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 "color" (by "color" I mean frequency or wavelength of the electromagnetic radiation) is actually seen most intensely when "looking" at a black body?

Disclaimer: By "seen", I mean detected by an optical instrument, not necessarily by the human visual system.

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 $\lambda_{\text{peak}} \neq \frac{c}{\nu_\text{peak}}$?

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 "color" (by "color" I mean frequency or wavelength of the electromagnetic radiation) is actually seen most intensely when "looking" at a black body?

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 "color" (by "color" I mean frequency or wavelength of the electromagnetic radiation) is actually seen most intensely when "looking" at a black body?

Disclaimer: By "seen", I mean detected by an optical instrument, not necessarily by the human visual system.