The effect of instrumental broadening on FWHM in Raman peaks

  • #26
Charles Link
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That's fine, but the problem that I have here is that you've schooled the OP on the peak width from the spectrometer, but you didn't describe how the OP can go from that and extract out the instrument broadening from the data.

Note that the broadened peak may also be due to other factors (finite temperature of the experiment, etc.). From many of the Raman papers and talks that I've come across, one has to perform some form of a devoncolution to remove the instrument broadening. This, you never covered.

I've performed similar method in extracting the instrument resolution from ARPES data. So this is not an unknown or unfamiliar method.

Zz.
@ZapperZ Yes, your input was very much needed and I agree, that is one thing that my input was lacking. The deconvolution process, although perhaps somewhat standard, is something that I myself have not previously encountered. (My career=I am now retired= was outside of academia and in the defense industry. My expertise is actually somewhat limited on this topic of Raman spectroscopy, although I do know the details of how a diffraction grating spectrometer works). ## \\ ## Additional item=I think perhaps only at most a handful of the Science Advisors and Homework Helpers on PF actually have any hands-on experience with Raman spectroscopy. I answered the question the best I could, having much familiarity with diffraction grating spectrometers, but my experience with Raman spectroscopy is very limited.
 
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  • #27
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Thank you Charles for the response. I have some questions regarding the formula you mentioned.
I use 1800gr/mm for the Raman measurement. Does that mean N will be equals 1800 here ? How will I determine the 'm' ? and how will I get the slit width 'd' ?

Thanks.
Hi Charles:

I would like to know the significance of 'd' in this equation. Is it the the same as the interplanner distance in Bragg's law ?

Thanks.

Regards,
Avra
 
  • #28
Charles Link
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Hi Charles:

I would like to know the significance of 'd' in this equation. Is it the the same as the interplanner distance in Bragg's law ?

Thanks.

Regards,
Avra
## d ## in a diffraction grating is similar to the interplaner distance ## d ## in Bragg's law. And ## d=\frac{1}{1800} \, mm ## for your grating. The derivation for constructive interference from a grating is similar to the derivation of Bragg's law for a crystal. (It's similar but not identical). Bragg's law actually uses constructive interference from adjacent planes, along with angle of incidence =angle of reflection ( to get constructive results from a single plane). Meanwhile, for a reflection grating ## m \lambda=d (\sin(\theta_i)+\sin(\theta_r)) ##. When you set ## \sin(\theta_i)=\sin(\theta_r) ##, you get Bragg's law (with the ## 2 ## in ## 2d \sin(\theta) ##. For an additional detail, Bragg's law can be written as ## m \lambda=2 d \cos(\theta) ## where ## \theta ## is measured from the normal to the crystal plane.). For a diffraction grating, you don't need to have ## \sin(\theta_i)=\sin(\theta_r) ## to get a maximum. ## \\ ## And note: These two things are somewhat different. For the lines on the grating to behave like the crystal planes that have angle of incidence =angle of reflection, the grating would need to be a transmissive type grating, so that it is a difficult comparison to make. And even in that case, the grating does not require angle of incidence=angle of reflection. There is also a sign convention in the equation. For a reflection grating with angle of incidence equal to angle of reflection(with the proper sign), this is the zeroth order maximum (## m=0 ##) for the grating. (For the way I wrote the equation ## m \lambda=d (\sin(\theta_i)+\sin(\theta_r)) ## with a "+" between the terms, this has ## \theta_r=-\theta_i ##). If you take a transmissive grating, the equation is also ## m \lambda=d (\sin(\theta_i)+\sin(\theta_r) ##. The zeroth order maximum there is for light that goes straight through (with ## \theta_i=-\theta_r ##). ## \\ ## The Bragg type interference occurs for ## m \lambda=2 d \sin(\theta) ##. There is no requirement on a diffraction grating maximum that ## \theta_i=\theta_r ##. If you took a case like this for the transmission grating where ## \theta_i=\theta_r ## and ## m \lambda=d(\sin(\theta_i)+\sin(\theta_r)) ##, then you would be finding the equivalent of a Bragg peak for the grating, but it wouldn't have higher intensity than a non-Bragg angle maximum. With this last case, you could let the lines in the grating become planes, and then you would have the equivalent of a Bragg peak.
 
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  • #29
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For a diffraction grating spectrometer, ## m \lambda=d (\sin(\theta_i)+\sin(\theta_r)) ## is the condition for constructive interference and getting a primary maximum. At first order (## m=1 ##), this means that ## \Delta \lambda \approx d \, \Delta \theta ##. A slit width of ## b ## will introduce a ## \Delta \theta=\frac{b}{f} ##. This results in a ## \Delta \lambda \approx \frac{d \, b}{f} ##. ## \\ ## There is a lower limit that can be achieved though by narrowing the slit widths: ## \frac{\Delta \lambda}{ \lambda}>\frac{1}{Nm } ## where ## N ## is the number of lines on the grating that are used in creating the spectral line and ## m ## is the order.
Hi Charles:

As you mentioned that a slit width b will introduce Δθ=b/f, can you provide me any schematic of the same how the angle is being calculated ? Like, I have searched many optical articles regarding it which is showing a incident ray and a reflected ray with the angles and planes with a separation of d, but I am unable to relate the slit width b, the focal length f and Δθ. I would really appreciate if you can show me how the angle is formed with b and f, or explain or upload any image. i have uploaded an image similar to it.

Best Regards,
Avra
 

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  • #30
Charles Link
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I couldn't find a good diagram either, but this is something that is used frequently in optics. Parallel rays coming in at angle ## \theta ## onto a focusing lens or mirror will focus in the focal plane at height ## h= f \,\theta ##. This should be somewhat obvious because if the rays are parallel, it means the source is far away. Using the lensmakers formula ## \frac{1}{f}=\frac{1}{s}+\frac{1}{m} ##, for object distance ## s =+\infty ##, image distance ##m= f ##. This means the parallel incident rays at angle ## \theta ## will converge and come to a focus in the focal plane. The ray through the middle of the lens goes straight through, so that the height is ## h=f \tan{\theta}= f \, \theta ## for small angles ## \theta ##. ## \\ ## Taking the rays in reverse, with the source on the focal plane, if the rays start at height ## h ##, they will come out parallel at angle ## \theta=\frac{h}{f} ##.
 
  • #31
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Thank you very much Charles.

Looking forward to discuss with you more. Have a great day !

Best Regards,
Avra
 
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  • #32
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Hi Charles:

Hope you are doing well. I would like to know whether the absorbance efficiency of 532 nm laser used in Raman spectroscopy is independent of temperature. If it is dependent on temperature, then how ?
 

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