# What is the numerical aperture of the James Webb Telescope?

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

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?

### 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 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 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 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 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.
• 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. ;) Jul 16, 2022 at 13:36
• @PM2Ring like this one? i.stack.imgur.com/LctBS.jpg :-)
– uhoh
Jul 16, 2022 at 13:55
• Beautiful! I knew a guy who had one, back in '73. I'm still using a Casio from the late 80s. Jul 16, 2022 at 14:00
• 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. Jul 16, 2022 at 14:20
• OMG... RPN logic and no = key!!! those were the days! Jul 16, 2022 at 14:25