Solar peak color in human eye

There are two forms of Wien's displacement law that can be derived from Planck's equation. They are: $\lambda_{m} T=\mbox{constant}$ (1) and $\frac{f_{m} }{T}=\mbox{constant}$- (2) Suppose that we have known a black body's temperature, then $\lambda_{\mathrm{m}}$ and $f_{\mathrm{m}}$ can be obtained from Eqs. (1) and (2). For example, the Sun's surface temperature, $T \quad =$ 5778 K, then according to Eq. (1) $\lambda _{\mathrm{m}} \quad =$ 500 nm which is green; but according to Eq. (2) $f_{\mathrm{m}} \quad =$ 3.40 $\times$ 10$^{\mathrm{14}}$ Hz which is near-infrared. While the inequality $\lambda_{m} f_{m} \ne c$ can be explained mathematically by substituting $\lambda f=c$ into Planck's radiation function, the question lingers: what color of sun light ``really'' peaks in human eye? The answer is that Planck's function, or Wien's law, can't answer this question. Planck's function, $I(\lambda )$, or $I(f)$, is the radiation intensity per $d\lambda $ (meter) or $df$ (Hz) and, $d\lambda $ and $df$don't have same interval. For a spectrometer, it peaks at green if $\lambda $ changes evenly; peaks at near-infrared if $f $changes evenly. For human eye, its peak's location depends on $I(\lambda )$, or $I(f)$, and, how much each type of cone is excited. We can naively represent any color as a triplet of numbers: (red, green, blue), where each is the degree to which the associated type of cone is excited. Then $\mbox{red}=\int {d\lambda I(\lambda )S_{r} (\lambda )} =\int {dfI(f)S_{r} (f)} $ -(3) where $S_{r} (\lambda )$/$S_{r} (\lambda )$ is the sensibility of red cone. For green and blue we have same equations. Then the peak is determined by the value of integral.