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Stargazing Solid angle acceptance of a muon telescope

  1. Nov 14, 2007 #1
    I am trying to work out the solid angle acceptance of a muon telescope. The telescope is comprised of two aligned square detector panels (of size x squared metres) set at a distance apart of L metres. The way I was initially working it out (by using the solid angle of a pyramid of base x squared and height 0.5L, containing [as I thought] all the possible muon tracks through the telescope) is completely different to that which my friend insists is the right way (a solid angle defined by [(x squared)/(L squared)]).

    Any help on how to find the solid angle from first principles and a resultant equation?
     
  2. jcsd
  3. Nov 14, 2007 #2

    mgb_phys

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    Remember it's all possible tracks through the distant square to a single point on the near detector.
    Assuming the detectors are much further apart than their size then x^2/L^2 sounds right.
     
  4. Nov 14, 2007 #3
    And how would you derive that?
     
  5. Nov 14, 2007 #4

    mgb_phys

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    From the definition of solid angle. And by assuming that sin(a) = a for small angles.
     
  6. Nov 23, 2007 #5
    Ok, so you have all possible tracks going through the first detector to a single point on the second detector (effectively a pyramid) but what about the tracks that hit the second detector outside of the point?
     
  7. Nov 23, 2007 #6

    mgb_phys

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    For any given point on the 2nd detector there is the same area of initial of detector 1 at the same distance and so the same solid angle.
    The corners of the detectors do see slightly different solid angles and so the corner-corner angle is larger - this is normally ignored if L is much greater than the size of the detector.
    In optical detectors it would correspond to the maximum angle vs the unvignetted angle.
     
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