# The effect of instrumental broadening on FWHM in Raman peaks

I would like to know how to extract the instrumental broadening effect in Raman spectrometer using solid angle of the objectives, slit width or spectrometer length. I am calculation the FWHM in Raman peaks and I would like to know the effect of instrumental broadening on Raman line width and how to get accurate measurement.

## Answers and Replies

Homework Helper
Gold Member
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.

AVRA
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.
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.

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.

What is the value 'f' here ?

Homework Helper
Gold Member
## d=\frac{1}{1800} ##mm. Meanwhile ## b ## is the slit width and should be adjustable=entrance and exit slits, usually set to be nearly equal width. For ## N ##, you need to know what the illuminated width ##w ## of the diffraction grating is. ## N=1800 \, w ## when ## w ## is in mm. ## \\ ## ## f ## is the focal length of the collimating optics that makes parallel rays incident onto the grating. The same ## f ## is usually also used to make the far field diffraction pattern image in the plane of the exit slit. ## f ## is often the "length of the spectrometer" when such a number is provided by the manufacturer.

AVRA
## d=\frac{1}{1800} ##mm. Meanwhile ## b ## is the slit width and should be adjustable=entrance and exit slits, usually set to be nearly equal width. For ## N ##, you need to know what the illuminated width ##w ## of the diffraction grating is. ## N=1800 \, w ## when ## w ## is in mm.
Thanks for the reply.

I would like to know, what is the formula of the instrument broadening here ? I mean is it
Δλ/λ ? and the value should be as small as possible for accurate linewidth measurement in Raman peaks ? and lastly, are 'w' and slit width 'b' instrument specific ?

Homework Helper
Gold Member
Please see additions to the post 5 above about ## f ##. ## \\ ## The instrument broadening will normally be ## \Delta \lambda \approx \frac{d \, b}{f} ##. Generally, if you narrow the slits to achieve minimum resolution ## \Delta \lambda=\frac{\lambda}{Nm} ##, (maximum resolving power), you get very little throughput=very low signal. ## \\ ## Spectrometers often have adjustable slit widths ## b ##. ## \\ ## Meanwhile, the ## w ## can be improved by optimally focusing the beam onto the entrance slit. The rays emerging from the entrance slit should nearly fill the collimating optic/diffraction grating rather than only illuminating a small portion across the grating. (We're referring here to filling it horizontally=vertical filling isn't necessary).

AVRA
Please see additions to the post 5 above about ## f ##. ## \\ ## The instrument broadening will normally be ## \Delta \lambda \approx \frac{d \, b}{f} ##. Generally, if you narrow the slits to achieve minimum resolution ## \Delta \lambda=\frac{\lambda}{Nm} ##, (maximum resolving power), you get very little throughput=very low signal. ## \\ ## Spectrometers often have adjustable slit widths ## b ##. ## \\ ## Meanwhile, the ## w ## can be improved by optimally focusing the beam onto the entrance slit. The rays emerging from the entrance slit should nearly fill the collimating optic/diffraction grating rather than only illuminating a small portion across the grating. (We're referring here to filling it horizontally=vertical filling isn't necessary).
Thank you again for the explanation. Actually I am new in this subject, so trying to connect the dots. I already measured the Raman peaks and calculated the FWHM from those measurements. Now, I am trying to figure out if the instrumental broadening is small enough not to influence the variation in FWHMs.
So, according to you Δλ≈db/f , is the formula of the instrumental broadening. In the equation, d is equal to 1/1800nm. What does 'f' represent here ? and how will I know the value of the slit width ? Can you give me any idea about the general values of d and f ?

Thanks.

Homework Helper
Gold Member
## d=\frac{1}{1800} ## mm. The slit widths ## b ## should be adjustable and a micrometer (hand turning to adjust) type control is often used. And see post 5 above for ## f ##. If you can open up the instrument, ## f ## is the distance of the concave mirror from the slit. Sometimes the same concave mirror is used both for entrance and exit slit=in other cases it will be two separate mirrors. The slits are always in the focal plane of these mirrors, so the focal length is readily measured by measuring the distance from slit to mirror. And a typical slit width (for low resolution work) might be around ## b=1 ## mm.

AVRA
## d=\frac{1}{1800} ## mm. The slit widths ## b ## should be adjustable and a micrometer (hand turning to adjust) type control is often used. And see post 5 above for ## f ##. If you can open up the instrument, ## f ## is the distance of the concave mirror from the slit. Sometimes the same concave mirror is used both for entrance and exit slit=in other cases it will be two separate mirrors. The slits are always in the focal plane of these mirrors, so the focal length is readily measured by measuring the distance from slit to mirror.
Thank you for the response. The concept is getting clearer for me. So the instrument broadening unit will be 1/nm according to the formula ? and if the value of instrumental broadening is very small, can I say that the FWHM measurement is nearly accurate ?

Homework Helper
Gold Member
## \Delta \lambda ## is in nm. ## \\ ## If ## \Delta \lambda =\frac{ d \, b}{f} ## << FWHM , e.g. a factor of 10 less, then the instrument broadening would be considered to be of little significance.

## \Delta \lambda ## is in nm.

Ok, I got it. There was a mistake in my calculation. So, if the value of instrument broadening is small, can I say that the linewidth measurement is correct ? And can you give me any reference of these formula, it will be helpful for me to dig deep into these. Thank you so much for helping me solving this problem.

Thanks.

Homework Helper
Gold Member
You will always have some error bars on your measured line widths, but those error bars can basically be ## \Delta \lambda=\frac{d \, b}{f} ##. I really don't have a good reference for this=you can try googling "Ebert Spectrometer" or "Czerny-Turner spectrometer". I can also look for a couple of "links" to PF homework questions, where I may have derived a couple of the equations: Let me see if I can find one or two...

AVRA
You will always have some error bars on your measured line widths, but those error bars can basically be ## \Delta \lambda=\frac{d \, b}{f} ##. I really don't have a good reference for this=you can try googling "Ebert Spectrometer" or "Czerny-Turner spectrometer". I can also look for a couple of "links" to PF homework questions, where I may have derived a couple of the equations: Let me see if I can find one or two...
Thank you very much for the reply.

One additional query I would like to know from you and that is
Thank you very much for the link and response. I would like to know mention here one thing that is that the Rama
You will always have some error bars on your measured line widths, but those error bars can basically be ## \Delta \lambda=\frac{d \, b}{f} ##. I really don't have a good reference for this=you can try googling "Ebert Spectrometer" or "Czerny-Turner spectrometer". I can also look for a couple of "links" to PF homework questions, where I may have derived a couple of the equations: Let me see if I can find one or two...
Thank you very much for the link and response. I would like to know mention here one thing that is that the Raman intensity is usually plotted against Raman shift whose unit is cm-1 and accordingly the FWHM of Raman peak is also expressed in cm-1. Now, does the formula of the instrument broadening applies same here as the unit there is nm and in case of Raman shift it is cm-1 ? I am asking these, because, if I want to plot the error bars for measuring raman shifts, then the units should be in cm-1, but using the formula you mentioned, the value will be in nm. and 1nm=10000000cm-1. So, the number will be huge compared with nm.

Homework Helper
Gold Member
In spectroscopy ## \nu=\frac{1}{\lambda} ##. This means, using a little calculus, that ## \Delta \nu=\frac{\Delta \lambda}{\lambda^2} ##. If ## \lambda=500 \, nm=.5 \, um ##, and ## \Delta \lambda=1 \, nm ##, that gives ## \Delta \nu=1 \cdot 10^{-7} cm/(.5 \cdot 10^{-4})^2 cm^2=40 \, cm^{-1} ##.

AVRA
In spectroscopy ## \nu=\frac{1}{\lambda} ##. This means, using a little calculus, that ## \Delta \nu=\frac{\Delta \lambda}{\lambda^2} ##. If ## \lambda=500 \, nm=.5 \, um ##, and ## \Delta \lambda=1 \, nm ##, that gives ## \Delta \nu=1 E-7 cm/(.5 E-4)^2 cm^2=40 \, cm^{-1} ##.
Thank you so much for the response. It really helped me clearing my concept regarding this. i will figure out the slit width and the length of the spectrometer to calculate the instrument broadening.

I would like to seek your help in resolving another query. Do you know how to determine the laser spot size in Raman spectrometer if I use a 10X objective with NA equal to 0.25 ?

Homework Helper
Gold Member
For the laser spot size, I'm not completely familiar with the terminology: What is a 10x objective? I'm more accustomed to using focal lengths and/or a schematic of what the optics involve. It sounds like the spot is being focused to near the diffraction limit. You might find this Insights article that I authored of interest: https://www.physicsforums.com/insights/diffraction-limited-spot-size-perfect-focusing/ I'm not sure if you might find it useful or perhaps it is too theoretical, but you can take a look at it...

AVRA
For the laser spot size, I'm not completely familiar with the terminology: What is a 10x objective? I'm more accustomed to using focal lengths and/or a schematic of what the optics involve. It sounds like the spot is being focused to near the diffraction limit. You might find this Insights article that I authored of interest: https://www.physicsforums.com/insights/diffraction-limited-spot-size-perfect-focusing/ I'm not sure if you might find it useful or perhaps it is too theoretical, but you can take a look at it...
Ok Thank you so much. I will look forward to read the article.

Hi Charles:

The formula that you mentioned for constructive interference, i.e., m λ = d ( sin ( θ i ) + sin ( θ r ) ) mλ=d(sin⁡(θi)+sin⁡(θr)) , can I use m λ = d sin θ for simplification following Bragg's law ?

Homework Helper
Gold Member
The angle of incidence ## \theta_i ## is measured from the perpendicular to the grating. Many introductory textbooks simply treat the case of ## \theta_i=0 ##. This is somewhat similar to Bragg's law for getting constructive interference between adjacent crystal planes, but don't confuse the two. Bragg scattering also has angle of incidence equal to angle of reflection from a single plane. ## \\ ## Here we are doing Fraunhofer diffraction=(the far field case), and the interference condition says that the path difference to the far field from adjacent grooves in the grating is an integer number of wavelengths. Depending on how the incident angle is measured, the formula can also read ## m \lambda=d (-\sin(\theta_i)+\sin(\theta_r) ) ##. Normally a diffraction grating spectrometer is wavelength calibrated using a source with a known spectrum with a lot of spectral lines. The formula ## m \lambda =d (\sin(\theta_i)+ \sin(\theta_r) ) ## etc. is useful for some simple calculations to estimate ## \Delta \lambda ## for a given slit width, as well as seeing how lines of different orders will overlap, but it is normally not used to do any wavelength calibration. It is still useful to learn the diffraction spectrometer using the equation which tells the locations of the primary maxima from the diffraction grating for a given wavelength. ## \\ ## The diffraction grating spectrometer has a textbook type solution in the way it is configured, with the entrance slit at the focal plane of the concave mirror creating a beam of parallel rays incident on the grating. The grating is rotated in order to change the wavelength that reaches the exit slit. ## \\ ## Meanwhile the far-field diffraction pattern is made to occur in the focal plane of the second concave mirror where the exit slit is located. This is done in the following manner: Parallel rays (a primary maximum at a given angle ## \theta_r) ## are brought to a focus at position ## x ## in the focal plane. (For small angles ## x=f \, \theta ##. Note that ## \theta ## here is the angle the parallel rays are incident on the concave mirror, and in general this ## \theta ## is related to, but is not equal to ## \theta_r ##). That way, instead of needing to go in the far field to observe the primary maximum from one wavelength (as well as the entire spectrum), e.g. 100 yards or more from the grating, the primary maximum can be observed as a focused spectral line in the focal plane of the second concave mirror. ## \\ ## (Note: When the parallel rays of one wavelength leave the grating at angle ## \theta_r##, they are a wide beam of parallel rays that may be 2 inches wide that gets focused by the concave mirror to a spectral line. The reason it is a focused line rather than a focused point is because it has the shape of the illuminated entrance slit. It's really quite remarkable that the far-field pattern is observed in this manner at the exit plane of the instrument. It is really a neat optical trick that results because parallel rays come to a focus in the focal plane of a concave mirror with the location ## x ## dependent on the angle ## \theta ## as ## x=f \, \theta ## ).

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ZapperZ
Staff Emeritus
I do not quite understand how involved this has gotten.

First of all, can't you find what the peak width should be theoretically at the temperature that the experiment was done? Then, you should either do

1. a deconvolution of that peak with another gaussian, and that FWHM of that gaussian is your instrument broadening, or

2. a convolution of your theoretical peak width with another gaussian, and then try and match your experimental peak width. When it matches as well as you can, the FWHM of that gaussian is your instrument broadening.

Zz.

Homework Helper
Gold Member
I do not quite understand how involved this has gotten.
I do think the OP will benefit by knowing the details of how a diffraction grating spectrometer works rather than treating it like a "black box" that gives you an answer. Some of the parameters of the spectrometer that determine the extent of the instrument broadening are user adjustable. (I specialized somewhat in diffraction grating spectroscopy, so my opinion will of course be somewhat biased).

Jahnavi and anorlunda
ZapperZ
Staff Emeritus
I do think the OP will benefit by knowing the details of how a diffraction grating spectrometer works rather than treating it like a "black box" that gives you an answer. Some of the parameters of the spectrometer that determine the extent of the instrument broadening are user adjustable. (I specialized somewhat in diffraction grating spectroscopy, so my opinion will of course be somewhat biased).

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.

Homework Helper
Gold Member
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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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

Homework Helper
Gold Member
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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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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