1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inverse square law with flat detector

  1. Dec 23, 2009 #1
    I recently did an experiment to measure the activity of a radioactive source, with the idea that the activity can be found by treating the detector area as part of the surface area of a sphere.

    In the actual experiment, we plotted
    [tex] u = \frac{n d^2}{\Delta t} [/tex]
    against d, where d is the distance between the source and detector, [tex]\Delta t[/tex] is the time over which the measurement was taken, and n was the number of counts recorded. The point of this was that if the inverse square law was obeyed, u would be approximately constant with any d (assuming the source was point-like, ignoring absorption of beta particles in air and detector dead time).

    The plot of the data looked more like an arctan graph, which made sense when setting the problem up geometrically:

    inversesquare.png

    With a detector of height h, distance d from the source, any particles emitted in one dimension with an angle less than
    [tex]\theta = 2 \arctan{\left(\frac{h}{2d}\right)}[/tex]
    would strike the detector, and the same for the other dimension. If there is an equal probability of particle being emitted from any angle, then the fraction hitting the detector is

    [tex]\left(\frac{\theta}{2 \pi}\right)^2[/tex]
    (possibly with a factor of 1/2, I wasn't sure if the second dimension should use [tex]\theta / 2 \pi[/tex] or [tex]\theta / \pi[/tex], as in spherical polars), so

    [tex]u = \frac{N d^2}{\Delta t} \left(\frac{2 \arctan{\left(\frac{h}{2d}\right)}}{2 \pi}\right)^2[/tex],
    where N is the activity of the source.

    When I showed this to my demonstrator, he told me I was talking bollocks and that any deviation from the expected inverse-square fit was because the source was line-like rather than point-like. I argued that because the detector is flat rather than curved as though on the surface of a sphere you wouldn't see an inverse-square fit even with a point source, but he was pretty adamant.

    Am I in fact talking bollocks, or does my argument make sense?
     
  2. jcsd
  3. Dec 23, 2009 #2

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    He's right
    Imagine a segment of the curve surface of the sphere.
    Now put a flat square in front of the curve section
    Any light that reaches the curved surface must go through the flat square - so the flux through both is the same.

    There might be small experimental effects with a real semconductor, the detector might be less sensitive to light not arriving at a normal (ie for the flat square light reaching the eges is at an angle to the surface an so more will reflect off)
     
  4. Dec 23, 2009 #3

    Borek

    User Avatar

    Staff: Mentor

    Actually I think there is a grain of truth to your idea. Imagine flat square detector of surface a2, put in the distance [itex]\frac a {\sqrt{4 \pi}}[/itex] from the point source. Surface of the detecor is identical to the surface of sphere of radius [itex]\frac a {\sqrt{4 \pi}}[/itex], so it should detect every emitted particle, for obvious reasons it detects less than half.

    However, that requires detector size to be comparable with the distance from the source. In normal experimental setups approximation of the part of the sphere surface with flat square is good enough.
     
  5. Dec 23, 2009 #4
    I forgot to mention that I was talking about the small distance case only -- the results were as expected from the inverse-square law when the source-detector distance was much greater than the detector size. We were specifically encouraged to look at the small distance case (with source-detector distance less than the height and width of the detector) and I was surprised when my reasoning was immediately dismissed. This is just something that's been bugging me for a while now because I couldn't see any hole in my reasoning, and it seemed to fit the data quite well, though I was discouraged from any in-depth analysis.

    I see your point, mgb_phys, but in the small distance case, the increase in distance from mapping a curved surface onto the square detector (distance between source and curved surface) would be comparable to the source-detector distance itself.

    Thanks for the quick replies :D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Inverse square law with flat detector
  1. Inverse square law (Replies: 5)

  2. Inverse Square Law (Replies: 9)

  3. Inverse square law (Replies: 8)

Loading...