= Solution
For unresolved, uniformly bright <blackbodies> at a common distance $d$, the monochromatic observed <radiative fluxes> are $F_{s,\lambda}=\pi B_\lambda(T_s)R_s^2/d^2$ and $F_{p,\lambda}=\pi B_\lambda(T_p)R_p^2/d^2$. Define the planet-star contrast $r_\lambda=F_{p,\lambda}/F_{s,\lambda}$. The <Planck law> gives
$$
r_\lambda=\left(\frac{R_p}{R_s}\right)^2\frac{e^{hc/(\lambda k_BT_s)}-1}{e^{hc/(\lambda k_BT_p)}-1}.
$$
Outside the <exoplanet secondary eclipse>, the system flux is $F_s+F_p$; during complete occultation it is $F_s$. Consequently the exact <thermal eclipse depth> normalized to the out-of-eclipse light is
$$
\boxed{d_\lambda=\frac{F_{p,\lambda}}{F_{s,\lambda}+F_{p,\lambda}}=\frac{r_\lambda}{1+r_\lambda}.}
$$
Only in the faint-planet limit is $d_\lambda\simeq r_\lambda$. Reflected starlight, partial occultation and spatial temperature variations are omitted under the stated <blackbody> model.
For a cooler <hot Jupiter> with $T_p<T_s$, the planet-star contrast is exponentially small at short wavelengths and increases towards a constant plateau. This is the <monotonicity of blackbody planet-star contrast>, rather than the peaked shape of the planet's absolute <blackbody> flux. An illustrative example uses $T_p=1500\,\mathrm K$, $T_s=5800\,\mathrm K$ and $R_p/R_s=0.1$:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-59-contrast.png]
{title=Blackbody hot-Jupiter planet-star contrast increasing towards its Rayleigh-Jeans plateau}
At wavelengths long enough for the <Rayleigh-Jeans law> to apply to both bodies, $B_\lambda\simeq2ck_BT/\lambda^4$. Therefore
$$
\boxed{r_\lambda\longrightarrow r_\infty=\left(\frac{R_p}{R_s}\right)^2\frac{T_p}{T_s},\qquad d_\lambda\longrightarrow\frac{r_\infty}{1+r_\infty}.}
$$
The common $\lambda^{-4}$ dependence cancels. In the illustrative case, the contrast plateau is about $2.6\times10^{-3}$, or $2600$ parts per million.
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