 meter diameter telescope)?</p>

CC BY-SA 4.0
2023-07-09T00:01:48.277Z
was my answer helpful? Is there more that you need? If you've made further progress and have a more complete or specific answer please feel free to add an answer of your own!
uhoh
CC BY-SA 4.0
2022-11-14T08:14:53.653Z
Could close voters provide a comment about what they think is unclear or could be improved? I don't get it...
Jean-Marie Prival
CC BY-SA 4.0
2022-11-21T08:32:45.270Z
@Jean-MariePrival unfortunately we have some random close-and-run voters who don't think twice nor leave any helpful comments.
uhoh
<observational-astronomy><photometry>
Calculating background photons from flux levels
Paul Blase
# Calculating background photons from flux levels

(Asked by: username_0 on 2022-11-13)

For a project, I'm trying to calculate the photon flux from the zenith galactic background that a sensor would see. From [ScienceDirect](https://www.sciencedirect.com/science/article/abs/pii/S0168900201012645) I get a value of $2.6 \times 10^{12}$ photons sr$^{−1}$ s $^{−1}$ m $^{−2}$ for 300nm < λ < 650nm. I want to make sure that I'm interpreting this properly: to get photons/second in the cited band, I multiply this by the FOV of the telescope in steradians and the telescope aperture in square meters (e.g. .785 m$^2$ for a 1 meter diameter telescope)?

**Tags:** observational-astronomy, photometry
> **Comments:**
> **username_2**: Could close voters provide a comment about what they think is unclear or could be improved? I don't get it...  
> **username_1**: @Jean-MariePrival unfortunately we have some random close-and-run voters who don't think twice nor leave any helpful comments.  
> **username_1**: was my answer helpful? Is there more that you need? If you've made further progress and have a more complete or specific answer please feel free to add an answer of your own!  

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## 1 answers

### Answer by username_1 on 2022-11-21. Score: 1

> For a project, I'm trying to calculate the photon flux from the zenith galactic background that a sensor would see. From [ScienceDirect](https://www.sciencedirect.com/science/article/abs/pii/S0168900201012645) I get a value of $2.6 \times 10^{12}$ photons sr$^{−1}$ s $^{−1}$ m $^{−2}$ for 300nm < λ < 650nm. I want to make sure that I'm interpreting this properly: to get photons/second in the cited band, I multiply this by the FOV of the telescope in steradians and the telescope aperture in square meters (e.g. .785 m$^2$ for a 1 meter diameter telescope)?

I'll venture a "yes" answer.

So for example if you have a 0.785 m2 aperture and a 1° diameter FOV (π/4 square degrees or (π/180)2 × π/4 sr, then you have a flux of $4.88 \times 10^{8}$ photons per second.

You may also want to do it per pixel.

In that case the solid angle is the pixel area divided by the focal length squared. For a 30 um pixel and a 10 meter focal length, you'd get $2.6 \times 10^{12} \times (3 \times 10^{-5} / 10)^2$ or about 26 photons per second.

The corresponding photoelectron production rate would be somewhat lower and the corresponding charge units collected and sent to the amplifier would be somewhat lower than that.


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