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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 Sep 5 '20 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 Sep 5 '20 at 17:49
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    $\begingroup$ Related: Frequency and wavelength of thermal radiation $\endgroup$ – G. Smith Sep 5 '20 at 18:01
  • $\begingroup$ Related: Wikipedia $\endgroup$ – G. Smith Sep 5 '20 at 18:09
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    $\begingroup$ see physics.stackexchange.com/questions/437769/… $\endgroup$ – Andrew Steane Sep 5 '20 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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