1. Lecture 3 Light Propagation In Optical Fiber
08/06/14
Lecture 3 Light Propagation In Optical Fiber
Course web page contains PowerPoint lecture notes, homework assignments, links to web sites, announcement and other course material.
By:
Mr. Gaurav Verma
Asst. Prof.
ECE, NIEC
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2. Introduction
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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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3. BASIC PRINCIPLE
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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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4. What Makes The Light Stay in Fiber
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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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5. The Light is Refracted Higher Refractive index Region 08/06/14
Lower Refractive index Region
This end travels further than the other hand
Higher Refractive index Region
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6. Refraction
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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 LASER Source
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7. Reflection
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Light inside an optical fiber bounces off the cladding - reflection
Reflection
LED or LASER Source
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9. Critical Angle
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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).
n1Sin X=n2Sin90o
Critical Angle
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10. Angle of Incidence Also incident angle Measured from perpendicular
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Also incident angle
Measured from perpendicular
Exercise: Mark two more incident angles
Incident Angles
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11. Angle of Reflection Also reflection angle Measured from perpendicular
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Also reflection angle
Measured from perpendicular
Exercise: Mark the other reflection angle
Reflection Angle
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12. Reflection
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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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13. Angle of Refraction Also refraction angle Measured from perpendicular
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Also refraction angle
Measured from perpendicular
Exercise: Mark the other refraction angle
Refraction Angle
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14. Angle Summary Three important angles
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Three important angles
The reflection angle always equals the incident angle
Refraction Angle
Incident Angles
Reflection Angle
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15. Index of Refraction n = c/v c = velocity of light in a vacuum
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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 away from normal - higher n to lower n
Light bends in towards normal - lower n to higher n
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16. Snell’s Law
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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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17. Snell’s Law
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The amount light is bent by refraction is given by Snell’s Law: n1sin1 = n2sin2
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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18. Snell’s Law 08/06/14 Normal Refraction Angle(2) Ray of light
Lower Refractive index(n2)
Ray of light
Higher Refractive index(n1)
Incidence Angle(1)
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19. Snell’s Law (Example 1)
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Calculate the angle of refraction at the air/core interface
Solution - use Snell’s law: n1sin1 = n2sin2
1sin(30°) = 1.47sin(refraction)
refraction = sin-1(sin(30°)/1.47)
refraction = 19.89°
nair = 1
ncore = 1.47
ncladding = 1.45
incident = 30°
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20. Snell’s Law (Example 2)
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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° °) = 1.45sin(refraction)
refraction = sin-1(1.47sin(70.11°)/1.45)
refraction = 72.42°
nair = 1
ncore = 1.47
ncladding = 1.45
incident = 30°
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21. Snell’s Law (Example 3)
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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 into cladding, it reflects back into the core (TIR)
nair = 1
ncore = 1.47
ncladding = 1.45
incident = 10°
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22. Critical Angle Calculation
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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 angles greater than the critical angle will reflect back into the core
n1 = Refractive index of the core
n2 = Refractive index of the cladding
Critical Angle, c
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23. NA Derivation
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24. Acceptance Angle and NA
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The angle of light entering a fiber which follows the critical angle is called the acceptance angle,   = sin-1[(n12-n22)1/2]
Numerical Aperature (NA) describes the light- gathering ability of a fiber NA = sin
n1 = Refractive index of the core
n2 = Refractive index of the cladding
Acceptance Angle, 
Critical Angle, c
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25. Numerical Aperture
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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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26. Acceptance Cone
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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 Angle, 
Acceptance Cone
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27. Acceptance Cone
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29. Formula Summary
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Index of Refraction Snell’s Law Critical Angle Acceptance Angle Numerical Aperture
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30. 08/06/14
Practice Problems
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31. Practice Problems (1)
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?

32. core/cladding interface
Practice Problems (2)
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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
nair = 1
ncore = 1.46
ncladding = 1.43
incident = 12°
Answers:
r = 8.2°
c = 78.4°
i = 81.8°
light will propagate
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33. Refractive Indices and Propagation Times
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34. Propagation Time Formula
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Metallic cable propagation delay
cable dimensions
frequency
Optical fiber propagation delay
related to the fiber material
formula t = Ln/c
t = propagation delay in seconds
L = fiber length in meters
n = refractive index of the fiber core
c = speed of light (2.998 x 108 meters/second)
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35. Temperature and Wavelength
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Considerations for detailed analysis
Fiber length is slightly dependent on temperature
Refractive index is dependent on wavelength
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