How to get Intensity over a range of wavelengths from a Total Intensity

This is a theory vs experiment problem for me I think in that - as usual - I think I know the theory, but when it comes time to run an experiment, I can't actually extract squat.

I am using the Cold White LED (MCWHL2). If you scroll down a bit, you'll see the LED normalized intensity vs. wavelength profiles for all the LEDs. The black line on this graph is my LED. As you can see, it meanders instead of being a simple Guassian-like thing.

I measure the intensity of this light by using this power meter: Scientech Astral AD30 meter - which displays power and energy - with an AC2501 detector. Specs are here . So I'm calculating total intensity by dividing by the area of the beam.

Now, here's my question: say I get a reading of $10\rm\frac{mW}{cm^2}$. If I want to know what is the intensity of the output only over the range 400nm to 455 nm, based on the Intensity vs Wavelength graphs and the characteristics of the meter and detector, what do I do?

Now, Planck's equation for a black body was where I first looked. My procedure was to find the total intensity for the light temperature (6500 K) and integrate from 0 to infinity. I then numerically integrated from 440 nm to 455 nm and divided the two. Multiplying this by the output on the detector gave a ridiculous solution. Given the intensity profile - not really black bodyish - the ridiculous results make sense.

It was at this point that I started searching and found nothing but Planck stuff. Thus, I humbly come before all of you. Please help! There must be some sort of standard procedure for those without specialized light analyzing equipment!

Thanks! Sam

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There is an Excel spreadsheet you can download on the page you linked. It has the numerical data that was used to produce that plot. Just download that spreadsheet, and integrate numerically. If you want to be really really insanely cautious, you could even interpolate the data and numerically integrate that.

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...or integrate using trapezoidal quadrature, etc. –  Colin K May 17 '11 at 17:24
In the future, you might want to know that Thor Labs has great customer service people, who have direct access to the engineers. If the spectral data wasn't on the web page, they would almost certainly send it to you if you called them up. –  Colin K May 17 '11 at 17:26
I'm becoming more and more of a fan of Thor as we use them. I'm also giving you the check because your answer supplies me with a direct solution. Can't believe I didn't notice that spec sheet! But Josh gave me a good way to treat this problem generally and Bjorn put my mind at ease about the detector being a non-factor. So thanks to all! –  pballjew May 17 '11 at 17:46
Being able to get through the customer service front end and contact an engineer or technician is a sign of a good technical supplier. And those guys generally understand your problem and want to help. –  dmckee May 17 '11 at 18:09

You are on the right lines with integrating, the problem lies in the fact that the intensity profile for your LED isn't particularly uniform. Clearly looking to the Planck function is going to give you problems because your LED does not emit like a blackbody.

My solution would be to produce an equation that roughly replicates what your LED does. Looking at the graph, you could do a sum of 3 or perhaps 4 Gaussian curves peaking at 450/530/560 ish (eyeballing). The curve would be something like:

$y = a(\exp(f_1(x))+0.4\exp(f_2(x))+0.3\exp(f_3(x))+ \cdots )$

where $f_n$ are the parameters for the Gaussian distributions for each peak. I'm using the fact that the 450nm peak is full intensity and the other peaks are scaled relatively by 0.4, 0.3 and so on. It's not too difficult to produce a peak with the correct x-shift and width. Playing around on Wolfram Alpha would do, or if you have access to Matlab/Mathematica then you could of course use them.

You can numerically integrate it for the area which will be a function of the scaling parameter, $a$. Find the value of $a$ that gives you an area of 10 and that's the equation you use. Then you can integrate between the wavelength values you're interested in. This is all without units, of course, but as it's all relative, it doesn't make a difference.

It seems like a little bit of a bodge, but unless you have the profile in equation form or in a table, I don't see why this wouldn't work.

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The white led is not a black-body source so you should just ignore the "Planck stuff" and use the specified profile.

Also, your intensity meter is (hopefully) calibrated so as long as your input to the calorimeter is within its range (266-1200 nm) you don't need to consider its spectral response either.

Just take the given spectral profile of the white LED and integrate the portion from 400 to 455 nm and divide by the complete integral. Multiply this ratio with your total intensity measurement. You can trace the curve in a suitable graphics program or something, or just approximate it as a trapezoidal shape for example by measuring some points with a ruler.

A spectroradiometer, as you imply, is a better tool for this but in your case you are saved by having a pre-measured spectral profile.

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To get fairly close, numerical integration appears to be the easiest solution. If you can give up a little accuracy, simply blow up the section from 400nm-700nm, overlay a grid, and enter height of the curve, from 420nm to 670nm, in an Excel spreadsheet for each 5nm slice. Then sum up the slices:

1.) w(0) = 420; I(0) = 0; Sum_Intensity(0) = 0; Sum_Target(0) = 0;

2.) For N = 1 to 50;

3.) w(N) = (420 + 5N); I(N) = < pick curve height off chart for wavelength = w(N) nm >

4.) Sum_Intensity(N) = Sum_Intensity(N-1) + 5 x ((I(N-1)+I(N))/2)

5.) Sum_Target(N) = Sum_Target(N-1)

6.) If 399 < W(N) < 456, Then Sum_Target(N) = Sum_Target(N-1) + 5 x ((I(N-1)+I(N))/2)

7.) Next N;

8.) 400-455nm power, as a percentage of total power = Sum_Target(50) / Sum_Intensity(50);

(Step 8 assumes intensities were measured in power or power/area units (e.g. mW/cm^2). Otherwise you need to square the intensity raio to get to the power ratio.)

This might look a little daunting at first, but should be easy to implement as a 52 row by 5 column-wide block in an Excel SS. Maybe take you two hours of blowing up the chart and picking off points and entering them into the SS. I would guess your accuracy to be better than +/- 3%.

To give you a check: my calibrated eyeball says you should expect Sum_Target(50) and Sum_Intensity(50) (which are the numbers you are interested in) to be ~ON THE ORDER OF~ 65 and 10, respectively. IF those numbers were actually 65 and 10, then your ratio of 400-455 power to total power would be 10/65 = 15.4%

Oh, and this all assumes that your receiver's response is fairly linear across the spectrum of 430-670nm. It appears that the manufacturer claims 1%, so you should be good to go.

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