I am looking for a lens for my Canon 550D for astrophotography. What lens would be suitable for it? Thank you
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14Possible duplicate of What lens characteristics are important in star photography? – Romeo Ninov Feb 10 '16 at 11:22
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3Adding to Romeo's comment above, since I see you are new here: We generally try to avoid "please recommend a specific product" types of questions, because what specific products are available changes very rapidly and we want to not just answer individuals' questions right now but also build a repository of knowledge for the future. Hence, it's better to ask questions about how to choose a product that fits your needs, which the suggested duplicate should help you with. – user Feb 10 '16 at 12:23
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Do you want to zoom in and photograph individual objects, or do you want to capture wide fields of stars? – Caleb Feb 10 '16 at 18:15
4 Answers
Look for a wide-angle lens like the Rokinon 14mm f2.8. It will let you capture a bigger section of the sky. F2.8 or smaller takes in more light in a shorter time so that you don't get star trails.
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What matters is the absolute aperture, not the aperture ratio (the "f/" value).
A 50mm f/5 (absolute aperture: 10mm) will give you more stars than a 14mm f/2.8 (absolute aperture: 5mm).
It is obvious after all: it has a bigger hole through which more light passes through.
Source: http://www.clarkvision.com/articles/low.light.photography.and.f-ratios/
This for more information, read the whole series of articles and you'll know enough :)
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And try here: http://www.lonelyspeck.com
There should be a link there to a spreadsheet with the work done for you calculating absolute apertures and lens width to give an index of that ought to work for you.
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Downvoting. While Lonelyspeck is a useful site, the link you provided is just a link to the domain's main page. I will happily reverse the downvote (and upvote) if you edit your answer to: 1) provide a direct link to the spreadsheet, or article explaining the takeaway of the spreadsheet; and 2) summarize or otherwise provide some details about the link (your answer is basically a link-only answer) – scottbb Feb 27 '16 at 16:30