According to high-school physics, light frequency and wavelength are inversely proportional, and their product is the speed of light. But when discussing spectral power, somehow sunlight peaks differently in these two spaces:


How is this possible?


3 Answers 3


This is all about the difference between "a quantity Q" and "the density of that quantity per unit X" where X is some other quantity. A good way to get a feel for this is to convert statements about density to statements about amount. For example, if $u$ is "the spectral energy density per unit wavelength range" then

(The amount of energy in the range $\lambda$ to $\lambda + d\lambda$) = $|u \, d\lambda|$.

Now let $\rho$ be "the spectral energy density per unit frequency range". Then

(The amount of energy in the range $\nu$ to $\nu + d\nu$) = $|\rho \, d\nu|$.

Next we need to realise that these two amounts of energy are the same. For, when the wavelength is $\lambda$ the frequency is $\nu$, and when the wavelength is $\lambda + d\lambda$ the frequency is $\nu + d\nu$, so the two energies we just referred to are one and the same energy, in the sense that they are two ways of referring to the same "chunk" out of the complete spectrum. Thus we find

$|u \, d\lambda| = |\rho \, d\nu|$

and therefore

$ u = \rho\, |d\nu/d\lambda| = \rho \, c / \lambda^2$

That's the algebra, but I hope I have also given a feel for what is going on. It does take some getting used to.

If you now take some distribution $\rho$ and multiply it by the function $1/\lambda^2$ you can see that any local peak in the distribution is going to be moved a bit: the long wavelength part gets reduced and the short wavelength part gets increased, so the peak moves to shorter wavelength. It is not because of any change in the relationship between frequency and wavelength, but because of the difference in "per unit frequency change" as compared to "per unit wavelength change". A similar shift will apply to a study of quantities involving other things related by an inverse, such as time and frequency, resistivity and conductivity, etc.


The units of the y-axis quantity spectral irradience, $\rm W \, m^{-2} \, nm^{-1}$ and $\rm W \, m^{-2} \, THz^{-1}$, differ in a subtle way.

They both are in terms of power $\rm W$ per unit area $\rm m^{-2}$ but the interval over which the power per unit area is measured is different, $\rm nm^{-1}$ and $\rm THz^{-1}$.
Noting that a frequency interval of $1\,\rm THz$ corresponds to a wavelength interval of $300,000\,\rm nm$ accounts for the difference in the spectral power graphs that you are referring to.


It's an interesting question!

If your graph is plotted as a function of wavelength,the solar irradiance is maximum in the visible and if it were plotted as a function of frequency the maximum is,interestingly,in the near infrared.

How is this possible?

This happens because the relationship between wavelength and frequency is not linear.

$ \vert d \nu\vert = \vert c/ \lambda^2 d\lambda \vert$

  • 3
    $\begingroup$ I would like to recommend the paper that is mentioned in the link. It does a great job at explaining clearly and precisely where the misconception comes from mathematically, and in what other situations a similar effect occurs. $\endgroup$
    – DK2AX
    Oct 30, 2018 at 16:39

Not the answer you're looking for? Browse other questions tagged or ask your own question.