= Oil lens capillary balance
{title2=$h^2=2\Gamma/((\rho_w-\rho_o)g)$}
For a broad, thin oil lens below a flat ice-water interface, define the effective interfacial energy $\Gamma=\gamma_{io}+\gamma_{ow}-\gamma_{iw}$ from the relevant <surface tensions>. At fixed oil volume $\mathcal V=Ah$, the energy is $E=\Gamma\mathcal V/h+\frac12(\rho_w-\rho_o)g\mathcal Vh$. Setting its <derivative> to zero gives $h^2=2\Gamma/[(\rho_w-\rho_o)g]$ for $\Gamma>0$, and the positive <second derivative> proves a minimum. A contact angle measured through the oil gives $\Gamma=\gamma_{ow}(1-\cos\vartheta)$. Oil-ice tension alone does not determine this effective quantity. Complete wetting or rough underside pockets invalidate this finite, flat-lens approximation. With densities $1025$ and $850\,\mathrm{kg\,m^{-3}}$, a lens below $1\,\mathrm{cm}$ requires $0<\Gamma<0.08584\,\mathrm{N\,m^{-1}}$.
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