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Planck's Law is commonly stated in two different ways:

$$ u_\lambda \left( \lambda, T \right) = \frac{2hc^2}{\lambda^5} \frac{1}{e^\frac{hc}{\lambda kT}-1} $$ $$ u_\nu \left( \nu, T \right) = \frac{2h\nu^3}{c^2} \frac{1}{e^\frac{h\nu}{kT}-1} $$ We can find the maximum of those functions by differentiating those equations with respect to $\lambda$ and to $\nu$, respectively. We get two ways to write Wien's Displacement Law: $$ \lambda_\text{peak} T = 2.898\cdot 10^{-3} m \cdot K $$ $$ \frac{\nu_\text{peak}}{T} = 5.879\cdot 10^{10} Hz \cdot K^{-1} $$ We see that $\lambda_{\text{peak}} \neq \frac{c}{\nu_\text{peak}}$. So what frequency or wavelength is actually detected by an optical instrument most intensely when analyzing a black body? If they are $\lambda_{\text{peak}}$ and $\nu_\text{peak}$, how is $\lambda_{\text{peak}} \neq \frac{c}{\nu_\text{peak}}$?

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    $\begingroup$ Continuing to change the question after it has been answered is not polite. $\endgroup$
    – G. Smith
    Commented Sep 5, 2020 at 17:48
  • $\begingroup$ Regarding your latest question about how the two peaks are not related by the relationship you expect... There are numerous questions and answers on this site about why the two distributions are different. Have you searched? $\endgroup$
    – G. Smith
    Commented Sep 5, 2020 at 17:49
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    $\begingroup$ Related: Frequency and wavelength of thermal radiation $\endgroup$
    – G. Smith
    Commented Sep 5, 2020 at 18:01
  • $\begingroup$ Related: Wikipedia $\endgroup$
    – G. Smith
    Commented Sep 5, 2020 at 18:09
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    $\begingroup$ see physics.stackexchange.com/q/437769 $\endgroup$ Commented Sep 5, 2020 at 18:52

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If you measure the frequency distribution the peak is at $\nu_\text{peak}$ and if you measure the wavelength distribution the peak is at $\lambda_\text{peak}$. They are two different distributions.

If your optical instrument measures something else, then you have to explain exactly what.

The two spectral densities are not related in the way you expect because $d\nu$ and $d\lambda$ do not have the same relation that $\nu$ and $\lambda$ do.

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