# The Resolving Power of a spectrometer

I can't understand one thing in the definition of the resolving power of a spectrometer:

Let the resolving power be defined as: $$R=\frac{\lambda}{\Delta \lambda}$$ where $$\Delta \lambda$$ is the separation of two resolved spectral lines and $$\lambda$$ is the average wavelength of the two lines.

I understand that if $$\Delta \lambda$$ is smaller then $$R$$ must be greater, because if the spectrometer can resolve two close spectral lines then its resolving power is greater but why do we need that $$\lambda$$ in the numerator?

If we have two lines with the same $$\Delta \lambda$$ but with smaller r wavelengths, why is the resolving power higher?

From my point of view I would simply define it as: $$R=\frac{1}{\Delta \lambda}$$, so why that numerator?

Because resolving power is unitless. By definition.

You can define something called the "resolution" of a spectrometer, which is just $$\Delta \lambda$$, which has units.

Why is a unitless resolving power useful? One reason is that it arises naturally in the properties of a simple diffraction grating, where the resolving power is just the order multiplied by the number of rule lines on the grating, and is independent of wavelength.

A further reason, especially in astronomy, is that it can be directly translated into a velocity resolution that is independent of wavelength $$\Delta v = c/R$$.

$$\dfrac{\Delta \lambda}{\lambda}$$ is the fractional change which can be resolved and the reciprocal is the resolving power which has no units.

So in terms of a fractional change $$\dfrac{1\,\rm nm}{400\,\rm nm}$$ is the same as $$\dfrac{2\,\rm nm}{800\,\rm nm}$$ ie you can say that the resolving power in terms of a fractional change is the same.

However that equivalence of resolution is not as obvious if you just compare $$(1/)\Delta \lambda$$ ie $$(1/)1\,\rm nm$$ vs $$(1/)2\,\rm nm$$ and note that a unit would need to be included.