= Solution
<Snell's law> is $n_a\sin\theta_a=n_b\sin\theta_b$, with both <angles> measured from the interface <normal vector>. Let $\psi$ be the <meridional ray>'s angle to the <optical axis> inside the core and let $n_0$ be the incident medium's <refractive index>. At the entrance face,
$$
n_0\sin\theta=n_{\rm inner}\sin\psi.
$$
At the cylindrical wall the <angle of incidence> is $i=\pi/2-\psi$. For $n_{\rm inner}>n_{\rm outer}$, the <critical angle for total internal reflection> satisfies $\sin i_c=n_{\rm outer}/n_{\rm inner}$. At the limiting accepted ray,
$$
\sin\psi_{\max}=\cos i_c=\sqrt{1-\frac{n_{\rm outer}^2}{n_{\rm inner}^2}}.
$$
Consequently the <optical fiber acceptance angle> obeys
$$
\boxed{n_0\sin\theta_{\max}=\sqrt{n_{\rm inner}^2-n_{\rm outer}^2}.}
$$
The requested expression follows for incidence from air, $n_0\simeq1$, provided the right-hand side is at most one. Strict <total internal reflection> occurs inside this limiting cone. The derivation uses the standard <meridional ray> model; skew launches need additional geometry.
If $\sqrt{n_{\rm inner}^2-n_{\rm outer}^2}>n_0$, every forward entrance angle allowed by this ideal model meets the side-wall condition, and the limiting half-angle is $\pi/2$, rather than an undefined <inverse sine>. For example, $n_{\rm inner}=1.6$, $n_{\rm outer}=1.1$, $n_0=1$ gives $\sqrt{1.35}>1$. Thus the unsaturated printed formula is not valid for every possible pair of <refractive indices>.
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