Technical and Application Notes

A Reference Guide to Optical Fibers and Light Guides

Optical Fiber Properties

Material

When selecting a fiber for a particular application, the user should know in advance what optical wavelength range will be required to transmit through the fiber. The material that makes up the core of the fiber can then be selected, based on it’s transmission characteristics. Generally speaking, if a material would not be acceptable as a lens for that wavelength range, it won’t be acceptable as a fiber core material, either.

For example, if you are working in the UV, around 300 to 400 nm, you will probably have to use a quartz or silica core fiber. Glass and plastic will not transmit very efficiently much below 400 nm. The cladding material is another consideration. Again, working in the UV, you want to be sure that the cladding of the fiber will not fluoresce. If the application calls for high energy levels, you must be aware of the effect on the cladding, which may be a polymer with a relatively low melting point. The cladding material has a direct effect on the acceptance angle of the fiber.

If you are working in the visible or Near Infra-Red, than you will probably be able to use a glass or a plastic fiber. These will transmit from about 400 nm to 1500 nm. They are considerable less expensive, easier to work with and consequently easier to obtain than UV. As with cladding materials, however, energy levels must be taken into account to avoid damage to the fiber.

Liquid Light Guides

Liquid Light Guides (LLGs) are becoming more and more popular for light collection and delivery roles. They are less expensive than glass, transmit well into the UV similar to quartz, and have a better acceptance angle than any optical fiber. Their downside is that they absorb the light that they do not transmit, and that can damage the liquid medium that they are filled with at high energy levels. Proper optical pre-filtering can eliminate this problem, however.

Back Reflection

Typically, approximately 4 % of the light striking a fiber is back reflected at the entrance and exit face of the fiber. This value can be reduced, over a fairly narrow bandwidth, to about 0.5 % with the help of Anti-Reflection coatings.

Attenuation

The transmission characteristics of a fiber are usually given in terms of attenuation for a given wavelength (or range), over a given distance (or length). You will probably read a specification that looks like this:

10.25dB/Km @ 0800 nm.

This reads as:

10.25 decibels per kilometer at 800 nanometers of wavelength.

There are three things you need to remember about attenuation:

    1. Attenuation is a loss per unit length.
    2. Attenuation can be converted to percent transmission for a given length
    3. Attenuation is wavelength specific.

To convert attenuation to transmission is simple. As an example, we will convert an attenuation specification of 10 dB/Km @ 500 nm to percent transmission over 1 meter.

We start by finding the attenuation in 1 meter.

10 [dB/Km] x 1/1000 [Km/m] = .01 dB/m

Now, the definition of a decibel of attenuation is:

dB = 10log(p2/p1)

where:
P2 = power level entering the fiber
P1 = power level exiting the fiber

This can be translated algebraically into:

(P1/P2) = 10–(dB/10)
or
Transmission (%) = 10–(dB/10)

so for our example:

(P1/P2) = 10–(.01/10) = .9977
or
99.77% transmission

You can use this formula to determine the transmission for your fiber at any wavelength you want to use it. Attenuation is wavelength specific and can change dramatically. Figure 2 illustrates a typical attenuation curve as a function of wavelength for a given fiber.

Packing Fraction

The packing fraction (or factor) of a fiber is the ratio of useable area (core) to total surface area (core + cladding + buffer).

This value determines what percentage of the light that is striking the face of the fiber will make it into the fiber core, and be transmitted. For instance, if you have a 100 micron diameter core and a 125 micron diameter fiber, your packing fraction will be:

Core Area

=

π (100/2)2

=

7853 µm2

= 64 %

Fiber Area

π (125/2)2

12272 µm2

In the above example, only a maximum of 64 % of the light striking the fiber has any chance of getting through, since only 64% of the end of the fiber is available for collecting light. This gets worse when the fibers are packed into bundles. The space between the individual fibers is also not transmissive, and contributes to a reduced packing fraction. This will obviously vary with fiber diameter, but can be conservatively estimated at approximately 20 %.

Minimum Bend Radius

The minimum bend radius of a fiber is the smallest bend that a fiber can undergo before it will fracture. The minimum bend radius for any fiber is determined by the diameter and the materials that it is fabricated from. The larger the fiber's diameter, the larger the minimum bend radius. Plastic fibers tend to have smaller minimum bend radii than glass. For information on the minimum bend radius of bundles, see the Technical Note Fiber Optic Bundles.

Input/Output Phenomenon

In theory, for a straight fiber whose end faces are perpendicular to its axis, a small collimated beam of rays entering the fiber at 25 degrees off axis will exit the fiber at 25 degrees off axis.

In practice, multiple internal reflections, fiber length, micro-bending (small external surface features) and manufacturing defects will spread collimated rays out to fill an annulus (or doughnut) of a cone whose total angle is twice that of the input angle (or 50 degrees, in our example). The thickness of the annulus will equal the diameter of the input beam. If the input beam is conical and focused on the face of the fiber, the output beam will still form an annulus of a cone whose nominal angle is twice the input angle, but the thickness of the annulus will now be determined by the angle of the input beam cone.

If the exit face is not perpendicular to the axis, the fiber becomes prism-like, and deflects the output off axis. The degree of deflection is based on the refractive indices of the fiber and the surrounding media (air, for instance). This deflection can be calculated as:

a - sin-1 [(n1 * sine b)/n0] - b

where:
a = angle of deflection, off-axis
b = angle of cut, from perpendicular
n1 = refractive index of fiber
n0 = refractive index of media surrounding fiber

For small angles, this can be approximated as:

a = b(n1 - n0)/n0

If the fiber is in air, then n0 is 1, and the approximation can be further reduced to:

a = b(n1 - 1)

Numerical Aperture

All fibers have a defined acceptance angle. The sine of the acceptance angle in air is known as the Numerical Aperture (NA). The NA of a fiber, and therefore the acceptance angle, is determined by the ratio of the refractive indices of the optical fiber core and its cladding. Rays entering a fiber at an angle greater than the NA will not be reflected internally, and will pass through, or be absorbed by, the cladding of the fiber. Either way, these rays are lost. Rays entering a fiber at an angle equal to or less than the acceptance angle will be reflected internally, and will propagate down the length of the fiber.

It is essential that the acceptance angle be carefully considered when designing with fiber optics. It is usually necessary to match the focal cone from the input or output device to the acceptance cone of the fiber to ensure optimum throughput through the system.

f/# = Y/X = beam dia./focal length

tan Θ = Y/2X

sine-1 (NA) = Θ = tan-1 [1/2(f/#)]

Figure 6 illustrates the relationship between the Numerical Aperture of optical fibers and the f/# of focusing systems. The chart below gives a quick reference to the relationship between some standard f/#s and NAs.

 

Acceptance angle (2Θ) 

NA

obb_fiber7.gif (2405 bytes)

f/1

51.13°

.45

f/2

28.07°

.24

f/4

14.25°

.12

f/8

7.15°

.06

A quick way to measure the acceptance angle of a fiber is back lighting as shown in Figure 7. If you send light into one end of a fiber using any illumination source that is larger than the fiber diameter, the only rays to pass through the fiber will be those within the acceptance angle. All rays will exit the fiber on the same angles at which they were input (see earlier section on input/output phenomenon). Therefore, all light exiting the fiber will be within a cone equal to the acceptance angle. The NA and the necessary f/# can be determined.

 

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