Help! calculating light intensity

  1. This is the question

    Find the intensity of a light source that produces an illumination of 7.50 lux at a distance of 30m from the source.

    This what I get but it's wrong
    84823

    This is the formula

    E = Illumination
    I = luminous intensity of the source ( in lumens)
    r= distance between the source and the illuminated surface.

    1 lux = 1lm/m^2

    Formula E = I/(4*pi*(r^2)

    So I did I = E*(4*pi*(r^2)

    I =?
    E=7.50
    r = 30m

    I = 7.50lux*(4*pi*(30m^2)


    and got I = 84823

    rounded in many different forms and I still get a wrong answer.. Any help would be appreciate it.

    and is wrong because the computer practice test tells me but doesnt give me an answer.
     
    Last edited: Feb 25, 2008
  2. jcsd
  3. Andy Resnick

    Andy Resnick 5,815
    Science Advisor
    Education Advisor

    Blech. Radiometry: that evil, dismal science. :)

    First, you need to know more about the source- is it an isotropic point source? Extended Lambertian radiator? It matters. Second, check the units: 'lux' is a photometric unit, meaning the conversion to radiometric units (intensity) requires knowledge of the wavelength (or waveband!)

    As written, the question is potentially meaningless, because of the (possible) mix of photometric and radiometric units. As it happens, there is a photometric term also called "luminous intensity" in units of lm/sr (candela [cd]). First, let's parse the question:

    7.5 lux at a distance of 30 m: the incidance of light onto a surface 30 m away from the source is 7.5 lm/m^2. This implies (to me), that we have an isotropic point source freely radiating into 4*pi sr.

    It's possible the question instead refers to an extended radiator, with an excitance of 7.5 lx. If the radiator is Lambertian, the radiated power does not depend on relative angle between the radiant surface and the detector, which simplifies things. The total radiated power is 7.5 *source area (lm), the received power is that fraction of solid angle (out of 2*pi) subtended by the detector.

    Maybe the idea is to teach you how to convert areas into steradians? For small (solid) angles,[tex]\Omega = A/r^{2}[/tex]

    thus, 7.5 lx at 30 m converts to 8.3*10^-3 cd, if I did that correctly.....

    Like I said, radiometry and photometry can be incredibly confusing. Check out William Wolfe's "Introduction to Radiometry" (Tutorial Texts, SPIE press, vol TT29) for a really clear discussion.
     
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