Besides spectral analysis of chemical substances, Raman spectroscopy can also be used as a non intrucive thermometer. From the raman-spectrum of a substance it is possible to determine its temperature or temperature changes using three different methode:
1) The ratio of stokes and anti-stokes intensities is determined by the ratio population of energy levels which intern are determined by the temperature of the sample.
2) The position of certain peeks might shift due to the temperature changing the strenght of the bands in molecules.
3) Line broughtening due to temperature increases.
Here are some works to examplify how raman spectroscopy is used to determine temperature:
https://www.spectroscopyonline.com/view ... hermometry
https://link.springer.com/article/10.10 ... 20-01356-z
I am currently trying to modify the cuvette holder to change and read out the temperature of the sample. For the reading out part I though about simply drilling a hole in the cover and sticking a temperature sensor through it. The temperature controling part is a little more difficult since I intende to use peltier elements to both cool and heat up the sample. If anybody else is interested in this topic I will gladely post my progress on here and discous with you.
Raman Thermometer
Re: Raman Thermometer
That looks very interesting -- thank you for popping it up
I'd be particularly interested in gaining more information on peak broadening with temperature because I'm pretty much convinced that the current resolution is limited by the experiment and not so much by the spectrometer itself.
nb: you won't be able to get anti-stokes peak with OpenRAMAN -- that would require a few changes
I'd be particularly interested in gaining more information on peak broadening with temperature because I'm pretty much convinced that the current resolution is limited by the experiment and not so much by the spectrometer itself.
nb: you won't be able to get anti-stokes peak with OpenRAMAN -- that would require a few changes
Re: Raman Thermometer
Glad you liked the idea. I am currently trying to modify the setup to get the anti-stokes lines aswell. I have removed the edgepass filter and put a mirror in place of the cuvette holder to get the laser light to the camera. Then I oriented the grating to have the 532 nm light hit right in the middle of the picture. This is what I recorded. As you can see there is a second wavelength to the left of the central wavelength that appears in the camera. Does this have anything to do with lasermodes? can I get rid of this in any way?
- Attachments
-
- Laserpointer.png (159 KiB) Viewed 11669 times
Re: Raman Thermometer
It's a bit oversaturated to tell for sure but it does indeed look like a main mode + some side lobes of the laser.
Also, if you want to see anti-stokes peaks you have to replace the dichroic mirror by a 50:50 beam splitter. The dichroic will prevent light below 540~550 nm from reaching the sensor and it will just deflect it back to the laser. You could use a dichroic that reflects above ~520 nm and let below pass through but then you won't see the stokes anymore.
Note that a beamsplitter will cut your signal by 75% and anti-stokes is already expected to be order of magnitudes weaker in intensity because it requires the emitter to be in the excited state first which is less likely than being at the non-excited state.
Also, if you want to see anti-stokes peaks you have to replace the dichroic mirror by a 50:50 beam splitter. The dichroic will prevent light below 540~550 nm from reaching the sensor and it will just deflect it back to the laser. You could use a dichroic that reflects above ~520 nm and let below pass through but then you won't see the stokes anymore.
Note that a beamsplitter will cut your signal by 75% and anti-stokes is already expected to be order of magnitudes weaker in intensity because it requires the emitter to be in the excited state first which is less likely than being at the non-excited state.
Re: Raman Thermometer
I managed to get a Beamsplitter with T=32%. It fit perfectly in the dichroic mirror holder. I also tried to activate the cameras color correction. I took two images. One with the ethanol filled cuvette and one with no cuvette in the holder but the laser running. I removed the laser runing image from the sample image and tried to find the stoke and antistoke signal. I am somewhat skeptical that those really go together, because it is after all a rather strong antistoke peek I chose. If I used the methode provided by Tuschel (the one where the device counts photons) I get a temperature of 364,4 K or about 91,2 °C. The ethanol was supposed to be room temperature however. When I get the middle between the two chosen peeks I get a wavelength of 527,7 nm. Since the laser operated in different modes it seems not unlikely that it would have such a the middle between the two frequencies might lie there. I will see if I get any other ideas but here are some of my first results for you to see.
- Attachments
-
- Herunterladen_28_06_23_2.png (24.21 KiB) Viewed 11661 times
Re: Raman Thermometer
It's pretty noisy and there is a large influence of the laser itself.
I checked to reproduce your experiment but using a notch filter to remove the laser line but unfortunately it's horribly expensive (>$500 at Thorlabs)...
Can't you use peaks that are farther away from the laser line ?
I checked to reproduce your experiment but using a notch filter to remove the laser line but unfortunately it's horribly expensive (>$500 at Thorlabs)...
Can't you use peaks that are farther away from the laser line ?
Re: Raman Thermometer
As I am using python for the data analysis I found a program that searches for peaks in a sample. Further more their is another function that calculates the prominence of a peak once it is found. When the program finds a peak I use its prominance as values and set everything else to 0. I hope this helps with the the strong background. I have here results for acetone. Once with the normal data(I removed the background and the laser contributions) and once with the peak prominences. The temperature of the acetone was determined to be around 304 K which is not unrealistic given that I recorded it on a hot day. If somebody has objections to my methode please let me know where I went wrong.
- Attachments
-
- Raman_pair_2023_7_4_9_2_27.png (22.71 KiB) Viewed 11644 times
-
- Intensity_spectrum_2023_7_4_9_39_33.png (27.89 KiB) Viewed 11644 times
Re: Raman Thermometer
I have found out that we had a notch filter available. It was a notch filter for 533 nm with 17 nm FWHM. I tested the filter by putting it at the place of the edge filter and recorded the camerapicture when there was only ambient sunlight. There was indeed a band of lower intensity on the camera (see image). Next I attempted to redo the measurement with ethanol. What I got was a tall peak at around 535 nm. I suppose that most of this light should be blocked. When I tried out an empty cuvette in the holder I found that it probably came from a reflection on the cuvette glass with the ethanol probably scattering or absorbing some of the light with lower wavelenght. However I was still surprised, that some light reached the camera. Especially since this light should be blocked by the filter. I am not sure if this might not be a secondary reflection of the beamsplitter. In any case I wanted to show it here in case anybody else has any idea why this light came through.
- Attachments
-
- Intensity_spectrum_2023_7_11_16_36_22.png (28.78 KiB) Viewed 11612 times
-
- Herunterladen_12_07_23_3.png (34.6 KiB) Viewed 11612 times
-
- notch_filter_2.jpg (222.88 KiB) Viewed 11612 times
Re: Raman Thermometer
That's to be expected.
Raman is pretty weak so we use very high exposure time and gain. But if you check regular scattering (not even to mention reflection) the order of magnitude is typically ~1:10^9 so an OD of ~9 or close to 9. Typical filters have rejection of OD 6 to 7 so even with the edge filter, or notch filter, you can expect to measure direct laser light about 100x more intense than the Raman peaks.
In the OpenRAMAN original (unmodified) version, you get and extra OD ~2 (slightly less actually) from the dichroic mirror and the premium edge pass filter is OD 7 so you reduce the elastically scattered laser light to about the same magnitude as the Raman signal.
Raman is pretty weak so we use very high exposure time and gain. But if you check regular scattering (not even to mention reflection) the order of magnitude is typically ~1:10^9 so an OD of ~9 or close to 9. Typical filters have rejection of OD 6 to 7 so even with the edge filter, or notch filter, you can expect to measure direct laser light about 100x more intense than the Raman peaks.
In the OpenRAMAN original (unmodified) version, you get and extra OD ~2 (slightly less actually) from the dichroic mirror and the premium edge pass filter is OD 7 so you reduce the elastically scattered laser light to about the same magnitude as the Raman signal.
Re: Raman Thermometer
I tried to get a signal with other substances. Since Isopropanol proved to have the stronges of all signals I figured that I might try out this. And as it turns out, I have finally gotten an Antistokes line. Some of the laser light still got through however I now have a really strong Stokes signal too. I needed to find the prominence of those peaks in order to determine their intensity compared to the baseline because after the removal of the background, I had negative values on the tail end of the spectrum. The program I utilized made use of the following algorythm:
1) Find a current peak
2) Draw a horizontal line starting from the peak and going to the left and to the right.
3) If you reach the end of the dataset or a predetermined window or some values that are higher than the current peak, stop.
4) To the left and the right find the minima in both reagions defined by the horizontal lines. Those are now the bases of the peak.
5) Calculate the prominence by subtracting the higher value base from the value of the peak.
This is a great methode to determine the location of certain signal, however for my goal I needed to know the prominence from the baseline after removal of the background. I therefore used the lower base to calculate the prominence. Using this I calculated the temperature again and got to 4,77 °C. I expected the result to be close to room temperature since the liquid was just sitting there with me in the same room. Also i noticed that the middle wavelength between the peaks (~537 nm) is not the same as the maximum wavelength of the laser (~533 nm). Could that be because of different laser modes? Is part of the Stokes signal maybe due to Fluorecence? I have set the camera to its maximum exposure time (about 30 s) and I have put down the gain to 0 dB. Help is as always appreciated.
1) Find a current peak
2) Draw a horizontal line starting from the peak and going to the left and to the right.
3) If you reach the end of the dataset or a predetermined window or some values that are higher than the current peak, stop.
4) To the left and the right find the minima in both reagions defined by the horizontal lines. Those are now the bases of the peak.
5) Calculate the prominence by subtracting the higher value base from the value of the peak.
This is a great methode to determine the location of certain signal, however for my goal I needed to know the prominence from the baseline after removal of the background. I therefore used the lower base to calculate the prominence. Using this I calculated the temperature again and got to 4,77 °C. I expected the result to be close to room temperature since the liquid was just sitting there with me in the same room. Also i noticed that the middle wavelength between the peaks (~537 nm) is not the same as the maximum wavelength of the laser (~533 nm). Could that be because of different laser modes? Is part of the Stokes signal maybe due to Fluorecence? I have set the camera to its maximum exposure time (about 30 s) and I have put down the gain to 0 dB. Help is as always appreciated.
- Attachments
-
- Herunterladen_19_07_23.png (31.34 KiB) Viewed 11570 times