Optics and resolution

  • Thread starter Niles
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  • #1
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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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  • #2
Andy Resnick
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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?
 
  • #3
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100 MHz is the FWHM. We are in the visible range, so λ ~ 500 nm (sorry, I should have specified that first).
 
  • #4
Andy Resnick
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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.
 
  • #5
Redbelly98
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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.)
 
  • #6
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Thanks to both of you. Redbelly98, isn't Δf/f =λ·Δf/c only valid in vacuum?
 
  • #7
Redbelly98
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Strictly speaking, yes. It's a reasonable approximation in air as well. What medium does your spectrometer operate in?
 
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In air. But what is the general version of λf =c?
 
  • #9
Redbelly98
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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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