The Forbes article Ask Ethan: How Does Very-Long-Baseline Interferometry Allow Us To Image A Black Hole? includes an example of optical interferometry from the Large Binocular Telescope, shown below.

The occultation of Jupiter's moon, Io, with its erupting volcanoes Loki and Pele, as occulted by Europa, which is invisible in this infrared image. The Large Binocular Telescope was able to do this owing to the technique of interferometry. Credit: LBTO

The LBTO consists of only two telescopes 8.4 m (330 inch) wide mirrors, with centres 14.4 m apart, so I would expect to see higher resolution in one direction than the other.

However in this GIF the resolution seems to be the same in both directions. While I see high spatial frequency ringing in the horizontal direction, I don't see any difference in resolution.

Since the event is so fast (probably minutes or seconds) there is no chance that the natural rotation of the baseline during the course of a night could have helped.

So I'd like to ask How does the Large Binocular Telescope resolve so well in both orthogonal directions simultaneously?

Large Binocular Telescope

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Large Binocular Telescope


The LBTI (green structure in the center of the frame) between the two 8.4m mirrors of the LBT. High resolution image Credit: LBTO - Enrico Sacchetti

  • $\begingroup$ You need to describe more what we are looking at and how the signals from the two telescopes were combined. It's difficult to guess. $\endgroup$ – Rob Jeffries Apr 14 at 7:36
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    $\begingroup$ @RobJeffries For "what we are looking at": The block quote says "The occultation of Jupiter's moon, Io, with its erupting volcanoes Loki and Pele, as occulted by Europa, which is invisible in this infrared image. The Large Binocular Telescope was able to do this owing to the technique of interferometry. Credit LBTO" which is immediately followed by When will the next series of mutual eclipses of Jupiter's moons begin? For "how the signals from the two telescopes were combined": if I knew that I could answer the question myself. I don't know!" $\endgroup$ – uhoh Apr 14 at 8:15

The simple answer is that it isn't resolving in both orthogonal directions equally well. The horizontal dimension is the binocular dimension, and from looking at your animation, it looks to have about ~3 times the resolution. The horizontal banding, I'm pretty sure is not ringing, and is in fact, representative of additional information.

This article does a nice job of explaining what is being seen, and it has a picture, both of a simulated depiction of Io from one of the two telescope mirrors, and the combined image. enter image description here The thing you should notice from the above image is that the 8.4 m telescope picture is at the same resolution as the vertical axis of your animation.

What is going on with the interferometer image?

By increasing the separation between telescopes, we increase the angular resolution. Just about everyone with a casual interest in astronomy will have learned that fact in this past week.

But the other thing you do, is introduce interference patterns. The two telescopes essentially act like a double slit experiment. In the vertical (non binocular) dimension, the diffraction can be treated like a single-slit experiment. The angular intensity formula for a single slit is as follows: $$ I(\theta) = I_0 sinc^2(\frac{\pi b sin(\theta)}{\lambda})$$ where $b$ is the diameter of each mirror, and $\lambda$ is the wavelength of the light. enter image description here

The horizontal (binocular) dimension acts as a double slit. The double slit formula is as follows: $$ I(\theta) = I_0 cos^2(\frac{\pi d sin(\theta)}{\lambda})sinc^2(\frac{\pi b sin(\theta)}{\lambda})$$ where $d$ is the distance between the centres of the two telescopes. enter image description here

When you combine these two together, using the telescopes dimensions that you quotes (b=8.4, d=14.4), you come up with a pattern remarkably close to what you actually see from the telescope.

enter image description here

On the left, screencapture from above animation, on the right, the predicted double slit intensity.

Fringe removal

It seems that the animation you saw is based off unprocessed images from the LBT. Obviously, they have methods for removing the fringe bands. As to how, I have no idea. I found a paper that discusses interferometry in depth and they mention:

What we therefore have is a series of fringes, whose amplitude is given by the Fourier transform of the source intensity distribution. In practice, steps are usually taken to get rid of the fringes using a phase rotation whose rate is known (as both B and s are known). This is done in optical interferometers by use of accurate delay lines to compensate for the path difference, and in radio interferometers by the insertion of electronic delays.

But I have no idea what that means. Perhaps some boffin from Physics.SE would be nice enough to answer.

  • $\begingroup$ I'll hold out for actual hard evidence that there is more resolution horizontally than vertically. I don't see it yet, and simply telling me I can see it, or that it should be there doesn't help. $\endgroup$ – uhoh Apr 14 at 11:56
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    $\begingroup$ @uhoh I think you misinterpret my post. I don't have the necessary mathematical background to prove that any signal convolved with the horizontal binocular fringe pattern provides more information than the same signal convolved with the vertical fringe pattern. I do have an illustrated explanation in the form of several graphs prepared, however. It's handwavey, but i'd find it sufficiently convincing, I don't know about you. $\endgroup$ – Ingolifs Apr 14 at 12:34
  • $\begingroup$ The quick explanation is: a blurry object (say, a gaussian distribution that approximately fits the sinc function) produces a blurry image in both dimensions with indistinct or nonexistent fringes, whereas a sharp object produces an image that is sharp (with fringes) in one dimension and blurry in the other. $\endgroup$ – Ingolifs Apr 14 at 12:38
  • $\begingroup$ okay thanks, I think this is a really interesting image and exercise in image processing. $\endgroup$ – uhoh Apr 14 at 14:18

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