Optics & Resolution: Can I See the Peak?

[Niles]

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?

 

[Andy Resnick]

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?

 

[Niles]

100 MHz is the FWHM. We are in the visible range, so λ ~ 500 nm (sorry, I should have specified that first).

 

[Andy Resnick]

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.

 

[Redbelly98]

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.)

 

[Niles]

Thanks to both of you. Redbelly98, isn't Δf/f =λ·Δf/c only valid in vacuum?

 

[Redbelly98]

Strictly speaking, yes. It's a reasonable approximation in air as well. What medium does your spectrometer operate in?

 

[Niles]

In air. But what is the general version of λf =c?

 

[Redbelly98]

The speed of light in a medium is c/n, so

λf =c/n​

where λ is the wavelength in the medium.