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Optical Fiber Construction and Attenuation, Lecture notes of Optics

The construction of optical fibers and the causes of intrinsic and extrinsic attenuation. It covers topics such as total internal reflection, numerical aperture, critical angle, and Rayleigh scattering. The document also mentions the use of Optical Time Domain Reflectometers (OTDR) to measure fiber loss.

Typology: Lecture notes

2021/2022

Uploaded on 05/11/2023

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w.wang
Multilayer Structure
Wei-Chih Wang
Department of Mecahnical Engineering
University of Washington
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Download Optical Fiber Construction and Attenuation and more Lecture notes Optics in PDF only on Docsity!

w.wang

Multilayer Structure

Wei-Chih Wang

Department of Mecahnical Engineering

University of Washington

Reading Materials

• Lecture notes• Additional reading provided in classw.wang

Optical Fiber Application

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Medicine

Communication

Fiber optic Technology has grown tremendously over the yearsand toda can be found in many places. Some of applications offiber optics are shown below

Sensors

Optical Fiber Construction

Single fiber constructionw.wang

Total Internal Reflection

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Critical angle: Angles 01 and 02 are related by:N sin 1 1

= N^ sin^2

(Snell’s Law)^2

When

= 90, 01 is called the Critical Angle^2

oN^ sin^1

^ = N^1

sin90, sin90 = 1 2

osin^ 

c = N

/ N^21

For incident angles greater that the critical angle, total internal reflection occurs

Light travels down the fiber in a pathway called a light guidew.wang

While discussing step-index fibers, we considered light propagation inside the fiber as aset of many rays bouncing back and forth at the core-cladding interface. There the anglew.wang

could take a continuum of values lying between 0 and cos

–1( n / n^2

), i.e., 1 Scientific and Technological Education

in Photonics

0 <^ ^ < cos

–1^ ( n / n^2

For^ n^2

= 1.5 and

 ^

= 0.01, we would get

n / n^^21

~^ and cos

-^

= 8.1°, so

0 <^ < 8.1°

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NA derivation

We know

and

Since Assume the

NA

is the half angle of the acceptance cone,

we get

sin

=(nNA

2 -n 12

2 1/2^ )

= n^1

sqrt(

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We define a parameter w.wang

through the following equations.

When

^ << 1 (as is indeed true for silica fibers where

n is very nearly equal to^1

n ) we may write^2

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Single mode fiber critical angle <

o

Multimode fiber critical angle <

o

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In a short length of an optical fiber, if all rays betweenw.wang

i^ = 0 and

i are launched, them^

light coming out of the fiber will also appear as a cone of half-angle

i emanating from m^

the fiber end. If we now allow this beam to fall normally on a white paper and measureits diameter, we can easily calculate the

NA^ of the fiber.

For a typical step-index (multimode) fiber with

n ^ 1.45 and^1

 ^ 0.01, we get

so that

i ^ 12°. Thus, all light entering the fiber must be within a cone of m^ half-angle 12°.

Example

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Attenuation w.wang

dB is a ratio of the power received verses the power transmitted Loss (dB) = 10log (power transmitted / power received)

Intrinsic attenuation is controlled by the fiber manufacturer Absorption caused by water molecules and other impurities Light strikes a molecule at the right angle and light energy is converted into heat Absorption accounts for 3-5% of fiber attenuation these is near the theoretical limitw.wang