<blockquote>
<p>Is he using a small angle approximation?</p>
</blockquote>
<p><em>Yes!</em></p>
<p>Consider this triangle:</p>
<p><a href="https://i.sstatic.net/H3aK5paO.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/H3aK5paO.jpg" alt="enter image description here" /></a></p>
<p>The bottom side which I labeled &quot;384403000 m&quot; as the Earth-Moon distance, and &quot;1 m&quot; is the diameter of the aperture. The angle <span class="math-container">$\theta$</span> is the angular separation of two rays, in your words, <em>the divergence of two light rays</em>. By trig function:</p>
<p><span class="math-container">$$\tan\theta = \frac{1}{384403000}.$$</span></p>
<p>Because the fraction is so small, when one expand the <span class="math-container">$\tan \theta$</span> function by Taylor series, all the quantities with high orders will be practically seen as 0, which gives <span class="math-container">$\tan \theta \approx \theta$</span>, in your words, the <em>small angle approximation</em>. So now we reduce <span class="math-container">$\tan  \theta$</span> to only <span class="math-container">$\theta$</span>.</p>
<p>Note <span class="math-container">$\theta$</span> is also extremely small, so we can approximate <span class="math-container">$\theta \approx 0$</span>. <em>Remember the unit of the angle is radian! Not degree!</em></p>
<p>That is the answer of your question, <span class="math-container">$\theta$</span> is just too small for astronomers to consider about it.</p>

CC BY-SA 4.0
2024-06-09T16:58:15.950Z
“Figure not to scale” :-)
jcaron
CC BY-SA 4.0
2024-06-10T03:00:20.743Z
There are some potential exceptions that prove the rule: [Did this satellite streak past the Hubble Space Telescope so close that it was out of focus? If so, how close was it?](https://astronomy.stackexchange.com/q/53064/7982) and [Has Hubble ever focused on something close enough that it had to move away from being focused at infinity?](https://space.stackexchange.com/q/50904/12102) For things closer than say 18,000 km for a 2 meter aperture (and further for larger), the parallel ray approximation gives the wrong answer.
uhoh
CC BY-SA 4.0
2024-06-10T14:13:34.760Z
Don't forget (@andro) that θ has to be in radians for the small angle approximation (something I think is always worth stating explicitly when the context is a self-described beginner question)
Chris H
CC BY-SA 4.0
2024-06-09T08:08:20.893Z
2024-06-10T16:47:41.363Z
2024-06-10T16:47:41.363Z
Polaris5744
<p>On this page:</p>
<p><a href="https://www.handprint.com/ASTRO/ae1.html" rel="nofollow noreferrer">https://www.handprint.com/ASTRO/ae1.html</a></p>
<p>the author gives this calculation to show light rays from very distant objects are essentially parallel:</p>
<blockquote>
<p>In astronomical applications, light sources are so distant that the
concentric wavefronts become a series of equally spaced parallel
planes across the width of any practical telescope aperture. To
illustrate: across the aperture of a 1 meter (39.4&quot;) telescope, light
rays from a single point on the Moon, the closest astronomical object
at 384,403 kilometers, diverge from parallel by no more than
1/384403000 of a radian or 0.0000026 millimeter, which is 0.0047 or
1/200 a wavelength of &quot;green&quot; light. Since the fabrication limits of
the highest quality astronomical optics are around λ/20 wave, or 10
times larger than the wavefront divergence, optical calculations can
assume perfectly flat and parallel wavefronts from a distant light
source.</p>
</blockquote>
<p>I don't follow the steps in this reasoning. Is he using a small angle approximation? I know it's very elementary, but could somebody explain it to me?</p>

<visible-light>
Divergence of light rays - parallel approximation
andro
# Divergence of light rays - parallel approximation

(Asked by: username_0 on 2024-06-09)

On this page:

<https://www.handprint.com/ASTRO/ae1.html>

the author gives this calculation to show light rays from very distant objects are essentially parallel:

> In astronomical applications, light sources are so distant that the
> concentric wavefronts become a series of equally spaced parallel
> planes across the width of any practical telescope aperture. To
> illustrate: across the aperture of a 1 meter (39.4") telescope, light
> rays from a single point on the Moon, the closest astronomical object
> at 384,403 kilometers, diverge from parallel by no more than
> 1/384403000 of a radian or 0.0000026 millimeter, which is 0.0047 or
> 1/200 a wavelength of "green" light. Since the fabrication limits of
> the highest quality astronomical optics are around λ/20 wave, or 10
> times larger than the wavefront divergence, optical calculations can
> assume perfectly flat and parallel wavefronts from a distant light
> source.

I don't follow the steps in this reasoning. Is he using a small angle approximation? I know it's very elementary, but could somebody explain it to me?

**Tags:** visible-light

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## 1 answers

### Answer by username_1 on 2024-06-09. Score: 9

> Is he using a small angle approximation?

*Yes!*

Consider this triangle:

[![enter image description here](https://i.sstatic.net/H3aK5paO.jpg)](https://i.sstatic.net/H3aK5paO.jpg)

The bottom side which I labeled "384403000 m" as the Earth-Moon distance, and "1 m" is the diameter of the aperture. The angle $\theta$ is the angular separation of two rays, in your words, *the divergence of two light rays*. By trig function:

$$\tan\theta = \frac{1}{384403000}.$$

Because the fraction is so small, when one expand the $\tan \theta$ function by Taylor series, all the quantities with high orders will be practically seen as 0, which gives $\tan \theta \approx \theta$, in your words, the *small angle approximation*. So now we reduce $\tan \theta$ to only $\theta$.

Note $\theta$ is also extremely small, so we can approximate $\theta \approx 0$. *Remember the unit of the angle is radian! Not degree!*

That is the answer of your question, $\theta$ is just too small for astronomers to consider about it.

> **Comments:**
> **username_2**: “Figure not to scale” :-)  
> **username_3**: There are some potential exceptions that prove the rule: [Did this satellite streak past the Hubble Space Telescope so close that it was out of focus? If so, how close was it?](https://astronomy.stackexchange.com/q/53064/7982) and [Has Hubble ever focused on something close enough that it had to move away from being focused at infinity?](https://space.stackexchange.com/q/50904/12102) For things closer than say 18,000 km for a 2 meter aperture (and further for larger), the parallel ray approximation gives the wrong answer.  
> **username_4**: Don't forget (@andro) that θ has to be in radians for the small angle approximation (something I think is always worth stating explicitly when the context is a self-described beginner question)  

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