Converting from Lux to Irradiance? (halp)

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Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 10:45 am

So I need to convert from lux to irradiance. I have my lux readings and I know all the wavelengths of light I am converting (only single wavelength lights). Wiki says that if you know your wavelengths you can multiply by a conversion factor http://en.wikipedia.org/wiki/Lux#Relati ... irradiance but I'm not sure how to get the conversion factor
I asked on yahoo answers and got a decent reply:
http://answers.yahoo.com/question/index ... 856AASvwhH
but yahoo answers isn't the most reliable thing and the guy explained lumen to irradiance conversion and just mentioned that you use the same process for lux but I want to be sure that I am doing it right.
Basically instead of using the 683lux/W I'm using .001464W/m2 per lux taken from wikipedia, but I don't think that is correct (the units don't feel right) and I'm not sure what else I could use instead.
My brain seems to have burnt itself out due to worry ("I really fucking hope that this number comes out like I want it, why isn't it coming out like I want it? This isn't right at all. My numbers are all off. The irradiance shouldn't be so high. Maybe I am doing something wrong? But I don't fully understand what I am doing. Let me try this instead. I hope this number comes out right..." repeat)

I also don't understand why this conversion (using the curve from the luminosity wiki) is based on the human visual sensitivity curve and how I would go about converting something below 400nm (I emailed the guy asking this but he hasn't gotten back to me yet).

So, it is probably obvious, I am getting a little desperate for some help or checking :( Any help would be nice. Also, if anyone is super helpful I will bake and mail you some cookies (any kind you want).

Edit: Okay so calming down and reading through the yahoo answer again, it says I just do literally the exact same thing and I get the correct units? I will trust that and just go with it and see what happens.
Last edited by mew on Fri Apr 09, 2010 10:55 am, edited 1 time in total.
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Re: Converting from Lux to Irradiance? (halp)

Postby Vanifae » Fri Apr 09, 2010 10:49 am

I don't know.
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Re: Converting from Lux to Irradiance? (halp)

Postby Arcand » Fri Apr 09, 2010 10:54 am

Vanifae wrote:I don't know.


Cookies otw. :)
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Re: Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 10:56 am

Arcand wrote:
Vanifae wrote:I don't know.


Cookies otw. :)

I know right D: He gets all the cookies he wants without being helpful.
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Re: Converting from Lux to Irradiance? (halp)

Postby Vanifae » Fri Apr 09, 2010 11:04 am

I stole more cookies last night.
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Re: Converting from Lux to Irradiance? (halp)

Postby Aubade » Fri Apr 09, 2010 11:16 am

mew wrote:
Arcand wrote:
Vanifae wrote:I don't know.


Cookies otw. :)

I know right D: He gets all the cookies he wants without being helpful.


Lucky bastard
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Re: Converting from Lux to Irradiance? (halp)

Postby Arcand » Fri Apr 09, 2010 11:30 am

Post one of your test calculations, and I'll see if I can follow it? (If you do, tell me what luminosity function you picked.)

Or it might be easier to wait for Theck, he's a laser guy.
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Re: Converting from Lux to Irradiance? (halp)

Postby Xenix » Fri Apr 09, 2010 11:52 am

I'm not a laser guy, but after a bit of reading, it doesn't look too complicated.

Edit: Had the constant in the numerator, not the denominator as it should have been.

First, grab a table of your lux values and your wavelengths.
Then, go look up the constants associated with your wavelengths (I'll call it K_lamba) from some standard Luminous Efficiency Function like this one. (The watt/meter^2 -> conversion is 100% of that conversion factor for 555 nm light, less for other wavelengths - that table gives you those percentages).

Then, sum up (0.001464 W/m^2/lux / K_lambda) / (lux measured) for each wavelength. That should give you the total irradiance in W/m^2.

As an example, if you had 1000 lux at 550 nm, that would contribute 1.464 W/m^2. If it was instead at 510 nm, where K_lambda = 0.5030000, you would have needed 1/.503 = 1.988 times as much irradiance to get that much Lux, or 1.464/.503 = 2.911 W/m^2.
Last edited by Xenix on Fri Apr 09, 2010 11:59 am, edited 1 time in total.
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Re: Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 11:57 am

Arcand wrote:Post one of your test calculations, and I'll see if I can follow it? (If you do, tell me what luminosity function you picked.)

Or it might be easier to wait for Theck, he's a laser guy.

I'm kind of just following that Yahoo answers guy's advice and following the graph on http://en.wikipedia.org/wiki/Luminosity_function
So one of my wavelengths is 525nm, which is about 80%. So I took 80% of 683 lum/W and it is 546.4 lum/W. My lux reading is 7.12lux, so I divided that by 546.4 and got .013W/m2 (at least I assume it is W/m2, according to what the guy said). Now I convert that to my units (uW/cm2) and it becomes 1.303uW/cm2. Another measurement I did just came out to .567uW/cm2.

The problem is that these measurements should be closer to .03uW/cm2. A while ago I did some readings with a spectrometer, took them to a physicist, and the physicist told me what equation to use and how to do the multiplication conversions, and it should have made the light intensity .03uW/cm2. But according to this I was actually at a .5 minimum

@Xenix: I have to afk for a minute but I will come back and take a few minutes to wrap my head around this and attempt something and give you some sort of reply :O
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Re: Converting from Lux to Irradiance? (halp)

Postby Talderas » Fri Apr 09, 2010 11:58 am

Vanifae wrote:I stole more cookies last night.


So did you get caught with your hand in the cookie jar?
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Re: Converting from Lux to Irradiance? (halp)

Postby theckhd » Fri Apr 09, 2010 12:13 pm

Theck to the rescue.

Lux is a photometric unit, meaning that it describes how things are perceived by the human eye. The contrasting type of unit (W/m^2) is a radiometric unit, which is a more "absolute" measurement since it's a direct measurement of electric field (and thus energy or power).

Most of us in Optics prefer to work with radiometric units so that we're working in SI units. The last time I used photometric units was when we first learned about them in a Radiation & Detectors course.

The relationship between them is the luminosity function, which describes how your eye responds to optical radiation. In other words, if your eye is very sensitive to a particular wavelength, the luminosity function is large; if your eye isn't very sensitive, the luminosity function is small.

Generally you would use the equation given in the Luminosity Function wikipedia article to convert:
Image

J here is the spectral power distribution of the source, and y is the normalized luminosity function. For a broadband source (with many wavelengths in some arbitrary composition) you'd need to actually perform this integral. However, you have a monocrhomatic source, which makes it much simpler. J is what's known as a Dirac Delta function centered at the wavelength of your light source. So the integral simplifies to

F=683.002 (lm/W) * y * P

with y being the value of the Luminosity function for your particular wavelength and P being the total power of your light source.

In your case you have F (in lux) and you want to find P. Technically you have Illuminance (lux/m^2) instead of Luminous Flux (lux), but the conversion from Luminous Flux (lux) to Power (W) is identical to the conversion from Illuminance (lux/m^2) to Intensity (W/m^2).

So in your case, you'd have:
P = F / (683.002 * y)

The last step is to find y, which you should be able to find online (this ucsd link from the wikipedia article should help you do that). This will require a little more detail about Luminosity functions though.

There are two different standard luminosity functions - a photopic function and a scotopic one. Photopic is for "light-adjusted" eyes, while scotopic is for "dark-adjusted" eyes. See, your eye responds to light differently when you're being bathed in light than it does when you're in the dark, partly due to our caveman/hunter predecessors. (Aside: It also takes time to go from one mode to the other, which is why it's hard to see anything for 10-20 seconds right after you turn the lights off in a room - your "dark vision" hasn't kicked in yet. Similarly, it's why turning on a light briefly in a dark room ruins your dark vision for a brief period.)

So you want to make sure you're using the right luminosity function. I'm assuming you want to use the photopic one, but you'll have to check with whatever device you used to take your Illuminance (lux/m^2) measurements - I would assume the manual will tell you what it's using.
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Re: Converting from Lux to Irradiance? (halp)

Postby Aubade » Fri Apr 09, 2010 12:16 pm

Theck, what have you done?!

MY BRAINNNNNNNNNNNNNNN
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Re: Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 12:19 pm

Xenix wrote:I'm not a laser guy, but after a bit of reading, it doesn't look too complicated.

Edit: Had the constant in the numerator, not the denominator as it should have been.

First, grab a table of your lux values and your wavelengths.
Then, go look up the constants associated with your wavelengths (I'll call it K_lamba) from some standard Luminous Efficiency Function like this one. (The watt/meter^2 -> conversion is 100% of that conversion factor for 555 nm light, less for other wavelengths - that table gives you those percentages).

Then, sum up (0.001464 W/m^2/lux / K_lambda) / (lux measured) for each wavelength. That should give you the total irradiance in W/m^2.

As an example, if you had 1000 lux at 550 nm, that would contribute 1.464 W/m^2. If it was instead at 510 nm, where K_lambda = 0.5030000, you would have needed 1/.503 = 1.988 times as much irradiance to get that much Lux, or 1.464/.503 = 2.911 W/m^2.

Sorry quick question, still working on this, do you mean .001464W/m^2? The wiki gives 1.464mW/m^2 (which is a bit odd since it's not the regular form).
I can understand the equation but the bottom part is messing with me a little bit. Why are you dividing 1.464 (irradiance per lux at 555nm) by .503 (the K_lambda for 510nm) and what exactly does the resulting 2.911 mean?

So I just tried the equation you gave me (0.001464 W/m^2/lux / K_lambda) / (lux measured) on the lux measurement that (using whatever method I was doing before) previously yielded 1.303 (supposedly in uW/cm2) and it is giving me .0259 uw/cm^2 :D This is much nicer, I will do this for the rest of them and see how it turns out.

Haha, Theck give me a few minutes now, I will try to understand that and get you a reply.

Edit: Quick reply to Theck real fast- I wanted a Gigahertz Optik radiometer for the lab but it would have been like $2000 and my supervisor didn't want to spend that much :( So I am settling with a 1980something photometer we had sitting around :V
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Re: Converting from Lux to Irradiance? (halp)

Postby Xenix » Fri Apr 09, 2010 12:22 pm

Wow - I should really proofread what I write to check units. Yeah, it's actually lux in the numerator (math is right, formula is wrong): (0.001464 W/m^2/lux / K_lambda) * (lux measured). More lux = more radiant energy incoming, not less.

Basically, one lux of whatever frequency of light is equal to (0.001464 / K_lambda) W/m^2.
Last edited by Xenix on Fri Apr 09, 2010 12:34 pm, edited 1 time in total.
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Re: Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 12:31 pm

Still working on going through your reply. Right now I'm working on the photopic v scotopic part. Check out this badass-mofo of a photometer (I am open to the possibility that all my measurements are complete BS based on the quality, or lack thereof, of this equipment):
https://docs.google.com/fileview?id=0B9 ... NDI1&hl=en

All I'm finding is that it "does not have a corrected photopic response" (in reference to its callibration) but it doesn't mention anything about scotopic. So I am assuming that means it is photopic?
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