<h3>Background</h3>
<p>Imagine you are a pixel sitting at the primary focus, looking up at the incoming light. What do you see? If the image of a star or something bright is falling on you, you'll see a big bright circle (or jagged hexagonalized circle-approximant for JWST) with a dark area in the center where the secondary, and hole in the mirror are.</p>
<p>Yes, you are looking towards the secondary, but it's a mirror so you're really looking at the primary via a reflection.</p>
<p>It doesn't matter how many mirrors or other optical elements are in the path, for the purposes of NA what matters is what the pixel at the focal plane sees.</p>
<p>Just fyi multi-instrument space telescopes like JWST and Hubble sometimes have multiple NAs (multiple f/numbers) to match different instruments.</p>
<h3>JWST's NA</h3>
<p><a href="https://en.wikipedia.org/wiki/Numerical_aperture#General_optics" rel="noreferrer">By definition</a> NA is the sine of the half-angle of the aperture as seen from the focal point.</p>
<p>Since this generally assumes a circular aperture we'll have to choose a radius for the average aperture. One easy way to do that is to find the aperture area <span class="math-container">$A$</span> and calculate the radius of a circle with the same area, but since so much of the JWST's aperture (and a lot of telescopes) is blocked, this won't be correct.</p>
<p>Luckily Wikipedia's <a href="https://en.wikipedia.org/wiki/James_Webb_Space_Telescope" rel="noreferrer">JWST</a> gives the following:</p>
<ul>
<li>Type:  Korsch telescope</li>
<li>Diameter: 6.5 m (21 ft)</li>
<li>Focal length:  131.4 m (431 ft)</li>
<li>Focal ratio:  f/20.2</li>
<li>Collecting area:  25.4 m<sup>2</sup> (273 sq ft)</li>
</ul>
<p>So the effective diameter <span class="math-container">$D$</span> is 6.5 meters, the effective radius <span class="math-container">$r$</span> is half that or 3.25 meters, and the primary focal length <span class="math-container">$f$</span> is 131.4 meters.</p>
<p>The f/no. or focal ratio which is the diameter divided by the focal length or <span class="math-container">$f/D$</span> is 20.2 (unitless).</p>
<p>The NA is really just another way to express the same ratio, but using trigonometry. It's the sine of the half-angle <span class="math-container">$HA$</span></p>
<p><span class="math-container">$$HA = \arctan\left( \frac{r}{f} \right)$$</span></p>
<p><span class="math-container">$$NA = \sin(HA) = \sin\left( \arctan\left( \frac{r}{f} \right)\right)$$</span></p>
<p>and in this case it will be 0.024726 and for small NA we can use the <a href="https://en.wikipedia.org/wiki/Small-angle_approximation" rel="noreferrer">small angle approximation</a> to show that it's very close to <span class="math-container">$r/f$</span> or 0.024734</p>
<h3>Optical Caveats</h3>
<ul>
<li>NA is used all the time in fields like photolithography and microscopy and fiber optics, and sometimes NA values are very large so the small angle approximation is not used.</li>
<li>In most of these situations people are working near the diffraction limit, so folks frequently work with <a href="https://en.wikipedia.org/wiki/Gaussian_beam" rel="noreferrer">Gaussian beam optics</a> instead of rays. In this case they may still sometimes mix concepts like NA with beam parameters like the <span class="math-container">$1/e^2$</span> intensity radius <span class="math-container">$\omega$</span>.</li>
<li>Microscopy and photolithography sometimes use immersion techniques where the business end of the lens pointed at the sample is immersed in a liquid with an index of refraction greater than air. In this case they <em>define</em> NA as the geometrical NA discussed above times the index of the fluid, which means values greater than 1.0 are sometimes found.</li>
</ul>

CC BY-SA 4.0
2022-07-16T13:36:14.057Z
FWIW, if $\tan\theta=r/f$, then $\sin\theta=r/\sqrt{r^2+f^2}$ (and $\cos\theta=f/\sqrt{r^2+f^2}$), which saves 2 trig calls. Of course, if you're doing this on a scientific calculator, it's probably easier to just do the division & press a couple of trig buttons. ;)
PM 2Ring
CC BY-SA 4.0
2022-07-16T13:55:40.677Z
@PM2Ring like this one? https://i.sstatic.net/LctBS.jpg :-)
uhoh
CC BY-SA 4.0
2022-07-16T14:00:13.200Z
Beautiful! I knew a guy who had one, back in '73. I'm still using a Casio from the late 80s.
PM 2Ring
CC BY-SA 4.0
2022-07-16T14:00:27.130Z
On a related note, I'd still love to know the angular sizes of these JWST images... https://physics.stackexchange.com/questions/717952/questions-about-first-jwst-image/717958#comment1606473_717958
PM 2Ring
CC BY-SA 4.0
2022-07-16T14:05:20.993Z
I grew up with my dad's HP then got a slightly programmable casio as well. But when my junior high school got a teletype and a 10 baud modem to log in to the high school's computer and write basic...
uhoh
CC BY-SA 4.0
2022-07-16T14:20:36.420Z
Those were the days... I know a guy (now in his early 80s) who has a pair of functional teletypes. But they're not in use, just sitting in his garage.
PM 2Ring
CC BY-SA 4.0
2022-07-16T14:25:13.790Z
OMG... RPN logic and no = key!!!   those were the days!
BradV
CC BY-SA 4.0
2022-07-16T09:27:44.533Z
2022-07-16T09:27:44.533Z
uhoh
<p>I was wondering what the NA of the James Webb telescope is:</p>
<p>Based on this eq:</p>
<p>R = λ / 2NA &gt;&gt; NA =  λ / 2R</p>
<p><a href="https://webb.nasa.gov/content/about/faqs/facts.html" rel="nofollow noreferrer">Key Facts - James Webb Space Telescope - NASA</a></p>
<p>λ = 0.6 - 28.5 microns = 0.6 - 28.5 x 10-6 m</p>
<p>R = ~0.1 arc-seconds = 4.848 × 10-7 radian</p>
<p>NA = 0.06 - 2.94 x 10m</p>
<p>But since NA is supposed to be dimensionless I would need a resolution in meters. Any Idea what a useful conversion would look like?</p>

<james-webb-space-telescope>
What is the numerical aperture of the James Webb Telescope?
rul30
# What is the numerical aperture of the James Webb Telescope?

(Asked by: username_0 on 2022-07-16)

I was wondering what the NA of the James Webb telescope is:

Based on this eq:

R = λ / 2NA >> NA = λ / 2R

[Key Facts - James Webb Space Telescope - NASA](https://webb.nasa.gov/content/about/faqs/facts.html)

λ = 0.6 - 28.5 microns = 0.6 - 28.5 x 10-6 m

R = ~0.1 arc-seconds = 4.848 × 10-7 radian

NA = 0.06 - 2.94 x 10m

But since NA is supposed to be dimensionless I would need a resolution in meters. Any Idea what a useful conversion would look like?

**Tags:** james-webb-space-telescope


---

## 1 answers

### Answer by username_1 on 2022-07-16. Score: 11 | Accepted answer

### Background

Imagine you are a pixel sitting at the primary focus, looking up at the incoming light. What do you see? If the image of a star or something bright is falling on you, you'll see a big bright circle (or jagged hexagonalized circle-approximant for JWST) with a dark area in the center where the secondary, and hole in the mirror are.

Yes, you are looking towards the secondary, but it's a mirror so you're really looking at the primary via a reflection.

It doesn't matter how many mirrors or other optical elements are in the path, for the purposes of NA what matters is what the pixel at the focal plane sees.

Just fyi multi-instrument space telescopes like JWST and Hubble sometimes have multiple NAs (multiple f/numbers) to match different instruments.

### JWST's NA

[By definition](https://en.wikipedia.org/wiki/Numerical_aperture#General_optics) NA is the sine of the half-angle of the aperture as seen from the focal point.

Since this generally assumes a circular aperture we'll have to choose a radius for the average aperture. One easy way to do that is to find the aperture area $A$ and calculate the radius of a circle with the same area, but since so much of the JWST's aperture (and a lot of telescopes) is blocked, this won't be correct.

Luckily Wikipedia's [JWST](https://en.wikipedia.org/wiki/James_Webb_Space_Telescope) gives the following:

* Type: Korsch telescope
* Diameter: 6.5 m (21 ft)
* Focal length: 131.4 m (431 ft)
* Focal ratio: f/20.2
* Collecting area: 25.4 m2 (273 sq ft)

So the effective diameter $D$ is 6.5 meters, the effective radius $r$ is half that or 3.25 meters, and the primary focal length $f$ is 131.4 meters.

The f/no. or focal ratio which is the diameter divided by the focal length or $f/D$ is 20.2 (unitless).

The NA is really just another way to express the same ratio, but using trigonometry. It's the sine of the half-angle $HA$

$$HA = \arctan\left( \frac{r}{f} \right)$$

$$NA = \sin(HA) = \sin\left( \arctan\left( \frac{r}{f} \right)\right)$$

and in this case it will be 0.024726 and for small NA we can use the [small angle approximation](https://en.wikipedia.org/wiki/Small-angle_approximation) to show that it's very close to $r/f$ or 0.024734

### Optical Caveats

* NA is used all the time in fields like photolithography and microscopy and fiber optics, and sometimes NA values are very large so the small angle approximation is not used.
* In most of these situations people are working near the diffraction limit, so folks frequently work with [Gaussian beam optics](https://en.wikipedia.org/wiki/Gaussian_beam) instead of rays. In this case they may still sometimes mix concepts like NA with beam parameters like the $1/e^2$ intensity radius $\omega$.
* Microscopy and photolithography sometimes use immersion techniques where the business end of the lens pointed at the sample is immersed in a liquid with an index of refraction greater than air. In this case they *define* NA as the geometrical NA discussed above times the index of the fluid, which means values greater than 1.0 are sometimes found.

> **Comments:**
> **username_2**: FWIW, if $\tan\theta=r/f$, then $\sin\theta=r/\sqrt{r^2+f^2}$ (and $\cos\theta=f/\sqrt{r^2+f^2}$), which saves 2 trig calls. Of course, if you're doing this on a scientific calculator, it's probably easier to just do the division & press a couple of trig buttons. ;)  
> **username_1**: @PM2Ring like this one? https://i.sstatic.net/LctBS.jpg :-)  
> **username_2**: Beautiful! I knew a guy who had one, back in '73. I'm still using a Casio from the late 80s.  
> **username_2**: On a related note, I'd still love to know the angular sizes of these JWST images... https://physics.stackexchange.com/questions/717952/questions-about-first-jwst-image/717958#comment1606473_717958  
> **username_1**: I grew up with my dad's HP then got a slightly programmable casio as well. But when my junior high school got a teletype and a 10 baud modem to log in to the high school's computer and write basic...  
> **username_2**: Those were the days... I know a guy (now in his early 80s) who has a pair of functional teletypes. But they're not in use, just sitting in his garage.  
> **username_3**: OMG... RPN logic and no = key!!!   those were the days!  

---

