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 $\PageIndex{1}$.

Figure $\PageIndex{1}$: Injecting light into a fiber optic cable. ( CC BY-SA 4.0; S. Lally)

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

$$\sin\theta_3 \ge \frac{n_c}{n_f} \label{m0192_eCAnm}$$

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

Now consider the constraint that Equation $\ref{m0192_eCAnm}$ imposes on $\theta^i$. First, we note that $\theta_3$ is related to $\theta_2$ as follows:

$$\theta_3 = \frac{\pi}{2} - \theta_2 \nonumber$$

therefore

$$\begin{align} \sin\theta_3 &= \sin\left(\frac{\pi}{2} - \theta_2\right) \\ &= \cos\theta_2 \end{align}$$

so

$$\cos\theta_2 \ge \frac{n_c}{n_f} \nonumber$$

Squaring both sides, we find:

$$\cos^2\theta_2 \ge \frac{n_c^2}{n_f^2} \nonumber$$

Now invoking a trigonometric identity:

$$1-\sin^2\theta_2 \ge \frac{n_c^2}{n_f^2} \nonumber$$

so:

$$\sin^2\theta_2 \le 1-\frac{n_c^2}{n_f^2} \label{m0192_e1}$$

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

$$\sin\theta_2 = \frac{n_0}{n_f}\sin\theta^i \nonumber$$

so Equation $\ref{m0192_e1}$ may be written:

$$\frac{n_0^2}{n_f^2}\sin^2\theta^i \le 1-\frac{n_c^2}{n_f^2} \nonumber$$

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

$$\sin\theta^i \le \frac{1}{n_0}\sqrt{ n_f^2 - n_c^2 } \nonumber$$

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:

$$\theta_a \triangleq \arcsin\left(\frac{1}{n_0}\sqrt{ n_f^2 - n_c^2 }\right) \nonumber$$

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 $\PageIndex{2}$.

Figure $\PageIndex{2}$: Cone of acceptance. ( CC BY-SA 4.0)

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

$$\mbox{NA} \triangleq \frac{1}{n_0}\sqrt{ n_f^2 - n_c^2 } \label{m0192_eNA}$$

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 $\PageIndex{1}$: 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 $\ref{m0192_eNA}$ 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.