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The optical fiber acceptance angle is the largest launch half-angle admitted by the specified guide model. For a flat entrance face perpendicular to the optical axis, external refractive index $n_0$, and a straight step-index optical fiber, Snell's law at the entrance givesAt the core–cladding interface, the critical angle for total internal reflection $i_c$ satisfiesFor a meridional ray, strict total internal reflection requires $i>i_c$. At the limiting geometry $i=i_c$, the refracted light ray is tangent to the interface, soThis identifies the numerical aperture with an entrance cone under these assumptions. With $n_0=1$ for air, the conventional inclusive boundary condition isfor a nonnegative launch angle $\theta$; a practical launch stays inside the boundary. If $\mathrm{NA}>n_0$, this ideal model imposes no further restriction within the forward hemisphere, and $\arcsin (\mathrm{NA}/n_0)$ must not be used. The light ray must also intersect the optical fiber core: a suitable angle alone does not correct a misplaced beam.

For an original example, take $n_1=1.50$, $n_2=1.48$ and air at the entrance. ThenThe internal limiting angle is only ${\varphi }_{\max }\simeq 9.37^{\circ }$, while the critical angle for total internal reflection is $i_c\simeq 80.63^{\circ }$. A $10^{\circ }$ launch has $\varphi \simeq 6.65^{\circ }$ and $i\simeq 83.35^{\circ }$, so it satisfies total internal reflection. A $20^{\circ }$ launch has $\varphi \simeq 13.18^{\circ }$ and $i\simeq 76.82^{\circ }$, so it fails this test.

This cone is a geometrical optics result for meridional rays. A single-mode optical fiber requires the spatial overlap of the incident electromagnetic wave with its guided normal mode; its numerical aperture does not impose an exact angular cutoff on arbitrary beams. Diffraction, finite cladding, and bends require a physical optics description.^[ref]

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,At the cylindrical wall the angle of incidence is $i=\pi /2-\psi$. For $n_{\mathrm{inner}}>n_{\mathrm{outer}}$, the critical angle for total internal reflection satisfies $\sin i_c=n_{\mathrm{outer}}/n_{\mathrm{inner}}$. At the limiting accepted ray,Consequently the optical fiber acceptance angle obeysThe requested expression follows for incidence from air, $n_0\simeq 1$, 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 $\sqrt{n_{\mathrm{inner}}^2-n_{\mathrm{outer}}^2}>n_0$, 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, $n_{\mathrm{inner}}=1.6$, $n_{\mathrm{outer}}=1.1$, $n_0=1$ gives $\sqrt{1.35}>1$. Thus the unsaturated printed formula is not valid for every possible pair of refractive indices.

In the Huygens-Fresnel principle construction, after a time $\Delta t$ the incident disturbance advances $v_1\Delta t$ while the secondary disturbance in the second medium advances $v_2\Delta t$. The two right triangles share the segment along the interface. HenceHere both angles are measured from the normal vector, and $v_j=c/n_j$ are phase velocities. This is Snell's law. For $n_1>n_2$, no real transmitted angle exists whenThis is total internal reflection; equality is the limiting grazing geometry. The statement concerns an ideal transparent interface, not material absorption or losses at a nearby second boundary.

Let $\beta$ be the internal angle to the optical axis in the straight step-index optical fiber. At its flat entrance, Snell's law gives $\sin {\theta }_i=n_1\sin \beta$. At the side wall a meridional ray has angle of incidence $\pi /2-\beta$. The limiting total internal reflection condition is $\cos \beta =n_2/n_1$. Thus the optical fiber acceptance angle satisfiesThe numerical aperture expression assumes the square root is at most one for air; otherwise all forward launch angles are admitted by this ideal ray test. Strictly confined meridional rays lie inside the limiting cone.

For $n_1=1.48$ and $n_2=1.46$, $\mathrm{NA}=0.242487$ and ${\theta }_{i,\max }=14.0334^{\circ }$. A light ray at internal angle $\beta$ travels distance $L/\cos \beta$ along a guide of axial length $L$. In the nondispersive approximation its travel time is $t=n_{1}L/(c\cos \beta )$. ThereforeHere $L=1000\mathrm{m}$ and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$; using $3.00\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$ gives $5.001\mu \mathrm{s}$ and $4.950\mu \mathrm{s}$. The halving applies to the external launch angle, not to $\beta$. For actual short pulses the group refractive index $n_{\mathrm{g}}$ replaces the prefactor $n_1$ in $t=n_{\mathrm{g}}L/(c\cos \beta )$; the source supplies no optical dispersion data. The difference calculated here is geometrical intermodal dispersion in a nondispersive geometrical optics model.

Bending lowers the outer-wall angle of incidence, allowing a previously guided light ray to escape into the cladding. To calculate a definite meridional ray bend threshold, continue the on-axis launch assumption and take a meridional ray in the plane of the circular bend. Its initial internal angle to the centreline tangent obeys $\sin \beta =\sin 10^{\circ }/n_1$. Its straight path has distance $b_{\mathrm{ray}}=R\cos \beta$ from the bend centre. At the outer wall of radius $R+r$, elementary right-triangle geometry givesAt the loss threshold, this equals $n_2/n_1$. Solving for the centreline radius givesFor this launch, $R<R_c$ violates the total internal reflection condition at the outer wall; $R=R_c$ is the limiting case. A real optical fiber has gradual bend loss governed by its electromagnetic wave modes, rather than a universal abrupt cutoff.

Launch dependence. The final sentence does not separately specify the entry position or orientation of the bend relative to the incident light ray. For a planar launch displaced by $u$ from the centreline toward the outer wall, replace the numerator of $\sin i_{\mathrm{out}}$ by $(R+u)\cos \beta$. For example, a launch at the inner edge, $u=-r$, instead givesThus $14.93\mathrm{mm}$ is the result for the stated on-axis continuation; it is not an entry-independent threshold. No missing figure is needed for these explicitly specified ray models.

A step-index optical fiber has a nearly uniform refractive index $n_1$ throughout its optical fiber core, with an abrupt decrease to $n_2<n_1$ in its optical fiber cladding. In a straight, ideal guide, meridional rays stay in a plane containing the optical axis and trace a zigzag through repeated total internal reflection. This is a geometrical optics approximation; it is particularly useful for a multimode optical fiber whose core diameter is large compared with the wavelength.

The upper panel shows an accepted meridional ray; the lower panel enlarges the limiting geometry. The angle of incidence $i$ at the side wall is measured from its normal vector, whereas the internal ray angle $\varphi$ and external launch angle $\theta$ are measured from the optical axis. Thus $i+\varphi =90^{\circ }$. The limiting ray is a boundary of the ideal model, rather than a complete normal mode solution.