"fig:"){width="\columnwidth"}]{}

The probability is $
{\cal P} = |f'(0)|^{-\frac 5 8} = |\frac \alpha{\beta^2}|^ {\frac 5 8}
$. Fig. \[f:2slitsprobaN\] shows a parametric plot of ${\cal P}$, and of the interaction energy, as function of $h/a$.

Consider now a small smooth object described by $z=x+ i y$, $0 < y \leq Y (x)$, away from the origin, i.e. $Y(0)=Y(\infty)=0$. If we find a function $f(t)$ with only positive fourier components $f_k$, such that $x=x(t)=\Re f(t)$, $Y(x)=\Im f(t)$ describes the boundary for $t$ real, then $f(z)=z+\int_{k>0} f_k e^{i k z}=z+{ \frac1\pi}\int_t \frac{Y(x(t))}{t-z} $ is the inverse uniformizing map. In an expansion in powers of $Y(x)$ [*and its derivatives*]{} one finds $f(z)=\tilde f(z)-\tilde f(0)$ with $\tilde f(z)=
z + {\frac1\pi}\int_t \frac{Y(t)}{t-z} - \frac{1}{{2} \pi^2} \int_{t,t'} Y'(t) Y'(t') (\frac{1}{t-z} + \frac{1}{t'-z}) \ln|t-t'| +...$. This yields the free energy $$\label{expansion}
{\cal F} = k T \frac{5}{8 \pi} \Big[ \int_t \frac{Y(t)}{t^2}  - \int_{t,t'} {\cal G}(t,t') Y'(t) Y'(t') +\ldots \Big]\ ,$$ where ${ 2} \pi {\cal G}(t,t')=(t^{-2}+t^{\prime -2}) \ln|t-t'| + { 1/(tt')}$. For a single object centered at position $a$, $Y(t)=h(t-a)$, the repulsive force $f_x=-\partial_a \cal F$ decays as $f_x \approx 5 k T A/(4 \pi a^3)$ at large distances, with a prefactor $A=\int_t Y(x(t))=\int_t h(t) 