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Lecture 3

Light Propagation In Optical Fiber

By:

Mr. Gaurav VermaAsst. Prof. ECE, NIEC

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Introduction

An optical fiber is a very thin strand of silica glass in geometry quite like a human hair. In reality it is a very narrow, very long glass cylinder with special characteristics. When light enters one end of the fiber it travels (confined within the fiber) until it leaves the fiber at the other end. Two critical factors stand out: Very little light is lost in its journey along the fiber Fiber can bend around corners and the light will stay within it and

be guided around the corners.

An optical fiber consists of two parts: the core and the cladding. The core is a narrow cylindrical strand of glass and the cladding is a tubular jacket surrounding it. The core has a (slightly) higher refractive index than the cladding. This means that the boundary (interface) between the core and the cladding acts as a perfect mirror. Light traveling along the core is confined by the mirror to stay within it - even when the fiber bends around a corner.

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BASIC PRINCIPLE

When a light ray travelling in one material hits a different material and reflects back into the original material without any loss of light, total internal reflection is said to occur. Since the core and cladding are constructed from different compositions of glass, theoretically, light entering the core is confined to the boundaries of the core because it reflects back whenever it hits the cladding. For total internal reflection to occur, the index of refraction of the core must be higher than that of the cladding, and the incidence angle is larger than the critical angle.

                                  

                    

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What Makes The Light Stay in Fiber

Refraction The light waves spread out along its beam. Speed of light depend on the material used called refractive

index.Speed of light in the material = speed of light in the free

space/refractive index

Lower refractive index higher speed

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The Light is Refracted

This end travels further than the other hand

Lower Refractive index Region

Higher Refractive index Region

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Refraction

When a light ray encounters a boundary separating two different media, part of the ray is reflected back into the first medium and the remainder is bent (or refracted) as it enters the second material. (Light entering an optical fiber bends in towards the center of the fiber – refraction)Refraction

LED or LASE

R Source

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Reflection

Light inside an optical fiber bounces off the cladding - reflection

Reflection

LED or LASER Source

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Critical Angle If light inside an optical fiber strikes the cladding too

steeply, the light refracts into the cladding - determined by the critical angle. (There will come a time when, eventually, the angle of refraction reaches 90o and the light is refracted along the boundary between the two materials. The angle of incidence which results in this effect is called the critical angle).

Critical Angle

n1Sin X=n2Sin90o

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Angle of Incidence

Also incident angle Measured from perpendicular Exercise: Mark two more incident angles

Incident Angles

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Angle of Reflection

Also reflection angle Measured from perpendicular Exercise: Mark the other reflection angle

Reflection Angle

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Reflection

Thus light is perfectly reflected at an interface between two materials of different refractive index if:

The light is incident on the interface from the side of higher refractive index.

The angle θ is greater than a specific value called the “critical angle”.

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Angle of Refraction

Also refraction angle Measured from perpendicular Exercise: Mark the other refraction angle

Refraction Angle

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Angle Summary

Refraction Angle

Three important angles The reflection angle always equals the incident angle

Reflection Angle

Incident Angles

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Index of Refraction

n = c/v c = velocity of light in a vacuum v = velocity of light in a specific

medium light bends as it passes from one medium

to another with a different index of refraction air, n is about 1 glass, n is about 1.4

Light bends in towards normal - lower n to higher n

Light bends away from normal - higher n to lower n

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Snell’s Law

The angles of the rays are measured with respect to the normal.

n1sin 1=n2sin 2

Where n1 and n2 are refractive index of two materials

1and 2 the angle of incident and refraction respectively

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Snell’s Law

The amount light is bent by refraction is given by Snell’s Law: n1sin1 = n2sin2

Light is always refracted into a fiber (although there will be a certain amount of Fresnel reflection)

Light can either bounce off the cladding (TIR) or refract into the cladding

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Snell’s Law

Normal

Incidence Angle(1)

Refraction Angle(2)

Lower Refractive index(n2)

Higher Refractive index(n1)Ray of light

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Snell’s Law (Example 1)

Calculate the angle of refraction at the air/core interface

Solution - use Snell’s law: n1sin1 = n2sin2

1sin(30°) = 1.47sin(refraction)

refraction = sin-1(sin(30°)/1.47)

refraction = 19.89°

nair = 1ncore = 1.47ncladding = 1.45incident = 30°

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Snell’s Law (Example 2)

Calculate the angle of refraction at the core/cladding interface Solution - use Snell’s law and the refraction angle from Example

3.1 1.47sin(90° - 19.89°) = 1.45sin(refraction)

refraction = sin-1(1.47sin(70.11°)/1.45)

refraction = 72.42°

nair = 1ncore = 1.47ncladding = 1.45incident = 30°

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Snell’s Law (Example 3)

Calculate the angle of refraction at the core/cladding interface for the new data below

Solution: 1sin(10°) = 1.45sin(refraction(core))

refraction(core) = sin-1(sin(10°)/1.45) = 6.88°

1.47sin(90°-6.88°) = 1.45sin(refraction(cladding))

refraction(cladding) = sin-1(1.47sin(83.12°)/1.45)= sin-1(1.0065) = can’t do

light does not refract intocladding, it reflects backinto the core (TIR)

nair = 1ncore = 1.47ncladding = 1.45incident = 10°

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Critical Angle Calculation

The angle of incidence that produces an angle of refraction of 90° is the critical angle n1sin(c) = n2sin(°)

n1sin(c) = n2

c = sin-1(n2 /n1)

Light at incident anglesgreater than the criticalangle will reflect backinto the core

Critical Angle, c

n1 = Refractive index of the coren2 = Refractive index of the cladding

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NA Derivation

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Acceptance Angle and NA

The angle of light entering a fiber which follows the critical angle is called the acceptance angle,

= sin-1[(n12-n2

2)1/2]

Numerical Aperature (NA)describes the light-gathering ability of a fiber

NA = sin

Critical Angle, c

n1 = Refractive index of the coren2 = Refractive index of the cladding

Acceptance Angle,

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Numerical Aperture

The Numerical Aperture is the sine of the largest angle contained within the cone of acceptance.

NA is related to a number of important fiber characteristics. It is a measure of the ability of the fiber to gather light at the

input end. The higher the NA the tighter (smaller radius) we can have

bends in the fiber before loss of light becomes a problem. The higher the NA the more modes we have, Rays can bounce

at greater angles and therefore there are more of them. This means that the higher the NA the greater will be the dispersion of this fiber (in the case of MM fiber).

Thus higher the NA of SM fiber the higher will be the attenuation of the fiber

Typical NA for single-mode fiber is 0.1. For multimode, NA is between 0.2 and 0.3 (usually closer to 0.2).

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Acceptance Cone

There is an imaginary cone of acceptance with an angle

The light that enters the fiber at angles within the acceptance cone are guided down the fiber core

Acceptance Cone

Acceptance Angle,

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Acceptance Cone

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Formula Summary

Index of Refraction

Snell’s Law

Critical Angle

Acceptance Angle

Numerical Aperture

n=cv

n 1 sin θ 1=n2 sin θ 2

θc=sin−1( n2n1 )

α= sin−1 (√n12−n22)

NA= sin α= √n12−n22

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Practice Problems

What happens to the light which approaches the fiber outside of the cone of acceptance? The angle of incidence is 30o as in Fig.1 (calculate the angle of refraction at the air/core interface, r/ critical angle, c/ incident angle at the core/cladding interface, i/) does the TIR will occur?

Practice Problems (1)

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Practice Problems (2)

Calculate:angle of refraction at the air/core interface, r

critical angle , c

incident angle at the core/cladding interface , i

Will this light ray propagate down the fiber?

air/core interface

core/cladding interface

Answers:r = 8.2°c = 78.4°i = 81.8°light will propagate

nair = 1ncore = 1.46ncladding = 1.43incident = 12°

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Refractive Indices and Propagation Times

0

1

2

3

4

5

6

Vacuum 1 3.336

Air 1.003 3.346

Water 1.333 4.446

Fused Silica 1.458 4.863

Belden Cable (RG-59/U)

N/A 5.551

Refractive Index

Propagation Time (ns/m)

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Propagation Time Formula

Metallic cable propagation delay cable dimensions frequency

Optical fiber propagation delay related to the fiber material

formula

t = Ln/c t = propagation delay in secondsL = fiber length in metersn = refractive index of the fiber corec = speed of light (2.998 x 108 meters/second)

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Temperature and Wavelength

Considerations for detailed analysis Fiber length is slightly dependent on temperature Refractive index is dependent on wavelength