# How does the JWST change focus when it goes from looking at a near subject to looking at a far subject? (Does it even change focus?)

After following the extensive steps in the process of aligning and phasing the optics and instruments of JWST... I now wonder if further focus adjustments would be made based on distance to subject. Perhaps that focus is already factored into the specific instruments. Perhaps big redshifts don't care about focus. Perhaps 'focused at infinity' is all that is needed.

• The JWST will only ever focus on objects that are at least hundreds of millions of km away. Most objects will be many lightyears (hundreds of trillions of km). So no need to change focus. Apr 5 at 1:24
• If you look at a camera with manually adjustable focus, you'll see the distance from the subject (in meters) printed as 0.3..............1.........2.....8..20∞ The difference between focus 100m away and 10km away is negligible. The difference in focus between 1AU and 1 parsec is none.
– SF.
Apr 5 at 9:47
• @hdhondt: Are you sure the JWST will never focus on Mars? Apr 5 at 13:10
• @TonyK A quick calculation shows that the image of the Mars would be around $100\rm\, nm$ out of the focal plane. Visible light has a wavelength of around $500\rm\, nm$, infrared even more. I don't know if JWST will even bother. Apr 5 at 15:06
• À propos Mars, Mark McCaughrean, senior advisor for science and exploration at the European Space Agency, says in an article about JWST: "We can look at planets like Mars, Jupiter, Saturn, Uranus and Neptune but also into the Kuiper Belt." Apr 5 at 15:48

It doesn't change focus. Not even to correct for which wavelengths are being measured, because that's needed by only refractors, not reflectors like JWST. In the terminology of handheld terrestrial cameras and lenses, everything that JWST or even a backyard Celestron sees is "at infinity."

Generally, focusing means aiming light rays from a subject, gathered by a big mirror or a big lens, to a point on the focal plane. (A pinhole camera is always in focus, because its "lens" is so small.) For a handheld camera pointing at a subject at arm's length, those rays diverge by the time they reach the face of the lens. A subject across the room has rays that diverge less, so you have to "change the focus." A subject in the next solar system has rays whose divergence is unmeasurably small: they're exactly parallel, for Alpha Centauri or for something at the edge of the observable universe. (For nearby Alpha Centauri, a bit of trig shows that across the 6.5 m width of JWST's mirror, its rays diverge by about a femtodegree.)

• So even looking a solar system objects like asteroids (which JWST will most certainly do) there is no need to refocus? Cool! Apr 5 at 6:54
• The rays from something in the asteroid belt, by the time they reach JWST, would diverge by - brace yourself - almost a nanodegree! Apr 5 at 14:52
• @TrySCE2AUX if my math is right, an object would have to be within about 3km of JWST to have a focal blur radius on the order of one NIRCAM pixel. Even if I messed that up, I didn't do so by enough orders of magnitude for solar system objects to be any worry. Anything at a distance that's millions of times the diameter of the mirror is effectively at infinity. Apr 5 at 16:56
• I assume they have some sort of a focusing-adjustment mechanism so that if the mirrors drift away from "infinity" it can be re-focuses, but that said mechanism does not depend on the distance away it's target is because even one AU is basically infinite. Apr 5 at 22:05
• "Basically infinite" starts at only 3 km, not 1 AU, hobbs calculated. Also, if the mirrors wander around and need to be shepherded back, that's called alignment, not focusing. Apr 5 at 22:12

The JWST focal length is 131.4 m. There is an equation for object distance $$u$$, image distance $$v$$ and focal length $$f$$:

$$\frac{1} { f} = \frac{1} {u} + \frac{1} {v}$$

If the object is in infinite distance then

$$\frac{1} { u} = 0$$

the equation simplifies to:

$$\frac{1} { f} = 0 + \frac{1} {v}$$

$$\frac{1} { f} = \frac{1} {v}$$

$$v = f$$

and $$v$$ is equal to $$f$$.

I set the image distance $$v$$ to 131.4 m + 10 nm (10 nanometers or 10-8 m or 0.00000001 m) and calculate the distance of the object $$u$$.

$$\frac{1} { 131.4 m} = \frac{1} {u} + \frac{1} { 131.4 m + 10 nm}$$

$$\frac{1} { u} = \frac{131.4 m} {u} - \frac{1} { 131.4 m + 10 nm}$$

$$\frac{1} { u} = \frac{10 nm} {131.4 m * (131.4 m + 10 nm)}$$

$$u = \frac{131.4 m * (131.4 m + 10 nm)} {10 nm}$$

$$u = 131.4 * (131.4 m + 10 nm) * 10^8$$

$$u = 131.4^2 * 10^8 m$$

$$u = 17.26 * 10^{12} m$$

$$u = 11.5 AU$$

The distance of Saturn to Sun is 9.5 AU.

So if the object is not at infinite distance but at 11.5 AU, the change of image distance (the distance to refocus properly) is 10 nm, at least sixty times smaller than the used IR wavelengths, which are 600 to 2500 nm.

The stars imaged by the JWST are much farther than Saturn, so the change of image distance is much smaller than 10 nm and very, very close to zero.