= Optical fiber acceptance angle
{title2=$\theta_{\max}$}
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 gives
$$
n_0\sin\theta=n_1\sin\phi.
$$
At the core–cladding interface, the <critical angle for total internal reflection> $i_c$ satisfies
$$
\sin i_c=\frac{n_2}{n_1},\qquad i=\frac{\pi}{2}-\phi.
$$
For 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, so
$$
n_0\sin\theta_{\max}
=n_1\cos i_c
=\sqrt{n_1^2-n_2^2}
\equiv\mathrm{NA}.
$$
This identifies the <numerical aperture> with an entrance cone under these assumptions. With $n_0=1$ for air, the conventional inclusive boundary condition is
$$
\sin\theta\leq\sqrt{n_1^2-n_2^2},
$$
for 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. Then
$$
\mathrm{NA}=\sqrt{2.25-2.1904}=\sqrt{0.0596}\simeq0.244131,
\qquad
\theta_{\max}=\arcsin(0.244131)\simeq14.13^\circ.
$$
The internal limiting <angle> is only $\phi_{\max}\simeq9.37^\circ$, while the <critical angle for total internal reflection> is $i_c\simeq80.63^\circ$. A $10^\circ$ launch has $\phi\simeq6.65^\circ$ and $i\simeq83.35^\circ$, so it satisfies <total internal reflection>. A $20^\circ$ launch has $\phi\simeq13.18^\circ$ and $i\simeq76.82^\circ$, 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.<When does NA provide a good estimate of the fiber's acceptance angle? by Thorlabs>[Insights into Optical Fiber, When does NA provide a good estimate of the fiber's acceptance angle?, Geometry Defines the Relationship and Single Mode Fibers are Different; revised January 20, 2020]{ref}
Back to article page
