Converting radiometric units (W/m^2) to photometric units (Lux)

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This approach may be more accurate and efficient compared to manually integrating Planck's law and using conversion factors for each individual wavelength.In summary, the conversation discusses the process of calculating the optical depth of the atmosphere using data collected by a lightmeter. The main challenge is converting illuminance values in lux to W/m², which becomes more complicated for a range of frequencies due to different conversion factors for each wavelength. Integrating a blackbody intensity curve or using a numerical integration method with the spectral energy distribution of solar radiation may be possible solutions.
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1. I am doing a lab about solar radiation. We collected data using a lightmeter which measures illuminance in lux. Now to calculate the optical depth of the atmosphere I either need to find an expression for the solar constant in lux or be able to convert illuminance values in lux to W/m². The conversion is straightforward for monochrome radiation but for a range of frequencies it becomes much more complicated because each wavelength has a different conversion factor. I've been told that integrating a blackbody intensity curve is a good place to start but any approximate methods for this conversion would also be welcome.

Homework Equations


I0 is the intensity of radiation at the source (Solar constant) and I is the observed intensity after a given path, then optical depth τ is defined by the following equation
[URL]http://upload.wikimedia.org/wikipedia/en/math/6/6/2/6622b2af6bba780c8a709106b0ec0f5b.png[/URL]
Solar constant: 1.361 kilowatts per square meter (kW/m²)
converting monochrome light:
1/683 watt of 555 nanometre green light provides one lumen: this is the peak of the weighting function and also the peak of the suns spectrum as a blackbody. All other wavelengths are "worth" progressively fewer lumens.

This website gives an idea of the problem but no solution for photometric to radiometric (and neither does the book he mentions in the last section)
http://www.optics.arizona.edu/palmer/rpfaq/rpfaq.htm

3. So far I have tried integrating Planck's law, by hand and using maple, between the visible light region limits to try to get the energy per second radiated by the sun but can't get the integral (integrating between 0 and ∞ gives the Stefan–Boltzmann law but other ranges end up with a reimann zeta function so cannot be done). I have assumed the sun is a blackbody at 5800K. I've been working on this for 2 full days but keep coming to dead ends no matter how i try to solve the problem.
 
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Homework EquationsPlanck's law:http://upload.wikimedia.org/math/e/d/f/edf8b2cba6ac45d6cfd5f7e9a0a61a3b.pngA possible solution is to use the spectral energy distribution of solar radiation (or other light source) and integrate it using a numerical integration method. This would involve using the spectral energy distribution in watts per square meter per nanometer to calculate the total energy in watts per square meter for a given wavelength range and then convert this to lux by multiplying it by the conversion factor for the desired wavelength range. For example, for the visible light region from 400nm to 800nm, you could use the conversion factor of 683 lm/W to convert the total energy in watts per square meter to lux.
 

1. What is the difference between radiometric units and photometric units?

Radiometric units measure the energy of electromagnetic radiation, while photometric units measure the perceived brightness of light by the human eye. This means that radiometric units take into account all wavelengths of light, while photometric units only consider the wavelengths visible to the human eye.

2. Why is it important to convert radiometric units to photometric units?

Converting radiometric units to photometric units allows us to measure the brightness of light in a way that is more relevant to human perception. This is especially important in fields such as lighting design, where the goal is to create a certain level of brightness that is comfortable for human eyes.

3. How do you convert radiometric units to photometric units?

The conversion between radiometric units and photometric units involves using a conversion factor called the luminous efficacy. This factor takes into account the sensitivity of the human eye to different wavelengths of light. The formula for conversion is: photometric units = radiometric units x luminous efficacy.

4. What is the luminous efficacy and how is it determined?

Luminous efficacy is a measure of how well a light source produces visible light relative to the amount of energy it consumes. It is determined by comparing the amount of visible light emitted by a light source to the amount of energy it consumes in watts. The higher the luminous efficacy, the more efficient the light source is at producing visible light.

5. Are there any tools or resources available for converting radiometric units to photometric units?

Yes, there are several online calculators and conversion tables available for converting between radiometric and photometric units. These tools use the standard luminous efficacy values for different light sources to make the conversion process easier. Additionally, most lighting design software programs have built-in conversion functions for easy conversion between units.

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