Spectral resolution of spectromter?

[andy]

Based on my measurements, it appears to be 0.4 - 0.8nm based on the smallest two peaks that I saw resolved for the performance edition (30um slit).

I found a theoretical equation online that looks like

lambda = ((resolution factor) * (spectral range) * (slit width)) / (number of pixels * width of pixels)

The resolution factor approximates to 1.5 in this situation according to the book pg 6-65 [2].

So that calculates to (1.5 * 120nm (i.e. 660-540 approx) * 30um) / (2048 pixels * 3.45um)

= .76nm.

I'm wondering how accurate this is?

I also saw on the website Luc mentions 35 cm-1 for the starter edition which would translate to about 1nm according to [3]


[1] https://www.ossila.com/en-us/pages/spec ... resolution
[2] Rabus, D. G. (2014). Optofluidics Systems Technology. Germany: De Gruyter.
[3] https://www.stellarnet.us/raman-shift-calculator/

[Luc]

There are detailed posts on thepulsar.be about resolution but information is a bit spread among the various posts.

In summary:

• resolution is defined as the ability to discriminate two nearby peaks provided the peaks broadness are not the limiting factor
• a good estimate of resolution is given by the FWHM of the slit image through the optical system multiplied by the dispersion of the system. So if the FWHM is 3 px and the dispersion is 0.1 nm/px, the resolution is 0.3 nm.
• a good approximation to the system FWHM is the square root of the sum of squared PSF FWHM and slit width. So a system with a 2 px PSF and a 3 px slit has a sqrt(2² + 3²) total FWHM (approximately). The formula can be modified in Raman to account for the laser FWHM too in the sum.
• beware of slit magnification effects due to grating angles because it modifies the apparent slit size.
• grating can become a limiting factor; maximum resolution power (theoretical) is the number of lines illuminated. With blazed grating, we often get 1/10th of that. 1200lp/mm × 20 mm × 0.1 is about 2400 so limiting resolution in openraman should be about 0.25 nm or a touch better.

Experimentation gives FWHM of ~12 cm-1 on a 3000 cm-1 (110 nm) range so 0.44 nm including all effects (laser broadness as well).

I tested the spectrometer separately and found 0.2-0.3 nm using the neon lamp depending on the slit size used

[andy]

This is hurting my brain a little bit.

I think in my head the slit was projected directly onto the sensor but that doesn't make sense because for a 30um slit and 400nm range this would be a 12,000um big. The detector is 3.45um * 2048 so 7066um. So that doesn't make any sense.

Okay so thinking of is as limiting factors, there are several. The number of detector pixels, the slit, grating, the width of the spectral lines.

Let's ignore everything but the detector pixels. The limiting factor of the pixels is 2048. The book above describes you need 3 pixels to resolve a peak (one low, one high, and one low) [1] [2] if each pixel is exactly resolving one wavelength. Two high pixels would just continue a single peak for example. So you have 682 resolving units. For a 400nm range this means .58nm because .58nm * 682 working units = 400 nm .

[1] I suppose if you did averaging and there was slight wiggling over time between the grating and the detector then you could get around this due to aliasing
[2] Another factor is in the current design if the slit is not perfectly straight and you are averaging columns of data, you might not need 3 pixels also due to aliasing

[Luc]

The total range on openraman is about 110nm (550 nm to 660nm approximately) :

1/550 - 1/532 = 600 cm-1
1/660 - 1/532 = 3650 cm-1

With a slit size of 20 um and a compression factor of cos45°/cos0° (grating configuration) the image on the camera is at best 14 um. In practice, it's a bit bigger due to image aberration.

14 um is 4.1 px and size the dispersion is 110 nm on 2048 px, you get a slit image of ~0.22 nm.

That also doesn't take account of the laser bandwidth. A 0.2 nm shift of the laser light results in a shift of 7 cm-1 for instance, so you blur your spectra by about that factor if you use a 0.2 nm wide source.

[andy]

Okay, that makes sense, thank you!