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All I know is its main mirror diameter; 2.40 m

I also know that the eye, which its pupil is 5mm in diameter can see up to an apparent magnitude of 6.

How can I calculate the apparent magnitude that the telescope can see?

I have used the equation of

$5⋅log(\frac{D_1}{D_0})$ However, it gave the answer of 13.40, while the answer should be 19.40

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    $\begingroup$ No, it's not. First of all, this is GROUND telescope, NOT a space one. Secondly, this doesn't have a transmission factor. $\endgroup$
    – Math Love
    Nov 30, 2018 at 13:12
  • $\begingroup$ What do you know about the magnitude scale, and what have you tried so far? $\endgroup$
    – Dr Chuck
    Nov 30, 2018 at 13:20
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    $\begingroup$ the other question starts with the Hubble but the answer directs us to the general formula in wikipedia, also talking about the background contrast, so it definitely applies to the ground observations $\endgroup$
    – szulat
    Nov 30, 2018 at 13:24
  • $\begingroup$ It gives the wrong answer, the answer should be +19.4 mag while the one I got is 13.4. There's 6 magnitudes difference, that's why I asked $\endgroup$
    – Math Love
    Nov 30, 2018 at 13:33
  • $\begingroup$ The question as it's currently written asks "How can I calculate the apparent magnitude that the telescope can see?", and the other question provides the answer, so I'm voting that this is a duplicate. However, if you reword your question to ask "I'm using this formula for my [details] telescope but I get +13.4 instead of +19.4 - where is my error?" it would be a different question and not a duplicate. And it would directly address the issue you're having :-) $\endgroup$ Dec 1, 2018 at 0:05

1 Answer 1

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You just need to add the 13.4 that you got to the 6 that you can see without a telescope and that's it. The formula is telling you the difference in magnitude that arises from the different apertures of the eye and the telescope.

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  • $\begingroup$ And that's because it's m - M = to that. I just had to make sure because the answer thingy that I got isn't particularly the best and has many mistakes in it $\endgroup$
    – Math Love
    Nov 30, 2018 at 20:37

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