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?


2 Answers 2


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.


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