8.2: Acceptance Angle
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In this section, we consider the problem of injecting light into a fiber optic cable. The problem is illustrated in Figure 8.2.1.

In this figure, we see light incident from a medium having index of refraction n0, with angle of incidence θi. The light is transmitted with angle of transmission θ2 into the fiber, and is subsequently incident on the surface of the cladding with angle of incidence θ3. For light to propagate without loss within the cable, it is required that
sinθ3≥ncnf
since this criterion must be met in order for total internal reflection to occur.
Now consider the constraint that Equation ??? imposes on θi. First, we note that θ3 is related to θ2 as follows:
θ3=π2−θ2
therefore
sinθ3=sin(π2−θ2)=cosθ2
so
cosθ2≥ncnf
Squaring both sides, we find:
cos2θ2≥n2cn2f
Now invoking a trigonometric identity:
1−sin2θ2≥n2cn2f
so:
sin2θ2≤1−n2cn2f
Now we relate the θ2 to θi using Snell’s law:
sinθ2=n0nfsinθi
so Equation ??? may be written:
n20n2fsin2θi≤1−n2cn2f
Now solving for sinθi, we obtain:
sinθi≤1n0√n2f−n2c
This result indicates the range of angles of incidence which result in total internal reflection within the fiber. The maximum value of θi which satisfies this condition is known as the acceptance angle θa, so:
θa≜arcsin(1n0√n2f−n2c)
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 θa with respect to the axis of the fiber.
The associated cone of acceptance is illustrated in Figure 8.2.2.

It is also common to define the quantity numerical aperture NA as follows:
NA≜1n0√n2f−n2c
Note that n0 is typically very close to 1 (corresponding to incidence from air), so it is common to see NA defined as simply √n2f−n2c. This parameter is commonly used in lieu of the acceptance angle in datasheets for fiber optic cable.
Typical values of nf and nc for an optical fiber are 1.52 and 1.49, respectively. What are the numerical aperture and the acceptance angle?
Solution
Using Equation ??? and presuming n0=1, we find NA ≅0.30_. Since sinθa= NA, we find θa=17.5∘_. Light must arrive from within 17.5∘ from the axis of the fiber in order to ensure total internal reflection within the fiber.