Spectrometer centered wavelengths

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  • Thread starter Tone L
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Summary:
Looking to determine the centerline of the wavelengths for a sensor.
I am looking at this Hamamatsu mini-spectrometer, C12880MA [https://www.hamamatsu.com/us/en/product/type/C12880MA/index.html]

The specs are,

Spectral response range: 340 to 850 nm

Spectral resolution (FWHM) (typ.): 12 nm

Number of total pixels: 288 pixels

Now the size of the band of sensor is 510 nm (850-340), how can we determine where the spectrometer's centered wavelengths, 510nm/12nm is a centered wavelength every 42.5nm? Surely, I am missing something. Any guidance will be helpful

Thanks!
 

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  • #2
hutchphd
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What are you looking for ? Please be more specific: "centerline of the wavelengths" is ambiguous. Give one specific wavelength and what you wish to know about the signal for instance.
 
  • #3
Charles Link
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If you just rely on the dial on a spectrometer, or the wavelength they designate as a function of position in a focal plane array, there may be slight errors. For more accurate wavelength determination, it helps to have wavelength calibration standards, such as atomic spectral lines, to determine the actual wavelength as opposed to the approximate number you get from the manufacturer.
 
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hutchphd
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Summary:: Looking to determine the centerline of the wavelengths for a sensor.

Now the size of the band of sensor is 510 nm (850-340), how can we determine where the spectrometer's centered wavelengths, 510nm/12nm is a centered wavelength every 42.5nm? Surely, I am missing something.

You need to understand the parameters. As you realize, the ratio you took is a dimensionless number and not very useful. @Charles Link is correct that the optical projection onto the pixels is usually quite linear and the best way is to use standard emission lines (historically mercury-argon gas emission) to calibrate wavelength positions by interpolation /extrapolation from the this test spectrum.
If you just believe Hamamatsu built it exactly right(not recommended), then pixel 0 would be 340nm and pixel 287 would be 850nm (or vice-versa) and each pixel would increment you by (510/288) nm/pixel along the way.
Spectrometers are extraordinary instruments for science and have been used since the time of Newton. There is a lot to know!
 
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  • #6
hutchphd
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Here is a moderately decent description of how they work. They don't seem to be linear:
The Hamamatsu gratings are usually concave reflective and holographic so they can make the optics do pretty much what they wish. My guess is they try to make it project linearly onto photodiode array but I have not used the part. Of course all of the geometrical nonlinearities can also n principal be mediated by calibration.
 
  • #7
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Thanks for the feedback. @hutchphd - Moving forward this is what I was truly trying to determine. Assuming I understand this correctly. I am trying to imagine what the output data would look like. Reading out the signal from the analog pin, would it give me 288 data points, where data[0] would be the signal at 340 nm which has a spectral resolution (typ) of 12nm, and so on? Maybe this isn't possible to answer.

So roughly every 1.77nm I would get a data point thus creating a spectra, if I were to plot it? Though at every 1.77nm the resolution would be 12nm.
 
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  • #8
hutchphd
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What they call the resolution is optical width on the sensor of a perfectly monochromatic source because of optics. Using our supposed calibration this signal would have a halfwidth of (12/1.77) pixels. This is because the inlet has a finite width slit aperture (otherwise you get no light) and this gets imaged onto the array so the tradeoff is signal strength vs resolution. They optimize this to the intended use. The resolution for this small device is not very good.
 
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  • #9
chemisttree
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I think they define resolution differently. I think they define resolution as the width in nm that the instrument can resolve a pair of closely spaced lines. Thus it isn’t the width in nm of a single monochromatic frequency at half height but the minimum separation, in nm, of a pair of monochromatic lines that can be resolved.

Maybe we’re saying the same thing, though...
 
  • #10
hutchphd
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Yes it is certainly measuring the same effect. I think the numbers would be very similar as well (but I don't know how you quantify "seeing two lines" .....probably ends up identical)
 
  • #11
chemisttree
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The IUPAC definition say something about the minimum separation that can be distinguished but I think the manufacturer implies that they be separate peaks at half height.
 

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