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Minimum angular separation for viewing stars

  • Thread starter grouper
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  • #1
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Homework Statement



What is the minimum angular separation an eye could resolve when viewing two stars, considering only diffraction effects?

Homework Equations



Rayleigh criterion: θ=(1.22*λ)/D

The Attempt at a Solution



This problem doesn't give very much to go on so I think there's a trick I'm missing. I tried estimating using λ=550 nm and D≈5.0 mm (which I got from the problem just before it), but this was incorrect. Our book also states that the best resolution of the human eye is 5e-4 rad, although I'm not sure how they came by this number and it is not the correct answer. I'm not really sure where to go with this one though, especially given so little information.
 

Answers and Replies

  • #2
1,506
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what did you get in your estimate?
I got 1.34 x 10^-4 radians
 
  • #3
52
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That is what I got as well but that is not the correct answer. Perhaps I should have included the following equations as well (derived from Rayleigh):

circular aperture diffraction:

dark rings: sinθ=1.22*(λ/D) or 2.23*(λ/D) or 3.24*(λ/D), etc.

bright rings: sinθ=1.63*(λ/D) or 2.68*(λ/D) or 3.70*(λ/D), etc.

Still doesn't help much though because it yields the same estimates. This problem must not be looking for estimates. I think it has something to do with treating the stars like diffraction points but I'm not really sure if that's correct or how to do that.
 
  • #4
1,506
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I can't see anything wrong with the way we have worked it out.... it is a standard text book exercise.
What has been given as the 'correct answer'?
 
  • #5
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It's an online thing so it tells me when I get it wrong but I can't see the right answer unless I want to give up and lose that point. I'll think about it some more.
 
  • #6
1,506
17
could you get the correct answer by using different (but reasonable) values for λ and D?
 
  • #7
52
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No, I tried that. I don't think estimation is what this problem is getting at. There's got to be some way to tease some of the variables out; perhaps by assuming the distance to be infinity, even though that's not correct. I'll keep playing around with it; it's due this weekend.
 

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