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 06/08/14 1 1

Introduction 08/06/14 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. 06/08/14 2 2

BASIC PRINCIPLE 08/06/14                                                        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. 06/08/14 3 3

What Makes The Light Stay in Fiber 08/06/14 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 06/08/14 4 4

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 06/08/14 5 5

Refraction 08/06/14 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 06/08/14 6 6

Reflection 08/06/14 Light inside an optical fiber bounces off the cladding - reflection Reflection LED or LASER Source 06/08/14 7 7

Critical Angle 08/06/14 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 06/08/14 9 9

Angle of Incidence Also incident angle Measured from perpendicular 08/06/14 Also incident angle Measured from perpendicular Exercise: Mark two more incident angles Incident Angles 06/08/14 10 10

Angle of Reflection Also reflection angle Measured from perpendicular 08/06/14 Also reflection angle Measured from perpendicular Exercise: Mark the other reflection angle Reflection Angle 06/08/14 11 11

Reflection 08/06/14 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”. 06/08/14 12 12

Angle of Refraction Also refraction angle Measured from perpendicular 08/06/14 Also refraction angle Measured from perpendicular Exercise: Mark the other refraction angle Refraction Angle 06/08/14 13 13

Angle Summary Three important angles 08/06/14 Three important angles The reflection angle always equals the incident angle Refraction Angle Incident Angles Reflection Angle 06/08/14 14 14

Index of Refraction n = c/v c = velocity of light in a vacuum 08/06/14 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 06/08/14 15 15

Snell’s Law 08/06/14 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 06/08/14 16 16

Snell’s Law 08/06/14 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 06/08/14 17 17

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) 06/08/14 18 18

Snell’s Law (Example 1) 08/06/14 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° 06/08/14 19 19

Snell’s Law (Example 2) 08/06/14 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 = 1 ncore = 1.47 ncladding = 1.45 incident = 30° 06/08/14 20 20

Snell’s Law (Example 3) 08/06/14 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° 06/08/14 21 21

Critical Angle Calculation 08/06/14 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 06/08/14 22 22

NA Derivation 06/08/14 23

Acceptance Angle and NA 08/06/14 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 06/08/14 24 24

Numerical Aperture 08/06/14 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). 06/08/14 25 25

Acceptance Cone 08/06/14 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 06/08/14 26 26

Acceptance Cone 08/06/14 06/08/14 27 27

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Formula Summary 08/06/14 Index of Refraction Snell’s Law Critical Angle Acceptance Angle Numerical Aperture 06/08/14 29 29

08/06/14 Practice Problems 06/08/14 30 30

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?

core/cladding interface Practice Problems (2) 08/06/14 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 06/08/14 32 32

Refractive Indices and Propagation Times 08/06/14 06/08/14 33 33

Propagation Time Formula 08/06/14 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) 06/08/14 34 34

Temperature and Wavelength 08/06/14 Considerations for detailed analysis Fiber length is slightly dependent on temperature Refractive index is dependent on wavelength 06/08/14 35 35