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Converting from Lux to Irradiance? (halp)

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Re: Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 12:35 pm

Xenix wrote:Wow - I should really proofread what I write to check units. Yeah, it's actually lux in the denominator (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.

Yeah, I got that part right :D Thanks. Hopefully that equation is correct.
I basically only have measurements for 525nm and 395nm light right now but I am going to be calculating intensities and lux values for 395, 440, 505, 525, 565, and 605nm. After I finish with Theck's post I am going to attempt to use the CIE table (thanks for that, it is amazing, I don't know how I didn't manage to find it before, I had printed off the graph and spent like 20-30 minutes with a ruler very carefully estimating, I was pretty close too) to attempt to do the 395. That is, if I am right in using photopic instead of scotopic (working on that part).
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Re: Converting from Lux to Irradiance? (halp)

Postby Xenix » Fri Apr 09, 2010 12:38 pm

NOTE: The selenium cell in the OS-9152B does not have a corrected photoptic response. The instrument is calibrated to a 2700° K tungsten filament lamp, and will indicate the correct intensity only for that source. For other light sources no calibration is specified. However, when that same source is used for all measurements the instrument will provide accurate relative intensities, and approximate absolute(lux) intensities.


Heh - what that PDF said is that it doesn't conform to any standard. That means those tables we gave you won't be accurate for it. You'll probably have to estimate values for K manually off that spectral response plot on page 1 like you're already doing. Or, you could use something like Digitizer to generate the numbers from a screenshot of the plot automatically.
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Re: Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 12:47 pm

theckhd wrote: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.

This was really helpful. That makes sense about the integral only being needed for multiple wavelengths. I had been wondering where that comes in to play (I was ready to call Vanifae in to my office to come help me). I like to brag about how I had through Cal III in high school, but that means I didn't do any math at all in undergrad, was not looking forward to figuring out the integrals (my plan was to just take the function to a graphing calculator though, like a real man).

So I understand pretty much all of this except the y. I was going strong and then I clicked that link and felt like Aubade's reply. I don't really understand what that online program/site is supposed to output, but I am assuming it isn't the same as the values from the CIE photopic curve (http://members.misty.com/don/photopic.html)?


@Xenix: I think the spectral response is just the device's relative sensitivity to different wavelengths (let me know if I am wrong on this though). So I have been taking my measurements and multiplying them up to 100% based on estimating from that curve (for example, with my 395nm ones I would double the lux reading before doing any calculations).
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Re: Converting from Lux to Irradiance? (halp)

Postby Xenix » Fri Apr 09, 2010 12:56 pm

Yep - what you're doing is the same thing as dividing by the value of the curve. Once you scale the readings by that number, you should be able to just multiply by the conversion factor.

All the photopic curve is is a standard human's eye's sensitivity to different wavelengths of light in daytime. It's basically a spectral response curve that lets everyone calibrate their sensors to the same readings. Yours, however, is not calibrated by one of those curves, it's response is given in that pdf you linked.

The link he gave you gives you values from a standard curve, either absolute or using a log scale.
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Re: Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 1:06 pm

Xenix wrote:Yep - what you're doing is the same thing as dividing by the value of the curve. Once you scale the readings by that number, you should be able to just multiply by the conversion factor.

All the photopic curve is is a standard human's eye's sensitivity to whatever wavelength of light it is. It's basically a spectral response curve that lets everyone calibrate their sensors to the same readings. Yours, however, is not calibrated by one of those curves, it's response is given in that pdf you linked.

The link he gave you gives you values from a standard curve, either absolute or using a log scale.

I'm trying to algebra it out with little success, so is the equation Theck gave and the one you gave just two different versions of the same equation, one using the irradiance/lux 555nm value and one using the lumens/W 555nm value? (If you say yes I will just take your word for it instead of mathing it out, I think it will be too much for me to handle, I still have a lot more maths to do once I solidify what equation to use)

I think I am good for now, unless either of you has anything else to add? I'm going to start making notes in my lab manual and calculating things out.
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Re: Converting from Lux to Irradiance? (halp)

Postby Xenix » Fri Apr 09, 2010 1:11 pm

We're both using the same conversion (and the same equation) - I'm multiplying by 0.001464 (W/m^2)/lux and in his, he's dividing by 683.002 lux/(W/m^2) - they're both the same thing - the 555 nm conversion between W/m^2 and Lux. One is just inverted.

To put it simpler: Take the reading from your meter (in Lux), then either multiply by .001464 or divide by 683.002. That will give you an irradiance value in W/m^2 consistent with a 100% spectral response at 555 nm (yours has it's 100% response at around 570 from the look of the plot, which might change that conversion slightly, but I'll assume it doesn't). Then, divide that number by the spectral response percentage (from the plot in the pdf) at whatever wavelength the light actually was at to get the actual irradiance for that wavelength and lux reading.
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Re: Converting from Lux to Irradiance? (halp)

Postby theckhd » Fri Apr 09, 2010 1:32 pm

Xenix wrote:
NOTE: The selenium cell in the OS-9152B does not have a corrected photoptic response. The instrument is calibrated to a 2700° K tungsten filament lamp, and will indicate the correct intensity only for that source. For other light sources no calibration is specified. However, when that same source is used for all measurements the instrument will provide accurate relative intensities, and approximate absolute(lux) intensities.


Heh - what that PDF said is that it doesn't conform to any standard. That means those tables we gave you won't be accurate for it. You'll probably have to estimate values for K manually off that spectral response plot on page 1 like you're already doing. Or, you could use something like Digitizer to generate the numbers from a screenshot of the plot automatically.

I don't think that's what it means, actually.

By saying it "doesn't have a corrected photopic response," I think it means that it won't give you correct values for an arbitrary input spectrum. Their statement about how it's calibrated seems to support this.

In other words, they calibrated it so that if you're using a 2700K tungsten filament lamp as your source, it gives you the correct Luminous Flux value. But if you use a monochromatic source it won't - it's just a selenium detector in there that's generating a photocurrent, but it is assuming that the photocurrent is generated by tungsten lamp, so it's using a different numerical constant to turn photocurrent into lux.

By saying that no calibration is specified for other sources, they basically say that any other measurement will be strictly approximate. However, relative measurements should be fine - if you use a 500nm source and measure the source itself and the transmitted intensity through an optical system, the relative transmission should be correct.

Have to head home now, I'll try and stop in later to post some more.
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Re: Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 1:36 pm

theckhd wrote:
Xenix wrote:
NOTE: The selenium cell in the OS-9152B does not have a corrected photoptic response. The instrument is calibrated to a 2700° K tungsten filament lamp, and will indicate the correct intensity only for that source. For other light sources no calibration is specified. However, when that same source is used for all measurements the instrument will provide accurate relative intensities, and approximate absolute(lux) intensities.


Heh - what that PDF said is that it doesn't conform to any standard. That means those tables we gave you won't be accurate for it. You'll probably have to estimate values for K manually off that spectral response plot on page 1 like you're already doing. Or, you could use something like Digitizer to generate the numbers from a screenshot of the plot automatically.

I don't think that's what it means, actually.

By saying it "doesn't have a corrected photopic response," I think it means that it won't give you correct values for an arbitrary input spectrum. Their statement about how it's calibrated seems to support this.

In other words, they calibrated it so that if you're using a 2700K tungsten filament lamp as your source, it gives you the correct Luminous Flux value. But if you use a monochromatic source it won't - it's just a selenium detector in there that's generating a photocurrent, but it is assuming that the photocurrent is generated by tungsten lamp, so it's using a different numerical constant to turn photocurrent into lux.

By saying that no calibration is specified for other sources, they basically say that any other measurement will be strictly approximate. However, relative measurements should be fine - if you use a 500nm source and measure the source itself and the transmitted intensity through an optical system, the relative transmission should be correct.

Have to head home now, I'll try and stop in later to post some more.

Fortunately I don't have to go below .5 relative sensitivity, so I'm pretty happy with that. I just want these light measurements to be over with D: It has been such a big hassle going through peoples' old work and realizing that they used crappy equipment and their measurements are completely false, then having to redo everything from the beginning multiple times because others' before me didn't know what they were dealing with.
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Re: Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 2:10 pm

Okay so confirming (.001464/K_lambda)*lux measured is correct?
I had been using the messed up one and getting closer to what I should have been D: So I guess I was doing it right to begin with (but now I understand it a lot better). Which means I really am super high over what I should be (I'm wanting .03uW/cm^2 but I am getting between .5 and 1 uW/cm^2). Sad times :(

I guess now what I am wondering is how the spectrometer measurements and calculations I did so many months ago went wrong...
I haven't done the 395nm light yet, so hopefully that might give some "better" results.
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Re: Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 2:49 pm

I have the conversions down, but if you guys want to go a bit further... (tl;dr=does this seem like a usable piece of equipment?)

What I am doing right now is using this photometer to check if my lights are the same intensity. I have one set of 525nm lights and one set of 395nm lights. A few months back I borrowed a spectrometer from a physicist to take measurements, then he gave me back calculations for intensity (it is possible that something went wrong). So I am checking those intensities to see if they are the same.
In theory I should be able to calculate a lux value for the intensity value I want and use the photometer (only equipment I have access to) to set the lights to that lux value (I can adjust the voltage of each light by use of a dial).

So good news and bad news. The good news is that the light intensities of the same wavelength are pretty consistent with each other (all the 395s are similar and all the 525s are similar). The bad news is that the intensities I am getting are completely higher than what they should be (.03uW/cm^2). 525s are at about 1.5 irradians and 395s range from 60 to 150 irradians. Is it possible that this huge difference in 395 could be due to the lack of sensitivity both by human vision (the CIE curve) and the photometer (only reading about 50%)? As in, do you think the device+conversion might not be accurate at such low wavelengths but more accurate/proper to use at higher ones around the 500 range?

I know that it's possible (quite probable actually) that these "absolute measurements" from the photometer are inaccurate (the thing has probably been sitting on a shelf for 10 years or more) but to what degree should the relative measurements still be alright? What I am trying to determine now is if this equipment is even good enough to be used. (So maybe if you guys have any experience working with various types of light measurement equipment and are familiar with their accuracy and precision)

Even if my current irradiance values are incorrect, if I were to pick an irradiance measurement, calculate the appropriate lux value, and used this photometer to set the lights to that lux measurement, would/should that be relatively correct (as in, the lights would at least all have the same intensity)?
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Re: Converting from Lux to Irradiance? (halp)

Postby mew » Fri Apr 09, 2010 3:31 pm

Sorry about posting on this again. One thing the Yahoo answers guy mentioned was that the LEDs are measured in candela. My lights are LEDs and I do have access to the candela values for each light. Would there be some better way to do these conversions using candela instead or in addition?
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Re: Converting from Lux to Irradiance? (halp)

Postby theckhd » Sat Apr 10, 2010 6:05 am

Ok, so there are a lot of things we need to discuss here.

1) In my haste, I made an error. The following statement is wrong:
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).


I mistyped the units just about everywhere there by accident. It should read:

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

2) The formula should be correct, yes. Intensity (in W/m^2) = lux/(683*y).

3) It looks like this device gives you Illuminance, but that's not what it measures. It's a selenium photovoltaic, so it's obviously measuring energy. It has to do some internal calculations in the electronics to convert it to lux. Again, their brief blurb about the calibration reinforces this.

This is good news for you, because it means that the device is actually measuring intensity, multiplying by 683 and some other value y' to get lux, and showing you that on the dial. The value it uses should be reverse-engineerable if you look up the power spectrum of a 2700K tungsten lamp (probably find-able online) and perform the integration I posted earlier. Once you find that value, you can compute the equivalent value for a monochromatic source at any wavelength and calculate the scaling factor needed to "correct" the approximate result the device is giving you. I can help you do that if you decide to go that route.

4) The fact that the intensity is high may not mean anything about the device. Intensity is power divided by area, so the farther away from the source you are, the weaker the intensity (because you're spreading the same amount of power over a larger area). So if you were taking your measurements with the LED very close to the detector, you would get a much higher reading than you would if it's far away. And you would get still different results if you measured with the "fiber optic" probe they mention (which I'm willing to bet isn't really an optical fiber at all, just a simple light pipe).

So the device should be just fine for determining relative intensities as long as the optical set-up is the same (LED in the same position, same distance from the detector, and pointing exactly the same direction). But to compare those results to the ones your physicist friend gave you, you'd need to know exactly how he took the measurements.
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Re: Converting from Lux to Irradiance? (halp)

Postby theckhd » Sat Apr 10, 2010 6:15 am

mew wrote:Even if my current irradiance values are incorrect, if I were to pick an irradiance measurement, calculate the appropriate lux value, and used this photometer to set the lights to that lux measurement, would/should that be relatively correct (as in, the lights would at least all have the same intensity)?

I think before I can give you a definite answer to this (or the candela question), I need to know more about what you're trying to do. If you want lights of different wavelengths to have the same intensity, you're going to need to use the calibration curve and/or do some numerical integration like I described earlier.

It might be easier if you describe what exactly you're trying to measure with this set-up. That way I can see if your procedure and this device will accomplish what you want, or if there's an easier/better way to take the measurements.
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Re: Converting from Lux to Irradiance? (halp)

Postby Ruex » Sat Apr 10, 2010 5:15 pm

Fuck the consistency: Use Foot-candles.
(random inflammatory blanket statement)
... Maybe furlong-candle would be more random?
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Re: Converting from Lux to Irradiance? (halp)

Postby Arcand » Sat Apr 10, 2010 9:18 pm

Ruex wrote:Fuck the consistency: Use Foot-candles.
(random inflammatory blanket statement)
... Maybe furlong-candle would be more random?


Cubit-candles.
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