Can anyone explain how Wien's distribution law seen in Wien's original "On the Division of Energy in the Emission-Spectrum of a Black Body":
$$ \phi_\lambda = \frac{C}{\lambda^5} e^{-\frac{c}{\lambda \theta}},$$
where (using Wien's notation) $\phi_\lambda$ is the intensity, $\lambda$ is the wavelength, $C$ and $c$ are constants and $\theta$ is the temperature, becomes in Planck's "On the Law of Distribution of Energy in the Normal Spectrum" (in Planck's notation)
$$ E.d\lambda = \theta^5 \psi(\lambda \theta).d\lambda, $$
where $E$ is the energy, $\lambda$ is the wavelength and $\theta$ is the temperature.
SOLUTION: We want to show that Planck's expression $E = T^5 \psi (\lambda T)$ is equivalent to Wien's distribution law, which has a general form $E = F(\lambda)e^{-\frac{f(\lambda)}{T}}$, but which in more concrete terms (as shown by Wien) must be $E = \frac{C}{\lambda^5}e^{-\frac{c_1}{\lambda T}}$
First off, we divide the r.h.s. by $c^4$ which, unlike Planck's case where he deals further with frequency, here, in the wave-length expression, $c^4$ is nowhere to be found. We'll do the division nevertheless, for the sake of the discussion and will get
$$E = C T^5 \psi (\lambda T),$$
where $C = \frac{1}{c^4}$. Then we will multiply and divide the r.h.s. by $\lambda^5$ and will get
$$ E = \frac{C}{\lambda^5} (\lambda T)^5 \psi (\lambda T) $$
From the above we get
$$F(\lambda) = \frac{C}{\lambda^5}$$
and
$$e^{-\frac{c_1}{\lambda T}} = (\lambda T)^5 \psi (\lambda T).$$
So, the function $\psi(\lambda T)$ in \S6 of Planck's paper must be
$$\psi (\lambda T) = \frac{e^{-\frac{c_1}{\lambda T}} }{(\lambda T)^5}$$
in order for Planck's and Wien's expressions to be equivalent. The function $\psi$, however, doesn't constitute the whole exponent part, which is $(\lambda T)^5 \frac{e^{-\frac{c_1}{\lambda T}}} {(\lambda T)^5} = e^{-\frac{c_1}{\lambda T}}$, and that, indeed, makes the two expressions equivalent.