= Solution
At the <thermodynamic critical point>, the two spinodal branches merge: $f''=0$ and $f^{(3)}=0$. The third <derivative> is
$$
\frac{f^{(3)}}{k_BT}=-\frac1{N\phi^2}+\frac1{(1-\phi)^2}.
$$
For $0<\phi<1$, its zero gives $1-\phi_c=\sqrt N\,\phi_c$. Substitution in the spinodal relation yields
$$
\boxed{\phi_c=\frac1{1+\sqrt N}\sim N^{-1/2},\qquad
\chi_c=\frac12\left(1+\frac1{\sqrt N}\right)^2
=\frac12+N^{-1/2}+\frac1{2N}.}
$$
An absolute <critical temperature> requires an interaction–<temperature> relation. Under the usual energetic approximation $\chi=C/T$, define the <theta temperature> $\Theta=2C$, at which an infinite chain has $\chi=1/2$. Then
$$
\boxed{T_c=\frac{\Theta}{(1+N^{-1/2})^2}
=\Theta\left[1-2N^{-1/2}+3N^{-1}+O(N^{-3/2})\right].}
$$
Thus $T_c$ tends to a nonzero constant, $T_c\sim N^0$, while $\Theta-T_c\sim N^{-1/2}$. It is the <temperature> deficit, not $T_c$ itself, that scales as $N^{-1/2}$.
More generally, if a smooth $\chi(T)$ obeys $\chi(\Theta)=1/2$ and $\chi'(\Theta)\ne0$, then
$$
T_c-\Theta=\frac{N^{-1/2}}{\chi'(\Theta)}+O(N^{-1}).
$$
This is the <critical point of a Flory–Huggins polymer solution>. The same leading chain-length exponent follows for a smooth <temperature> law, while its coefficient depends on the omitted <solvent> thermodynamics. The lattice interpretation connects the chain-length dependence of mixing <entropy> to the theta condition $\chi=1/2$.https://ocw.mit.edu/courses/3-063-polymer-physics-spring-2007/resources/lec3_07/[MIT Polymer Physics, lecture 3: mean-field lattice theory]{ref}.
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