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Photometry / radiation question

  1. Aug 28, 2004 #1
    Hey guys (and girls)

    I've got a rather annoying question.

    If you were given the radiant power (45 mW) and luminous flux (22 l), and then was told it was spread uniformly spread over a hemisphere. (a) how do find the luminous intensity??, (b) the illuminance 1.8m away, and (c) illumance 5.0 cm straing out, and 3.0cm to the side.

    The first questioj is really annoying me, any help??

    Thanks
     
  2. jcsd
  3. Aug 28, 2004 #2

    HallsofIvy

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    What about the rest of us?
     
  4. Aug 29, 2004 #3

    Galileo

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    Phew, all those funky optical jargon units are pretty annoying sometimes.
    Let's see:
    Radiant energy is [itex]Q_e[/itex], so radiant power is probably [itex]\frac{dQ_e}{dt}[/itex], measured in Joules and J/s (Watts).
    While luminous energy is measured in lm-s (talbot)
    and the luminous flux is [itex]\frac{dQ_v}{dt}[/itex] where [itex]Q_v[/itex] is the luminous energy. The unit of flux is lm (lumen).
    the Luminous Intensity (candlepower) [itex]I_v[/itex] is lm/sr (lumen per steradian) or cd (candela).

    You have to have some formula for switching from photometric units to radiometric units. According to my opicts book it's:
    [tex]photometric unit = K(\lambda)\times radiometric unit[/tex]
    where [itex]K(\lambda)[/itex] is the luminous efficacy.

    The first question is not so tough. Since the luminous flux is 22 lm and the luminous intensity is lm/sr and a hemisphere has [itex]2\pi[/itex] steradian, the luminous intensity is [itex]11\pi[/itex] lm/sr.

    You really don't need the radiant power after all. :grumpy:
     
    Last edited: Aug 29, 2004
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