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Engineering LibreTexts

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.

m0192_fAA.png
Figure 8.2.1: 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 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

sinθ3ncnf

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=π2θ2

therefore

sinθ3=sin(π2θ2)=cosθ2

so

cosθ2ncnf

Squaring both sides, we find:

cos2θ2n2cn2f

Now invoking a trigonometric identity:

1sin2θ2n2cn2f

so:

sin2θ21n2cn2f

Now we relate the θ2 to θi using Snell’s law:

sinθ2=n0nfsinθi

so Equation ??? may be written:

n20n2fsin2θi1n2cn2f

Now solving for sinθi, we obtain:

sinθi1n0n2fn2c

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:

θaarcsin(1n0n2fn2c)

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.

m0192_fAcceptanceAngle.png
Figure 8.2.2: Cone of acceptance. ( CC BY-SA 4.0)

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

NA1n0n2fn2c

Note that n0 is typically very close to 1 (corresponding to incidence from air), so it is common to see NA defined as simply n2fn2c. This parameter is commonly used in lieu of the acceptance angle in datasheets for fiber optic cable.

Example 8.2.1: Acceptance angle

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.


This page titled 8.2: Acceptance Angle is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Steven W. Ellingson (Virginia Tech Libraries' Open Education Initiative) via source content that was edited to the style and standards of the LibreTexts platform.

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