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Intensity by cylindrical light source

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data
    The intensity produced by a long cylindrical light source at a small distance r from the source is proportional to
    a)[itex]\frac{1}{r^2}[/itex]
    b)[itex]\frac{1}{r^3}[/itex]
    c)[itex]\frac{1}{r}[/itex]
    d)None of these


    2. Relevant equations
    I=[itex]\frac{dF}{dω}[/itex]
    I is the intensity, dF is the luminous flux of the radiation emitted by the source in a small cone of solid angle dω constructed around the given direction.

    3. The attempt at a solution
    I have absolutely no idea on where to begin with. I know this would involve integration but i am clueless on how to form the integral. Any help is appreciated!
     
  2. jcsd
  3. Nov 7, 2012 #2

    tiny-tim

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    Hi Pranav-Arora! :smile:

    Why integrate? :confused:

    Just draw the field-lines! :wink:
     
  4. Nov 7, 2012 #3
    Hi tiny-tim! :smile:

    My book doesn't mention about the field lines but i suppose they would look like as shown in the attachment. Sorry, if that's wrong because my book has a very small chapter on Photometry and i have learned this on my own.
     

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  5. Nov 7, 2012 #4

    tiny-tim

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    Hi Pranav-Arora! :smile:
    Yup! :biggrin:

    ok, so how much do they spread out at distance r?

    (btw, do the same for the field lines of a point source, just for comparison :wink:)
     
  6. Nov 7, 2012 #5
    The intensity should decrease with the distance r but you ask about field lines here, i can't really visualize.
     
  7. Nov 7, 2012 #6

    tiny-tim

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    Sorry, I don't know why I called them the field lines, I meant the rays. :redface:

    Intensity is proportional to rays per area.

    For a point source, the same rays cover a sphere of area 4πr2 for any distance r,

    and so the intensity (rays per area) is proportional to 1/r2 (usual result!)​

    Similarly, for a line source, use a cylinder, and you get … ? :smile:
     
  8. Nov 7, 2012 #7
    Thanks for that link. :smile:

    I guess I get it now. The rays will cover a cylinder of area [itex]2\pi rh[/itex] (h is the height of cylinder) and therefore the intensity is proportional to 1/r. Right? :smile:

    What if the cylindrical source is of finite length?
     
  9. Nov 7, 2012 #8

    tiny-tim

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    S'right!:smile:
    Then the rays are a sort of mixture of point source at the ends, and cylinder source in the middle.

    The closer you are, and the closer you are to the middle, the nearer the intensity is proportional to 1/r. :wink:

    (but in that case there's no easy way, and you certainly would need to integrate!!)
     
  10. Nov 7, 2012 #9
    I expected it. But i am clueless on how to consider a small part and find its intensity.
     
  11. Nov 7, 2012 #10

    tiny-tim

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    consider a small length dx, find the intensity from that (at a given point P), and integrate over all values of x from 0 to L
     
  12. Nov 7, 2012 #11
    Haha, i guess my question wasn't clear enough. :biggrin:
    That was the first thought came into my mind but then i got stuck at finding intensity. I know i have completed the chapter on Photometry but i am still a dumb at applying the formulas. How should i find the intensity? Should i use the "relevant equations" posted by me? :smile:

    Even if i use the equation posted by me, i am having difficulty finding F as function of solid angle, ω.
     
  13. Nov 8, 2012 #12
  14. Nov 8, 2012 #13

    tiny-tim

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    the intensity, from a point source, at distance r and solid angle ω is proportional to ω/r2 :smile:
     
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