8.2 Acceptance Angle

In this section, we consider the problem of injecting light into a fiber optic cable. The problem is illustrated in Figure 8.3.

A plane wave travels through a medium (n subscript zero) at an angle theta superscript i before entering a fiber optic cable at its center. The wave travels through the fiber (n subscript f) at an angle theta subscript 2 and hits the edge at an angle of theta subscript 3 measured from the vertical. The wave reflects back into the fiber. Area outside the fiber is n subscript c.
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Figure 8.3. Injecting light into a fiber optic cable. © S. Lally CC BY-SA 4.0

In this figure, we see light incident from a medium having index of refraction \(n_0\), with angle of incidence \(\theta^i\). The light is transmitted with angle of transmission \(\theta_2\) into the fiber, and is subsequently incident on the surface of the cladding with angle of incidence \(\theta_3\). For light to propagate without loss within the cable, it is required that

\begin{equation} \sin\theta_3 \ge \frac{n_c}{n_f} \tag{8.5}\end{equation}

since this criterion must be met in order for total internal reflection to occur.

Now consider the constraint that Equation 8.5 imposes on \(\theta^i\). First, we note that \(\theta_3\) is related to \(\theta_2\) as follows:

\begin{equation} \theta_3 = \frac{\pi}{2} - \theta_2 \tag{8.6}\end{equation}

therefore

\begin{flalign} \sin\theta_3 &= \sin\left(\frac{\pi}{2} - \theta_2\right) \tag{8.7} \\ &= \cos\theta_2 \tag{8.8}\end{flalign}

so

\begin{equation} \cos\theta_2 \ge \frac{n_c}{n_f} \tag{8.9}\end{equation}

Squaring both sides, we find:

\begin{equation} \cos^2\theta_2 \ge \frac{n_c^2}{n_f^2} \tag{8.10}\end{equation}

Now invoking a trigonometric identity:

\begin{equation} 1-\sin^2\theta_2 \ge \frac{n_c^2}{n_f^2} \tag{8.11}\end{equation}

so:

\begin{equation} \sin^2\theta_2 \le 1-\frac{n_c^2}{n_f^2} \tag{8.12}\end{equation}

Now we relate the \(\theta_2\) to \(\theta^i\) using Snell’s law:

\begin{equation} \sin\theta_2 = \frac{n_0}{n_f}\sin\theta^i \tag{8.13}\end{equation}

so Equation 8.12 may be written:

\begin{equation} \frac{n_0^2}{n_f^2}\sin^2\theta^i \le 1-\frac{n_c^2}{n_f^2} \tag{8.14}\end{equation}

Now solving for \(\sin\theta^i\), we obtain:

\begin{equation} \sin\theta^i \le \frac{1}{n_0}\sqrt{ n_f^2 - n_c^2 } \tag{8.15}\end{equation}

This result indicates the range of angles of incidence which result in total internal reflection within the fiber. The maximum value of \(\theta^i\) which satisfies this condition is known as the acceptance angle \(\theta_a\), so:

\begin{equation} \theta_a \triangleq \arcsin\left(\frac{1}{n_0}\sqrt{ n_f^2 - n_c^2 }\right) \tag{8.16}\end{equation}

This leads to the following insight:

In order to effectively launch light in the fiber, it is necessary for the light to arrive from within a cone having half-angle \(\theta_a\) with respect to the axis of the fiber.

The associated cone of acceptance is illustrated in Figure 8.4.

A cross-section of optical fiber with exterior cladding. The cone of acceptance appears as a triangle centered around the axis of the cable. The maximum angle from center is theta subscript a.
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Figure 8.4. Cone of acceptance. © S. Lally CC BY-SA 4.0

It is also common to define the quantity numerical aperture NA as follows:

\begin{equation} \mbox{NA} \triangleq \frac{1}{n_0}\sqrt{ n_f^2 - n_c^2 } \tag{8.17}\end{equation}

Note that \(n_0\) is typically very close to \(1\) (corresponding to incidence from air), so it is common to see NA defined as simply \(\sqrt{ n_f^2 - n_c^2 }\). This parameter is commonly used in lieu of the acceptance angle in datasheets for fiber optic cable.

 

Example 8.2

Acceptance angle.
 
Typical values of \(n_f\) and \(n_c\) for an optical fiber are 1.52 and 1.49, respectively. What are the numerical aperture and the acceptance angle?

Solution. Using Equation 8.17 and presuming \(n_0=1\), we find NA \(\cong \underline{0.30}\). Since \(\sin\theta_a =\) NA, we find \(\theta_a=\underline{17.5^{\circ}}\). Light must arrive from within \(17.5^{\circ}\) from the axis of the fiber in order to ensure total internal reflection within the fiber.

Additional Reading: