 Solution

With $f=3/4$,Hydrophobicity requires ${\theta }^{\ast }>90^{\circ }$, henceEquality is the ${\theta }^{\ast }=90^{\circ }$ boundary. A smooth-solid angle between about $70.53^{\circ }$ and $90^{\circ }$ can therefore become apparently hydrophobic if the suspended state persists.

For general $0<f\le 1$, the criterion is $\cos \theta <1/f-1$. For $1/2<f\le 1$, the threshold is ${\cos }^{-1}(1/f-1)$, reaching $90^{\circ }$ at $f=1$. For $f=1/2$, all $\theta >0$ satisfy it, with $\theta =0$ marginal. For $f<1/2$, the ideal equation predicts hydrophobicity for every original angle; formally ${\theta }^{\ast }\to 180^{\circ }$ as $f\to 0$.

These conclusions require retention of gas beneath the liquid. On intrinsically wettable textures the Cassie–Baxter state can be metastable. Pressure, condensation or defects may trigger invasion; penetration tests and contact-angle hysteresis distinguish that failure from the ideal suspended state. Roughness alone does not guarantee stable water repellency.