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I am trying to measure the transmittance spectrum through a liquid sample using a visible-NIR spectroscopy (a spectrometer connected to a fiber optics and a collimating lens at the end), with tungsten-halogen light source (through another collimating lens) passing through the sample before reaching the lens connected to the spectrometer. From what I understand, transmittance can be calculated by dividing the measured spectrum with the light source spectrum.

enter image description here

I am able to measure the spectrum (intensity vs. wavelength) of light passing through the sample. However, when I tried to measure the source spectrum without the sample, the intensity exceeds the maximum that can be detected at all wavelengths (so the plot just shows the maximum intensity).

There is also an attenuator (screw) on the light source, which I can use to lower the intensity of light. But if I attenuate it further so that the light source spectrum without sample is measurable, the intensity through the sample will be very low that it is very close to 0.

Is there any suggestion on how to get the light source spectrum at high intensity?

I am thinking of getting the light source spectrum with high attenuation, so the intensity will be within the range that can be measured, and then scale the spectrum up using a constant based on the attenuation levels. For example, turn the attenuation screw by half-full rotation, measure the spectrum, then turn it more by half-rotation and measure the spectrum, and then find the scaling constant by dividing the two spectra. But I tried to measure two spectra at different attenuation levels, but the scaling is not constant across different wavelengths.

Am I on the right track? Or the spectra could not be simply scaled up by some constants?

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    $\begingroup$ This isn't an engineering question, it's a question about experimental design. As an (ex) experimental physicist this is exactly the sort of issue I had to deal with and is on topic for this site. $\endgroup$ Nov 22, 2023 at 8:08

4 Answers 4

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Many spectrometers can change measurement time. Sometimes labeled as "integration time". So, suppose you set light at minimum intensity suitable for measurement without sample. Then increase time with sample to say 10 seconds instead of 100 ms. Longer time generally comes with more noise, but measurement could be still good enough and is a button press away.

The second way is to use a known attenuator. If you don't have any, various ND filters aren't very expensive and come with measured transmittance spectrum. You can also measure attenuator's transmittance yourself, using the approach you are already using. You use unfiltered light compared to ND1 filter to get ND1 filter's response. For ND2, if spectrometer range is unsuitable for direct measurement you could use ND1 as the reference to measure ND2 and so on. Errors increase as does effort; it is easier and probably better to just trust the values you are given.

It doesn't really matter how flat filter's spectral response is - just that it transmits suitable amount of light in parts of spectrum of interest. It might be even better if its transmittance is inversely proportional to the amount of light in spectrum - eg for halogen lamp you might want something that transmits blue but attenuates green and red.

You can also stitch different wavelength ranges - eg take 400 to 700 from one measurement and 700-1000 from another (preferably with some overlap for validation). Using different filters, this can improve signal to noise ratio. However, if any part of spectrum is over-saturated, entire measurement might be wrong, not just the over-saturated part. Sure, it might also be just fine, but better be safe than sorry.

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  • $\begingroup$ “Longer time generally comes with more noise” — that's interesting.  Why does that happen?  (I'd naively expect the reverse: for noise to average out over longer timescales.  In visual photography, for example, it's the short exposures with ‘faster’ film that give more visual noise.) $\endgroup$
    – gidds
    Nov 22, 2023 at 17:10
  • $\begingroup$ @gidds I don't know. Dark current maybe? Maybe I was doing something wrong and let some stray stuff in that wasn't noticeable before? Note this is a different effect to shot noise - I am comparing say signal of 1000 units collected over 0.1 s or same signal collected over 10 s. 1000 units at 0.1s vs 2000 units at 0.2s would have the latter with less noise. It is just that long times (seem to) have somewhat more noise at same signal level. $\endgroup$ Nov 22, 2023 at 17:17
  • $\begingroup$ Thanks! The integration time and ND filters are new to me. I am able to get lower intensity with shorter measurement time, and will use the units counts/secs (instead of previously just counts) to calculate the transmittance. I also found a glass slide with measured transmittance spectrum and used it to lower the intensity further with known attenuation across wavelengths. Will consider to buy ND filters for future. $\endgroup$
    – smk
    Nov 22, 2023 at 19:29
  • $\begingroup$ Changing the integration time is often not a good idea in my experience, if you want to compare two measurements. You get more noise, but also you dont know what part of the background spectrum scales with time and which not. So you should not just compare two spectra with different integration time, or you should at least be very careful with that. $\endgroup$ Nov 22, 2023 at 21:34
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I think you are on the right track. What I did in a similar situation was to use some filter with a known attenuation. In my experiment I used glass plates with a finely ground surface. This should just scatter the light in a wavelength independent manner so for example if the transmission through the plate was 1% it would be 1% at all wavelengths. Then you can insert the plate, measure the intensity, and just multiply by 100.

Note however ground glass plates don't work with coherent sources like lasers as you get a speckle interference pattern that is pretty but useless for measuring intensity. You may have to experiment to find a suitable "standard absorber". Thin metal films (thin = 10 to 50nm) would probably work and should be fairly frequency independent as long as you stay in the visible part of the spectrum.

I'm a bit surprised using the adjusting screw caused changes in the spectrum, but it you were also adjusting the intensity of the source while measuring the effect of the screw you will probably find the spectrum changes with the intensity of the source.

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In my opinion your problem is not that the light source is too strong, but that your sample absorbs too much light.

Use a cuvette with a shorter lightpath or a lower concentration. I do not know what kind of cuvette you are using, but assuming you are using a standard 10 mm lightpath cuvette, a 0.2 mm lightpath cuvette will make your spectrum easier to obtain by magnitudes.

This is what I would do in our lab. If your sample is not in a cuvette, try to change its geometry. As the transmitted light decreases exponentially with pathlength, this is one of the easiest and most reliable ways to get a better spectrum.

There are also flow cuvettes, if this is what you need.

Whenever you change something of the light you use, this will affect the spectrum. I tested this today with a quartz cuvette and know this very well from mirrors. Optical density filters and all glass optics will alter the spectrum, too. Lowering the output of the source would alter the spectrum in most of the cases, too, except your source is exceptionally good, which it is often not for those applications (as you don’t need absolute, but relative intensities).

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I'd try replacing your sample with a plate with a hole in it. As long as the hole is more than a few wavelengths in diameter it should make a fine achromatic attenuator.

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    $\begingroup$ Besides chromatic effects of diffraction, a single small aperture (when sub-optimally placed) can make the system more susceptible to drift and enhance the chromatic effects of the condenser lenses which are almost certainly not achromats over the full range of the UV-vis system. Instead, ND filters based on arrays of dots (or holes) in a metal thin film on glass over a large area might be a better choice. However I'm having trouble finding a commercial example now in 2023, probably because they are useless in real imaging systems due to diffraction. I guess the continuous film ND filters rule $\endgroup$
    – uhoh
    Nov 22, 2023 at 23:44
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    $\begingroup$ @uhoh At the same time, I'm uncomfortable about ND filters because of the possibility that there could be nonlinearities which affect their transmittance at high intensities. It might be worth exploring whether an aperture placed in a broadly-colimated beam could be used to verify the performance of an ND filter for the final apparatus. $\endgroup$ Nov 23, 2023 at 8:18
  • $\begingroup$ @MarkMorganLloyd yep yep $\endgroup$
    – uhoh
    Nov 23, 2023 at 8:48
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    $\begingroup$ @uhoh Lots of small holes will diffract more than one larger one, and a regular array of them will be very obviously chromatic: it's a diffraction grating! A random array will still be chromatic, although not so obviously. $\endgroup$
    – John Doty
    Nov 23, 2023 at 13:48
  • $\begingroup$ @MarkMorganLloyd Yes. Use multiple methods and checks. $\endgroup$
    – John Doty
    Nov 23, 2023 at 13:49

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