1. unit 1 INTRODUCTION TO OPTICAL FIBERS MODE THEORY OF OPTICAL FIBER COMMUNICATION

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5. r

9. ELECTROMAGNETIC MODE THEORY Propagation of light in optical waveguide: The Ray theory: to get a clear picture of the propagation of light inside the cable. The Mode theory: to understand the behavior of the light inside the cable (comprehending of the properties of absorption, attenuation and dispersion). MODE: EM WAVES TRAVELS IN A WAVEGUIDE WITH DIFFERENT SPEED

10. TE,TM & TEM MODES Transverse Electric mode (TE): Electric field is perpendicular to the direction of propagation, (Ez = 0), but a corresponding component of the magnetic field H in the direction of propagation(Z). Transverse Magnetic (TM) mode: A component of E field is in the direction of propagation(Z), but Hz=0. Modes with mode numbers; TEm and TMm Transverse Electro Magnetic (TEM) : Total field lies in the transverse plane both Ez and Hz are zero.

11. MODE THEORY FOR CIRCULAR WAVEGUIDES 7 To understand optical power propagation in fiber it is necessary to solve Maxwell’s Equation subject to cylindrical boundary conditions WhensolvingMaxwell’sequationsforhollowmetallicwaveguide,onlytransverse electric (TE) and transverse magnetic (TM) modes are found Inopticalfibers,thecorecladdingboundaryconditionsleadtoacoupling between electric and magnetic field components. This results in hybrid modes. Hybrid modes EH means (E is larger) or HE means H is large r

13. The equations 7 and 8 are known as standard wave equations. The modes can be found by solving the wave equation subjected to the core - cladding boundary condition.

20. Modes in Cylindrical Fibers –Weakly Guided Approximation

21. Linearly Polarized modes Fibers are constructed so that n 1 -n 2 << 1. The field components are called linearly polarized (LP) modes and are labeled LP jm where j and m designate mode solutions.

22. Linearly Polarized modes

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35. Single Mode fibers Only one mode of propagation Core diameter 8-12 μm and V = 2.4 Δ varies between 0.2 and 1.0 percent Core diameter must be just below the cut off of the first higher order mode LP 01 mode alone exists 0 <V<2.405

38. λ

43.  Total number of guided modes in graded index fiber is given by  M = ( α / α + 2 ) ( n 1 k a) 2 Δ  Normalized frequency  V = n 1 k a (2 Δ) 1/2  M = ( α / α + 2 ) (V 2 / 2)  For a parabolic refractive index profile α = 2  M = V 2 / 4

44. 10/7/2 020 optical fiber communication- session-3 44

45. μ μ 10/7/2 020 optical fiber communication- session-3 45

46. λΔ 10/7/2 020 optical fiber communication- session-3 46

49. μ 10/7/2 020 optical fiber communication- session-3 49

50. λπ√Δ) )= 1214 λ 10/7/2 020 optical fiber communication- session-3 50

51. 1. An optical fiber in air has an NA of 0.4. Compare the acceptance angle for meridional rays with that for skew rays which change direction by 100° at each reflection. Solution: The acceptance angle for meridional rays is given by NA = n 0 sin θ a = (n1 2 − n2 2 ) 1/2 For Air medium n 0 = 1 NA = θ a = sin −1 NA = sin −1 0.4 = 23.6° The skew rays change direction by 100° at each reflection, therefore γ = 50°.The acceptance angle for skew rays is: θ as = sin −1 (NA/Cos γ) = sin −1 (0.4/Cos 50°) = 38.5° In this example, the acceptance angle for the skew rays is about 15° greater than the corresponding angle for meridional rays PROBLEM

52. Determine the cutoff wavelength for a STEP INDEX fiber to exhibit SINGLE MODE operation when the core refractive index and radius are 1.46 and 4.5 μm with relative index difference being 0.25%

53. Exercise (1) A GI fiber with parabolic RI profile core has a RI at the core axis of 1.5 and RI difference Δ=1%.Estimate the maximum core diameter which allows single mode operation at λ = 1.3μm (2)Calculate the NA, Cut off parameter and number of modes supported by a fiber having n 1 = 1.54,n 2 = 1.5.core radius 25μm and operating wavelength 1300nm (3)Single mode step index fiber has core and cladding refractive indices of 1.498 and 1.495 respectively. Determine the core diameter required for the fiber to permit its operation over the wavelength range 1.48 and 1.6μm. (4)Silica optical fiber has a core index of 1.5 and cladding index of 1.47.Determine the acceptance angle in water(RI is 1.33)of the fiber (5)A step index fiber has n 1 = 1.5 and n 2 = 1.47.Determine the solid acceptance angle