<p>For <em>any</em> type of imaging telescope: optical, radio, interferometric etc, the focal plane is at infinity.</p>
<p>Consider a simple optical telescope, looking at a nearby planet (Jupiter). And suppose the objective lens of the telescope has a focal length of one metre (so an object at infinity would be brought to sharp focus at one metre, <span class="math-container">$f=1$</span>).  Then where would the focus for Jupiter be?  The formula is quite simple: if <span class="math-container">$s_i$</span> and <span class="math-container">$s_o$</span> are the distances of the image and object from the lens, then <span class="math-container">$1/s_i+1/s_o=1/f$</span>.</p>
<p>In metres, Jupiter at opposition is <span class="math-container">$900\, 000\, 000\, 000m$</span> from Earth, so the focal distance for Jupiter <span class="math-container">$s_i$</span> is 1.000000000001 m. The difference is unmeasurable. When focussing for Jupiter, you focus at infinity.</p>
<p>A similar calculation could be done for the 50million light year distance to M81.  The conclusion would be that you don't change the focal length.  The focal length is fixed at infinity and it is not possible to change it finely enough to make any difference to the sharpness of he image.</p>
<p>Practical telescopes do have the ability to change the focus. This is so you can focus on nearby objects on Earth, and to adjust for things like thermal expansion of the telescope. It is easier to create a telescope that is adjustable than to create one with a permanently fixed and perfect focus at infinity.</p>

CC BY-SA 4.0
2022-03-10T08:40:08.823Z
The EHT could certainly look at Jupiter and it would definitely be way way way "out of focus" at that short distance. With a 10,000 km aperture and 1 mm wavelength it's a huge effect; about 15 meters or 15,000 $\lambda$ wavefront error!
uhoh
CC BY-SA 4.0
2022-03-09T06:37:09.783Z
2022-03-09T06:37:09.783Z
James K
<p>The EHT is an array of radio telescopes, so it doesn't have a &quot;focal plane&quot; or &quot;focal length&quot; or &quot;depth of field&quot; like optical telescopes do.</p>
<p>Instead, the EHT measures radio waves at many different geographically diverse locations and then mathematically processes them together to make conclusions about the emitters.  The &quot;picture&quot; of M87 is the result of an algorithm incorporating the processed data gathered by the radio telescopes, rather than a snapshot with a classical optical telescope.</p>
<p>Consider a triangle formed by a distant galaxy and two radio telescopes here on Earth.  That triangle would have an incredibly small interior angle at the distant galaxy.  That means small timing errors in the radio telescopes' measurements translate into large errors in a distance estimate.  This prevents accurate direct trigonometric distance estimates for far away objects with radio telescopes.</p>
<p>However, radio telescopes can make incredibly fine measurements of Doppler shift. This can help determine rotation rates of gaseous disks near the center of other galaxies, which can indirectly allow very accurate distance estimates for some far-away galaxies.  e.g. <a href="https://www.nrao.edu/pr/1999/distance/" rel="nofollow noreferrer">https://www.nrao.edu/pr/1999/distance/</a></p>

CC BY-SA 4.0
2022-03-09T00:53:59.357Z
I knew that the EHT used interferometry not lenses to take its picture, but given that the method is all about using the very acute angles in that triangle to get information I figured that it would nevertheless have the same issues as a lenses do with their angles, just with all the information imbedded in the interference pattern instead of any "'visible'" change in the "focus". Maybe I shouldn't have used so much metaphor in a science question.  By "focal length" I meant any distance estimate that was used to produced the image, or was no such estimate needed by the telescopes or algorithm?
Caston
CC BY-SA 4.0
2022-03-09T17:39:10.297Z
@Caston Perhaps distance estimates were used to produce the image, but the geometry of each radio telescope assumes that photons from an emitter arrive parallel to one another across the dish.  One could think about this as the emitter being infinitely far away.
Connor Garcia
CC BY-SA 4.0
2022-03-08T23:42:49.973Z
2022-03-08T23:42:49.973Z
Connor Garcia
<blockquote>
<p>Can the Event Horizon Telescope been used to find intergalactic distances?</p>
</blockquote>
<p><strong>tl;dr:</strong> No, but you could confirm the approximate distance to a radio source in the Oort cloud!</p>
<hr />
<p><a href="https://en.wikipedia.org/wiki/Event_Horizon_Telescope" rel="nofollow noreferrer">EHT</a> uses <a href="https://en.wikipedia.org/wiki/Very-long-baseline_interferometry" rel="nofollow noreferrer">VLBI</a> and for several ground stations we can think of the baseline as its aperture.</p>
<p>Absolute phases (sub-millimeter wavefront errors) are hard to establish across an aperture the size of Earth but if you record signals from both your target of interest and a calibration source you could estimate the curvature of one wavefront with respect to the other.</p>
<p>The image released a few years ago was at <a href="https://astronomy.stackexchange.com/a/30364/7982">230 GHz or 1.3 mm wavelength</a> though it will get smaller <a href="https://astronomy.stackexchange.com/q/35757/7982">Is there any work underway to push the long baseline capabilities of the Event Horizon Telescope to sub-millimeter wavelengths?</a> and reference therein. For the sake of argument let's call the half-baseline (radius of aperture) 5,000 km or <span class="math-container">$1 \times 10^{10}$</span> half-wavelengths.</p>
<p>For aperture radius <span class="math-container">$r$</span> and a finite distance <span class="math-container">$D$</span> the wavefront error at the edge is given by <span class="math-container">$\frac{1}{2} r^2/D$</span> and that will be a wavelength at about <span class="math-container">$10^{13}$</span> km or about 70,000 AU.</p>
<p>So theoretically, in principle, if you had a good sharp calibration source in the same interferometric field of view, you could use <em>focus range finding</em> to detect the presence of a self-luminous, radio-bright object in the Oort cloud and get a rough distance for it.</p>
<p>Of course it's pretty cold out there and so folks would be really surprised if one was found; perhaps a tiny, sizzling black hole eating an icy comet?</p>
<p>Now if you had millimeter wave radio telescopes in heliocentric orbit <span class="math-container">$r$</span> is now <span class="math-container">$10^4$</span> larger and so the resulting distance <span class="math-container">$D$</span> is then  <span class="math-container">$10^8$</span> larger or more than the <em>astronomical</em> 10^{13} AU!</p>
<hr />
<p>Currently unanswered in Space SE: <a href="https://space.stackexchange.com/q/50904/12102">Has Hubble ever focused on something close enough that it had to move away from being focused at infinity?</a></p>

CC BY-SA 4.0
2022-03-10T08:35:27.420Z
2022-03-10T08:43:08.283Z
2022-03-10T08:43:08.283Z
uhoh
<p>Now that we have the Event Horizon Telescope (EHT), why can't we just guess and check intergalactic distances by changing the EHT's focal length?  For instance why don't we take another picture of M87, but assume it is actually 50 lightyears further away, and see if the image is any clearer?  If so than M87 is actually closer to being 50 light years further away than it is where we currently assume.  Try again enough times and we shouldn't we be able to determine the galaxy's position to the same precision we can change the focal point of the EHT?  I assume the blurriness of the M87 picture was mostly due to the long exposure time, but it couldn't be the only factor.  Obviously the depth of field increases for galaxies further away (if I'm using &quot;depth of field&quot; correctly), but I would think that measuring even the closest galaxies to such immense precision would render standard candles nearly obsolete, and cut out the bottom rungs of the <a href="https://youtu.be/dsCjRjA4O7Y?t=395" rel="noreferrer">cosmological distance ladder</a>.</p>
<p>Why haven't I heard of using the EHT in this way?</p>

CC BY-SA 4.0
2022-03-09T13:30:25.013Z
The difference in focal distance is based on the direction of light rays in a similar way as parallax measurements are. But where the baseline for focal distance is limited to the telescope size, the baseline for parallax measurements is the diameter of Earth's orbit.
jpa
<distances><expansion><event-horizon-telescope>
Can the Event Horizon Telescope been used to find intergalactic distances?
Caston
# Can the Event Horizon Telescope been used to find intergalactic distances?

(Asked by: username_0 on 2022-03-08)

Now that we have the Event Horizon Telescope (EHT), why can't we just guess and check intergalactic distances by changing the EHT's focal length? For instance why don't we take another picture of M87, but assume it is actually 50 lightyears further away, and see if the image is any clearer? If so than M87 is actually closer to being 50 light years further away than it is where we currently assume. Try again enough times and we shouldn't we be able to determine the galaxy's position to the same precision we can change the focal point of the EHT? I assume the blurriness of the M87 picture was mostly due to the long exposure time, but it couldn't be the only factor. Obviously the depth of field increases for galaxies further away (if I'm using "depth of field" correctly), but I would think that measuring even the closest galaxies to such immense precision would render standard candles nearly obsolete, and cut out the bottom rungs of the [cosmological distance ladder](https://youtu.be/dsCjRjA4O7Y?t=395).

Why haven't I heard of using the EHT in this way?

**Tags:** distances, expansion, event-horizon-telescope

> **Comments:**
> **username_1**: The difference in focal distance is based on the direction of light rays in a similar way as parallax measurements are. But where the baseline for focal distance is limited to the telescope size, the baseline for parallax measurements is the diameter of Earth's orbit.  

---

## 3 answers

### Answer by username_2 on 2022-03-09. Score: 6

For *any* type of imaging telescope: optical, radio, interferometric etc, the focal plane is at infinity.

Consider a simple optical telescope, looking at a nearby planet (Jupiter). And suppose the objective lens of the telescope has a focal length of one metre (so an object at infinity would be brought to sharp focus at one metre, $f=1$). Then where would the focus for Jupiter be? The formula is quite simple: if $s\_i$ and $s\_o$ are the distances of the image and object from the lens, then $1/s\_i+1/s\_o=1/f$.

In metres, Jupiter at opposition is $900\, 000\, 000\, 000m$ from Earth, so the focal distance for Jupiter $s\_i$ is 1.000000000001 m. The difference is unmeasurable. When focussing for Jupiter, you focus at infinity.

A similar calculation could be done for the 50million light year distance to M81. The conclusion would be that you don't change the focal length. The focal length is fixed at infinity and it is not possible to change it finely enough to make any difference to the sharpness of he image.

Practical telescopes do have the ability to change the focus. This is so you can focus on nearby objects on Earth, and to adjust for things like thermal expansion of the telescope. It is easier to create a telescope that is adjustable than to create one with a permanently fixed and perfect focus at infinity.

> **Comments:**
> **username_3**: The EHT could certainly look at Jupiter and it would definitely be way way way "out of focus" at that short distance. With a 10,000 km aperture and 1 mm wavelength it's a huge effect; about 15 meters or 15,000 $\lambda$ wavefront error!  

---

### Answer by username_4 on 2022-03-08. Score: 4

The EHT is an array of radio telescopes, so it doesn't have a "focal plane" or "focal length" or "depth of field" like optical telescopes do.

Instead, the EHT measures radio waves at many different geographically diverse locations and then mathematically processes them together to make conclusions about the emitters. The "picture" of M87 is the result of an algorithm incorporating the processed data gathered by the radio telescopes, rather than a snapshot with a classical optical telescope.

Consider a triangle formed by a distant galaxy and two radio telescopes here on Earth. That triangle would have an incredibly small interior angle at the distant galaxy. That means small timing errors in the radio telescopes' measurements translate into large errors in a distance estimate. This prevents accurate direct trigonometric distance estimates for far away objects with radio telescopes.

However, radio telescopes can make incredibly fine measurements of Doppler shift. This can help determine rotation rates of gaseous disks near the center of other galaxies, which can indirectly allow very accurate distance estimates for some far-away galaxies. e.g. <https://www.nrao.edu/pr/1999/distance/>

> **Comments:**
> **username_0**: I knew that the EHT used interferometry not lenses to take its picture, but given that the method is all about using the very acute angles in that triangle to get information I figured that it would nevertheless have the same issues as a lenses do with their angles, just with all the information imbedded in the interference pattern instead of any "'visible'" change in the "focus". Maybe I shouldn't have used so much metaphor in a science question.  By "focal length" I meant any distance estimate that was used to produced the image, or was no such estimate needed by the telescopes or algorithm?  
> **username_4**: @username_0 Perhaps distance estimates were used to produce the image, but the geometry of each radio telescope assumes that photons from an emitter arrive parallel to one another across the dish.  One could think about this as the emitter being infinitely far away.  

---

### Answer by username_3 on 2022-03-10. Score: 2

> Can the Event Horizon Telescope been used to find intergalactic distances?

**tl;dr:** No, but you could confirm the approximate distance to a radio source in the Oort cloud!

---

[EHT](https://en.wikipedia.org/wiki/Event_Horizon_Telescope) uses [VLBI](https://en.wikipedia.org/wiki/Very-long-baseline_interferometry) and for several ground stations we can think of the baseline as its aperture.

Absolute phases (sub-millimeter wavefront errors) are hard to establish across an aperture the size of Earth but if you record signals from both your target of interest and a calibration source you could estimate the curvature of one wavefront with respect to the other.

The image released a few years ago was at [230 GHz or 1.3 mm wavelength](https://astronomy.stackexchange.com/a/30364/7982) though it will get smaller [Is there any work underway to push the long baseline capabilities of the Event Horizon Telescope to sub-millimeter wavelengths?](https://astronomy.stackexchange.com/q/35757/7982) and reference therein. For the sake of argument let's call the half-baseline (radius of aperture) 5,000 km or $1 \times 10^{10}$ half-wavelengths.

For aperture radius $r$ and a finite distance $D$ the wavefront error at the edge is given by $\frac{1}{2} r^2/D$ and that will be a wavelength at about $10^{13}$ km or about 70,000 AU.

So theoretically, in principle, if you had a good sharp calibration source in the same interferometric field of view, you could use *focus range finding* to detect the presence of a self-luminous, radio-bright object in the Oort cloud and get a rough distance for it.

Of course it's pretty cold out there and so folks would be really surprised if one was found; perhaps a tiny, sizzling black hole eating an icy comet?

Now if you had millimeter wave radio telescopes in heliocentric orbit $r$ is now $10^4$ larger and so the resulting distance $D$ is then $10^8$ larger or more than the *astronomical* 10^{13} AU!

---

Currently unanswered in Space SE: [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)


---

