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Making a guess based on my knowledge of photography: probably not. It's taking vast starscapes with a wide lens with a focus range (or fixed focus) made for sho

by Kye 5y ago

Making a guess based on my knowledge of photography: probably not. It's taking vast starscapes with a wide lens with a focus range (or fixed focus) made for shooting at infinity. That means the closer something is, the less in focus it'll be. Try looking at something right in front of your face. At its furthest, JWST will be at a proportional distance relative to what Hubble is equipped to focus on. Even stitching images together, it's always going to have similar focus. No amount of resolution will get past the physics of optics. Just a guess. I could be wrong!

_moof 5y ago
I'll add that Hubble has an angular resolution of 0.05 arcseconds, and Earth-Sun L2 is roughly 1,500,000 km away, which comes out to about 350 m resolution at that distance--fifteen times larger than JWST.
- smingo 5y ago
  Well that actually makes it almost plausible for Hubble to capture an image of JWST According to Wikipedia, the order is: Sun, L1, Earth, L2. So L2 is always in the night sky. I don't see issues with focus. 1.5M km is effectively infinity. Indeed for my camera on earth, 10m resolves as pretty much as well as infinity, f stops notwithstanding. JWST is about 20m by 14m in size. And it's out of the earth + moon shadow because of the nature of its orbit. It has a sun shield, and radio dish in the shield. The resolving power of 0.05 arcseconds at 1,500,000 km apogee is indeed 350 m (ish), which is much bigger than the JWST. The situation is worse at perigee. An image of JWST would take up less than 1% of Hubble 'pixel'. So there may be a faint blob, depending on the albedo of the JWST. (I don't know how Hubble's the photodetector arrays' resolution compares to the maximum resolving power of the lens.) However, Hubble's guidance systems have permitted astronomy at 0.0003 arcseconds, which comes out at a little over 2 m at JWST's range. So after a lot of computation, stacked images and fine-guidance shifting, presumably there could be an image of around 10 x 7 pixels (or whatever they are after the computation). And that's a result!

by Kye 5y ago

_moof 5y ago

I'll add that Hubble has an angular resolution of 0.05 arcseconds, and Earth-Sun L2 is roughly 1,500,000 km away, which comes out to about 350 m resolution at that distance--fifteen times larger than JWST.

smingo 5y ago
Well that actually makes it almost plausible for Hubble to capture an image of JWST According to Wikipedia, the order is: Sun, L1, Earth, L2. So L2 is always in the night sky. I don't see issues with focus. 1.5M km is effectively infinity. Indeed for my camera on earth, 10m resolves as pretty much as well as infinity, f stops notwithstanding. JWST is about 20m by 14m in size. And it's out of the earth + moon shadow because of the nature of its orbit. It has a sun shield, and radio dish in the shield. The resolving power of 0.05 arcseconds at 1,500,000 km apogee is indeed 350 m (ish), which is much bigger than the JWST. The situation is worse at perigee. An image of JWST would take up less than 1% of Hubble 'pixel'. So there may be a faint blob, depending on the albedo of the JWST. (I don't know how Hubble's the photodetector arrays' resolution compares to the maximum resolving power of the lens.) However, Hubble's guidance systems have permitted astronomy at 0.0003 arcseconds, which comes out at a little over 2 m at JWST's range. So after a lot of computation, stacked images and fine-guidance shifting, presumably there could be an image of around 10 x 7 pixels (or whatever they are after the computation). And that's a result!