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- Thread starter AVRA
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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.

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What is the value 'f' here ?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.

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Thanks for the reply.## 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.

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 ?

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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.

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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 ?## 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.

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

Thanks.

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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

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.

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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.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} ##.

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 ?

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Ok Thank you so much. I will look forward to read the article.

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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 ?

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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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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.

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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).I do not quite understand how involved this has gotten.

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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.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).

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.

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