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06/08/141 Lecture 3 Light Propagation In Optical Fiber By: Mr. Gaurav Verma Asst. Prof. ECE, NIEC.

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Presentation on theme: "06/08/141 Lecture 3 Light Propagation In Optical Fiber By: Mr. Gaurav Verma Asst. Prof. ECE, NIEC."— Presentation transcript:

1 06/08/141 Lecture 3 Light Propagation In Optical Fiber By: Mr. Gaurav Verma Asst. Prof. ECE, NIEC

2 06/08/142 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.

3 06/08/143 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.

4 06/08/144 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

5 06/08/145 The Light is Refracted This end travels further than the other hand Lower Refractive index Region Higher Refractive index Region

6 06/08/146 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 LASER Source

7 06/08/147 Reflection  Light inside an optical fiber bounces off the cladding - reflection Reflection LED or LASER Source

8

9 06/08/149 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 90 o 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 n 1 Sin X=n 2 Sin90 o

10 06/08/1410 Angle of Incidence  Also incident angle  Measured from perpendicular  Exercise: Mark two more incident angles Incident Angles

11 06/08/1411 Angle of Reflection  Also reflection angle  Measured from perpendicular  Exercise: Mark the other reflection angle Reflection Angle

12 06/08/1412 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”.

13 06/08/1413 Angle of Refraction  Also refraction angle  Measured from perpendicular  Exercise: Mark the other refraction angle Refraction Angle

14 06/08/1414 Angle Summary Refraction Angle  Three important angles  The reflection angle always equals the incident angle Reflection Angle Incident Angles

15 06/08/1415 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

16 06/08/1416 Snell’s Law  The angles of the rays are measured with respect to the normal.  n 1 sin  1 =n 2 sin  2  Where  n 1 and n2 are refractive index of two materials   1 and  2 the angle of incident and refraction respectively

17 06/08/1417 Snell’s Law  The amount light is bent by refraction is given by Snell’s Law: n 1 sin  1 = n 2 sin  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

18 06/08/1418 Snell’s Law Normal Incidence Angle(  1 ) Refraction Angle(  2 ) Lower Refractive index(n 2 ) Higher Refractive index(n 1 ) Ray of light

19 06/08/1419 Snell’s Law (Example 1)  Calculate the angle of refraction at the air/core interface  Solution - use Snell’s law: n 1 sin  1 = n 2 sin  2  1sin(30°) = 1.47sin(  refraction )   refraction = sin -1 (sin(30°)/1.47)   refraction = 19.89° n air = 1 n core = 1.47 n cladding = 1.45  incident = 30°

20 06/08/1420 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° °) = 1.45sin(  refraction )   refraction = sin -1 (1.47sin(70.11°)/1.45)   refraction = 72.42° n air = 1 n core = 1.47 n cladding = 1.45  incident = 30°

21 06/08/1421 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 into cladding, it reflects back into the core (TIR) n air = 1 n core = 1.47 n cladding = 1.45  incident = 10°

22 06/08/1422 Critical Angle Calculation  The angle of incidence that produces an angle of refraction of 90° is the critical angle  n 1 sin(  c ) = n 2 sin(  °)  n 1 sin(  c ) = n 2   c = sin -1 (n 2 /n 1 )  Light at incident angles greater than the critical angle will reflect back into the core Critical Angle,  c n 1 = Refractive index of the core n 2 = Refractive index of the cladding

23 06/08/1423 NA Derivation

24 06/08/1424 Acceptance Angle and NA  The angle of light entering a fiber which follows the critical angle is called the acceptance angle,   = sin -1 [(n 1 2 -n 2 2 ) 1/2 ]  Numerical Aperature (NA) describes the light- gathering ability of a fiber NA = sin  Critical Angle,  c n 1 = Refractive index of the core n 2 = Refractive index of the cladding Acceptance Angle, 

25 06/08/1425 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).

26 06/08/1426 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, 

27 06/08/1427 Acceptance Cone

28 06/08/1428

29 06/08/1429 Formula Summary  Index of Refraction Snell’s Law Critical Angle Acceptance Angle Numerical Aperture

30 06/08/1430 Practice Problems

31  What happens to the light which approaches the fiber outside of the cone of acceptance? The angle of incidence is 30 o 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)

32 06/08/1432 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 n air = 1 n core = 1.46 n cladding = 1.43  incident = 12°

33 06/08/1433 Refractive Indices and Propagation Times

34 06/08/1434 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 seconds L = fiber length in meters n = refractive index of the fiber core c = speed of light (2.998 x 10 8 meters/second)

35 06/08/1435 Temperature and Wavelength  Considerations for detailed analysis  Fiber length is slightly dependent on temperature  Refractive index is dependent on wavelength


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