# Optics and resolution

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?

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).

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.

Staff Emeritus
Homework Helper
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?

Staff Emeritus
Homework Helper
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?

Staff Emeritus