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Minimun lens size (in meters) that will permit just resolving 2 stars

  1. May 17, 2006 #1

    I have really no clue on how to this:cry: please help!
     
    Last edited: May 17, 2006
  2. jcsd
  3. May 17, 2006 #2

    nrqed

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    The first minimum occurs when [itex] a sin (\theta) = \lambda [/itex].



    In these problems, often the angles are so small that you may replace sin and tan by the angle itself at the condition of working in radians.

    Then [itex] a \theta \approx \lambda [/itex].

    They give you lambda and theta. Just convert theta in radians and solve for "a" which will be the slit width "a"
     
  4. May 17, 2006 #3
    how to calculate theta? thanks you so much!
     
  5. May 17, 2006 #4

    nrqed

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    If one is given the distance between the sources of light (perpendicular to the line of sight) and the distance between the sources and the lens, there is a formula in terms of a tangent. But here you are in luck, they give you the value of theta in the problem!
    They are 1.1 arcsecond apart (if they said second, they really meant arcsecond). Do you know what an arc second is? It is one degree divided by 3600 (this comes from the fact that one degree is 60 arcminutes and one arcminute is 60 arcsenonds). So your angle is 1.1/3600 degrees. Now convert that in radians, plug that in the equation and you are done.

    Patrick
     
  6. May 17, 2006 #5
    thx for your answer!
     
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