Step-index optical fiber

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.

The dimensionless waveguide parameter $V$ combines radius, vacuum wavelength and the numerical aperture factor. In the weak-guidance scalar model the first additional spatial family appears at $V=j_{0,1}=2.404825\dots$, the first positive zero of a Bessel function of the first kind. The fundamental family has no positive cutoff. Its two polarizations are not two distinct spatial profiles.

For a light ray launched on the centreline in the plane of a circular optical fiber bend, let $R$ be centreline radius, $r$ core radius and $\beta$ the internal launch angle from the tangent. The straight segment's distance from the bend centre is $R\cos \beta$. Thus its outer-wall angle of incidence satisfies $\sin i=R\cos \beta /(R+r)$. Equating this to $n_2/n_1$ gives the limiting total internal reflection threshold in the title, provided $n_1\cos \beta >n_2$. Radially displaced or nonplanar launches need different geometry; this is not a universal wave-optical loss cutoff.

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]