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Solid angle subtended by disc

  1. Aug 6, 2015 #1
    The problem is to find out the solid angle ω subtended by a disc of radius a at a point P distant z from its centre along its axis. α is the semi-vertical angle of the disc at the point P in question.

    The answer is supposed to be ω = 2π (1 - cos α), according to an online text. However, I find that ω = (Area) / (perpendicular distance)2 = (πa2)/z2 = π tan2α.

    I mark my answer in red, in contrast to the "correct" answer in blue.

    Any help?
  2. jcsd
  3. Aug 6, 2015 #2
    "Area" is not the area of the disc, but the area delimited by the disk on the surface of the sphere with the center in P.
  4. Aug 6, 2015 #3
    Thanks mate.

    Now how do I find this area?
  5. Aug 6, 2015 #4
    I think it is a good exercise to calculate it:

    800px-Spherical_Cap.svg.png 800px-Spherical_Cap.svg.png
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