Optics & Resolution: Can I See the Peak?

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Discussion Overview

The discussion revolves around determining whether a spectral peak at 100 MHz can be resolved given a wavelength resolution of Δλ 0.5 nm. Participants explore the relationship between frequency and wavelength in the context of optics and resolution, focusing on the implications of the full width at half maximum (FWHM) and the medium in which the measurements are taken.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether the 100 MHz refers to the peak frequency or the FWHM, seeking clarification on the mean wavelength.
  • Another participant confirms that 100 MHz is indeed the FWHM and specifies that the mean wavelength is approximately 500 nm.
  • A suggestion is made to convert the FWHM into a wavelength range by first converting the mean wavelength to frequency, adjusting for the FWHM, and then converting back to wavelengths to compare with the spectrometer's resolution.
  • It is proposed that the ratio Δf/f should be compared with Δλ/λ to determine if the peak can be resolved, with Δf representing the FWHM and Δλ the spectrometer resolution.
  • A participant raises a concern about the validity of the ratio Δf/f = λ·Δf/c in non-vacuum conditions, prompting a discussion about the medium of operation for the spectrometer.
  • Clarification is provided that the speed of light in a medium is given by c/n, where n is the refractive index, affecting the relationship between wavelength and frequency.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the medium for the calculations and the validity of certain approximations, indicating that the discussion remains unresolved regarding the best approach to determine peak visibility.

Contextual Notes

There are limitations regarding the assumptions made about the medium's effect on light speed and the applicability of certain formulas in different conditions. The discussion does not resolve these issues.

Niles
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Hi

Say I have a peak of about 100 MHz, and my apparature has a resolution of about Δλ 0.5 nm. How can I find out whether I will see the peak or not?
 
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It's not clear from your post: is the peak located at 100MHz, or is that the full width half maximum? You have a certain wavelength resolution, but what is the mean wavelength?
 
100 MHz is the FWHM. We are in the visible range, so λ ~ 500 nm (sorry, I should have specified that first).
 
Ok.

Here's what I would do: convert the FWHM into a wavelength range: do this by converting 500nm into frequency, then add +/- 50 MHz, and convert back to wavelengths. Then you can compare directly with the wavelength resolution of your spectrometer.

Something to keep in mind- detecting the peak is different than resolving the peak width.
 
I would compare Δf/f (=λ·Δf/c) with Δλ/λ (=0.001).

You need Δf/f > Δλ/λ in order to resolve the peak.

(In case it's not clear from the context of this thread, Δf is the FWHM of the peak, and Δλ is the spectrometer resolution.)
 
Thanks to both of you. Redbelly98, isn't Δf/f =λ·Δf/c only valid in vacuum?
 
Strictly speaking, yes. It's a reasonable approximation in air as well. What medium does your spectrometer operate in?
 
In air. But what is the general version of λf =c?
 
The speed of light in a medium is c/n, so
λf =c/n
where λ is the wavelength in the medium.
 

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