Spectrometer: bandpass and resolution

In summary, according to the Ridchardson grating handbook, the FWHM of a spectrometers instrumental profile can be calculated using the following equation: -The limiting resolution dλ(resolution)-The bandpass dλ(slits)-The natural line width dλ(line)
  • #1
Ecthe
10
2
Hello there,

I am trying to improve my understanding regarding spectrometers bandpass, resolution etc. For that, I went through the Ridchardson grating handbook and on Horiba’s website where they have a nice explanation (http://www.horiba.com/us/en/scientific/products/optics-tutorial/monochromators-spectrographs/). I will use this link for reference to explain my issue.

In sec. 2.12.4 Determination of the FWHM of the Instrumental Profile, they give an approximation to calculate the FWHM (eq. 2-20) which is a function of
-The limiting resolution dλ(resolution)
-The bandpass dλ(slits)
-The natural line width dλ(line)

For the bandpass dλ(slits) I recalculated it from scratch to obtain the eq. 2-21 and I would think that going through the calculations and reading the definition made me understand what it means.

However, I am a bit confused by what they mean by the limiting resolution dλ(resolution). I went through sec. 2.12 Bandpass and Resolution and have a hard time understanding the difference between both terms.

But then, I read again sec. 2.12.4 where they write “Bandpass (BP) = FWHM”. Thus, is the Bandpass (BP) ≠ bandpass dλ(slits)? Or just that since they write “In general, most spectrometers are not routinely used at the limit of their resolution so the influence of the slits may dominate the line profile.”, it means that dλ(slits) ≫ dλ(resolution) and dλ(slits) ≫ dλ(line) so in the end, their FWHM = BP = dλ(slits)?

So I am a bit puzzled there and would appreciate any hints. Furthermore, if someone has a nice book reference on all that, I would be happy to know it to check if they have it in our library.
 
Physics news on Phys.org
  • #2
The diffraction intensity pattern of a monochromatic plane wave source of wavelength ## \lambda ## from a grating with ## N ## lines/grooves is given by ## I(\theta)=I_o \frac{sin^2(N \phi/2)}{sin^2(\phi/2)} ## where ## \phi=\frac{2 \pi}{\lambda} d sin(\theta) ##. In the limit of a very narrow slit, this gives a finite width ## \Delta \theta ##, (due to the diffraction pattern), to the primary maxima (which are the spectral lines) from a monochromatic source of wavelength ## \lambda ## given by ## \Delta \theta \approx \frac{ \lambda}{Nd} ##. This results in a linewidth ## \Delta x ## at the exit slit with the focusing optic, (which puts the far-field diffraction pattern in the focal plane which is also the plane of the exit slit), as ## \Delta x=f \Delta \theta ##. ## \\ ## (In the above equation, the principal maxima occur when the denominator equals zero , (with ## \phi/2=m \pi ## for some integer ## m ##, giving the well-known result ## m \lambda =d \sin(\theta) ##) , with the numerator also equal to zero. In the limit that the denominator is zero, ## I=N^2 I_o ##. The width ## \Delta \theta ## of the maximum can be found by finding the first zero of the numerator next to this peak.) ## \\ ## If a wider entrance slit is used, with slit width ## \Delta x_{slit} ##, it will result in an image from a monochromatic line of with ## \Delta x \approx \Delta x_{slit} ## at the plane of the exit slit, even in the case of a perfectly monochromatic source with minimal spread due to the width of the diffraction pattern. ## \\ ## Normally slit widths are used that are somewhat wider than that which would allow for getting close to the diffraction limit of the system because a slit that is this narrow usually does not allow enough light to pass through the spectrometer to be able to read the signal well at the exit slit. Normally entrance and exit slit widths are set to be nearly equal in size. ## \\ ## Additional comment: To make the incident source onto the grating into a plane wave, a collimating optic is used that has its focal point at the entrance slit off the spectrometer. A similar optic is used to focus the outgoing far-field plane waves (diffraction pattern) onto the plane of the exit slit. This way, the textbook diffraction grating equations apply, which in complete detail have ## \phi=\frac{2 \pi}{\lambda} d (\sin(\theta_i)+\sin(\theta_r)) ##.
 
Last edited:
  • Like
Likes blue_leaf77 and Ecthe
  • #3
Thanks a lot for your answer.

Still a few questions. Do you have a textbook reference for all of that? I would like to have a deeper look at it.
Also, from what I understood, while it is interesting to know the device resolution, in the end, spectrometers are monstly limited slits width. Is that correct?
 
  • #4
The formula and derivation for the interference pattern of of a diffraction grating with ## N ## lines I believe can be found in Hecht and Zajac Optics, but also in Halliday-Resnick's Physics II book if I'm not mistaken. The spectrometer configuration is the Ebert type spectrometer, but I don't have a good reference for the topic. Additional items are for optimal results that the focusing onto the entrance slit with a lens needs to be such that with the F#, you fill most of the collimating optic with the incident light in order to use most of the ## N ## lines on the grating. If you send in a narrow beam onto the grating, you may use far less than ## N ## lines and get lower resolution. ## \\ ## And additional item is most of the time, a slit width of ## b= 1 ## mm or thereabouts is used in order to get sufficient throughput. In these cases, the resolution is far less than what the instrument is capable of. As I recall, it usually takes a slit width of around ## b= 5 ## microns to start to achieve the full instrument resolution. (These are ballpark numbers). ## \\ ## And one additional item: The diffraction gratings that are presented in the textbook are usually transmission type gratings, while a spectrometer uses a reflective type grating, but the principles are the same. The lines on the grating can be thought of as Huygens mirrors that scatter the light in all directions, just like a narrow slit is a Huygens source.
 
Last edited:
  • Like
Likes Ecthe
  • #6
Thanks a lot. I will need a bit of time to go through all of that.
 
  • Like
Likes Charles Link
  • #7
For the interference theory/diffraction pattern of ## N ## lines or slits=where ## N ## may be 30,000 or more for a diffraction grating, (600 lines/mm with a grating 5 cm wide gives you ## N=30000 ## ), suggestion is to review the case of a 2 slit interference pattern with spacing ## d ##, and then go to ## N=3 ## equally spaced slits, etc., and then ## N=4 ##... ## \\ ## The primary maxima always occur at ## m \lambda=d \sin(\theta) ##. ( ## m ##=integer, usually small. Oftentimes, a spectrometer is operated using ## m ##=1. Not ##m ##=0 though, because every wavelength has ## m##=0 at the same place. Especially for higher orders ## m ##, order sorting filters are often necessary, because e.g. the ## m=3 ## blue (shorter wavelength) can overlap the ##m=2 ## and ## m=1 ## of longer wavelengths, etc.) ## \\ ## As ## N ## increases, the primary maxima begin to narrow and get higher., and the secondary maxima get much lower. When ## N =1000 ## or more, (even for ## N=100 ## or more), most of the energy from the pattern goes into the primary maxima which become almost like lines rather than wide peaks(bright spots).
 
Last edited:
  • #8
besides going through the theory for a diffraction grating, it is advisable to plot out the diffraction pattern as a function of the number of slits. Completely illumination (pun intended)...and don't just do it for normal incidence, do a parameter sweep in incident azimuth also.
 
  • Like
Likes Charles Link

What is a spectrometer?

A spectrometer is a scientific instrument used to measure the properties of light, such as intensity, wavelength, and polarization. It works by separating light into its component wavelengths and measuring their properties using various techniques.

What is bandpass in relation to spectrometers?

Bandpass refers to the range of wavelengths or frequencies of light that a spectrometer can measure. It is determined by the design and components of the spectrometer, and can vary depending on the specific instrument.

How does bandpass affect the accuracy of a spectrometer?

The larger the bandpass of a spectrometer, the more wavelengths it can measure simultaneously. This can improve speed and efficiency, but may also decrease the resolution and accuracy of the measurements. A smaller bandpass allows for more precise measurements, but can be slower and less efficient.

What is resolution in a spectrometer?

Resolution in a spectrometer refers to the ability to distinguish between closely spaced wavelengths or frequencies of light. It is typically measured as the smallest difference in wavelength that the instrument can detect. Higher resolution spectrometers can provide more precise and accurate measurements.

How can bandpass and resolution be optimized in a spectrometer?

The bandpass and resolution of a spectrometer can be optimized by adjusting its design, such as the size and spacing of its components, and choosing the appropriate wavelength range for the specific application. Additionally, using advanced techniques such as spectral deconvolution can improve the resolution of a spectrometer without sacrificing its bandpass.

Similar threads

  • High Energy, Nuclear, Particle Physics
Replies
1
Views
2K
Replies
1
Views
1K
  • Optics
Replies
9
Views
2K
Replies
2
Views
739
Replies
2
Views
2K
  • Quantum Physics
Replies
33
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
920
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
9
Views
20K
Back
Top