Шокиран от огромната разлика, реших да ви покажа пълния отговор на ИИ1
на въпроса на английски и да ви запитам: Кой ИИ сбърка и защо
-----------------------------
The James Webb Space Telescope (JWST) has an impressive angular resolution of approximately 0.1 arcseconds (or 100 milliarcseconds) at a wavelength of 2 μm[5]. This resolution is comparable to or slightly better than the Hubble Space Telescope in the overlapping wavelength range[4].
To calculate the size of one pixel at a distance of 4.2 light years, we need to convert the angular resolution to a linear size at that distance:

Therefore, at a distance of 4.2 light years, one pixel in an image from the James Webb Space Telescope would represent an area approximately 19.2 million kilometers across. This is about 1.5 times the diameter of the Sun, which is roughly 1.4 million kilometers[1].
It's important to note that this calculation assumes ideal conditions and doesn't account for various factors that might affect the actual resolution in practice, such as the specific wavelength being observed or any image processing techniques applied.
Citations:
[1] https://en.wikipedia.org/wiki/James_...pace_Telescope
[2] https://www.reddit.com/r/jameswebb/c...of_james_webb/
[3] https://astronomy.stackexchange.com/...inted-at-earth
[4] https://astronomy.stackexchange.com/...ames-webb-have
[5] https://sci.esa.int/documents/34594/...ntific_cap.pdf
[6] https://www.cloudynights.com/topic/6...e-pixel-scale/
[7] https://astronomy.tools/calculators/ccd_suitability
[8] https://ww2.ac-poitiers.fr/math/site...james_webb.pdf
на въпроса на английски и да ви запитам: Кой ИИ сбърка и защо

-----------------------------
The James Webb Space Telescope (JWST) has an impressive angular resolution of approximately 0.1 arcseconds (or 100 milliarcseconds) at a wavelength of 2 μm[5]. This resolution is comparable to or slightly better than the Hubble Space Telescope in the overlapping wavelength range[4].
To calculate the size of one pixel at a distance of 4.2 light years, we need to convert the angular resolution to a linear size at that distance:

Therefore, at a distance of 4.2 light years, one pixel in an image from the James Webb Space Telescope would represent an area approximately 19.2 million kilometers across. This is about 1.5 times the diameter of the Sun, which is roughly 1.4 million kilometers[1].
It's important to note that this calculation assumes ideal conditions and doesn't account for various factors that might affect the actual resolution in practice, such as the specific wavelength being observed or any image processing techniques applied.
Citations:
[1] https://en.wikipedia.org/wiki/James_...pace_Telescope
[2] https://www.reddit.com/r/jameswebb/c...of_james_webb/
[3] https://astronomy.stackexchange.com/...inted-at-earth
[4] https://astronomy.stackexchange.com/...ames-webb-have
[5] https://sci.esa.int/documents/34594/...ntific_cap.pdf
[6] https://www.cloudynights.com/topic/6...e-pixel-scale/
[7] https://astronomy.tools/calculators/ccd_suitability
[8] https://ww2.ac-poitiers.fr/math/site...james_webb.pdf
Коментар