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JWST reveals its first direct image discovery of an exoplanet

(www.smithsonianmag.com)
342 points divbzero | 1 comments | 27 Jun 25 17:44 UTC | HN request time: 0.212s | source
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GMoromisato ◴[27 Jun 25 23:00 UTC] No.44401068[source]▶
In case anyone is wondering, we are (sadly) very far from getting an image of this planet (or any extra-solar planet) that is more than 1 pixel across.

At 110 light-years distance you would need a telescope ~450 kilometers across to image this planet at 100x100 pixel resolution--about the size of a small icon. That is a physical limit based on the wavelength of light.

The best we could do is build a space-based optical interferometer with two nodes 450 kilometers apart, but synchronized to 1 wavelength. That's a really tough engineering challenge.

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nico ◴[27 Jun 25 23:06 UTC] No.44401110[source]▶
How big would the telescope/mirror/lens need to be to get a picture of something in the Alpha Centauri system, 4.37 light years away?

Also, could the image be created by “scanning” a big area and then composing the image from a bunch of smaller ones?

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1. GMoromisato ◴[27 Jun 25 23:38 UTC] No.44401269[source]▶
It's linear, so if it is 25 times closer then the telescope can be 25 times smaller. At 4.37 light-years we'd need an 18 kilometer telescope to image at Jupiter-sized planet at 100x100 pixel resolution.

If you only wanted 10x10 resolution you could get by with a 1.8 kilometer telescope.

Wikipedia has more: https://en.wikipedia.org/wiki/Angular_resolution. The Rayleigh criterion is the equation to calculate this.