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The sun in the Voyager 1 photo here from 1990 appears 13x bigger than its diameter, diffraction of course. On Earth, photographs of the sun do not make it appear 13x bigger. Why?

A description of the photo from jpl.nasa.gov,

"The wide-angle was taken with the camera's darkest filter (a methane absorption band), and the shortest possible exposure (5 thousandths of a second) to avoid saturating the camera's vidicon tube with scattered sunlight. "

enter image description here

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    $\begingroup$ Where do you get the value "13" times bigger from? Why do you say "diffraction of course" $\endgroup$ – James K Oct 26 '18 at 22:37
  • $\begingroup$ distance to Venus, 77 solar diameters, see position of planets and Voyager 1 here, en.wikipedia.org/wiki/Pale_Blue_Dot#/media/… $\endgroup$ – user24662 Oct 26 '18 at 22:46
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    $\begingroup$ Can you show your working? $\endgroup$ – James K Oct 26 '18 at 22:56
  • $\begingroup$ its known data, the distance to Venus from the sun is on average 77 solar diameters, Venus is in the composite closest to the sun, and Earth the box next to it $\endgroup$ – user24662 Oct 26 '18 at 23:12
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    $\begingroup$ Apparently the Sun also has 8 spikes coming out of it and an off-centre corona with 2 distinct zones with different brightness. $\endgroup$ – Rob Jeffries Oct 27 '18 at 22:01
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Don't base the size of the Sun on what you see in that image because the Sun is over-exposed. The big spot in the center of the diffraction spikes are saturated pixels because they all appear to have the same value (the spot looks uniform) so you don't know what their real values are; you would need to take a shorter-exposure image (or put filters in front of the lens) if you wanted to take a proper picture of the Sun. Since Venus is a lot dimmer than the Sun, they set the exposure long enough to see it and accepted the fact that the Sun would be over-exposed. You see the same thing in some of the Hubble images where to bring out the background stars, the brighter ones are over-exposed.

Another way of looking at it, if you see an image where you can very clearly see the diffraction spikes, that part of the image is over-exposed.

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  • $\begingroup$ the image where Venus is visible was a separate photo, it is composited into the photograph, same with the image of Earth $\endgroup$ – user24662 Oct 27 '18 at 14:48
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    $\begingroup$ Either way, you can't resolve the Sun in that picture because it is washed out. For instance, look at these streetlight images. Clearly the streetlights aren't as physically large as the images suggest they are. $\endgroup$ – Dave Oct 27 '18 at 15:00
  • $\begingroup$ what do you mean either way, the images are composited. there is no either way, there is one way since the image is what it is. most images of the sun, if you Google, are not 13x larger, they're less than 2x larger, why would the Voyager 1 image capture it as 13x larger, if it was making a composite why overexpose the image for the sun? $\endgroup$ – user24662 Oct 27 '18 at 15:13
  • $\begingroup$ By either way, I meant that it doesn't matter whether it was a composite image or not because the image of the Sun is overexposed just like the image of some of the stars in the Hubble image are, and the images of the streetlamp are. I don't know the source of the image so I can't tell you what the Voyager image team was trying to look, but I would be surprised if they were trying to image the Sun. Does the image not look overexposed to you? Here is another example of overexposed stars. $\endgroup$ – Dave Oct 27 '18 at 15:30
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    $\begingroup$ From what you said, it seems they were trying to get the "family picture", which in itself is really cool. But also from what you said, they used their darkest filter and their shortest exposure, so they did all they could do for the Sun. Even with those settings the Sun was just too bright (but not so bright as to wash out the whole picture). Voyager wasn't meant to image the Sun. If it was, they would have included an appropriate filter, like an H-$\alpha$ filter. Check out SOHO for a mission designed for that. $\endgroup$ – Dave Oct 27 '18 at 16:09