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 I N A WAVEGUIDE W I T H
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 T H E O RY FOR C I R C U L A R
WAVEGUIDES
7
• To understand optical power propagation in fiber it is necessary to solve Maxwell’s
Equation subject to cylindrical boundary conditions
• When solving Maxwell’s equations for hollow metallic waveguide, only transverse
electric (TE) and transverse magnetic (TM) modes are found
• In optical fibers, the core cladding boundary conditions lead to a coupling
between electric and magnetic field components. This results in hybrid modes.
• Hybrid modes EH means (E is larger) or HE means H is larger
12.
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.
21. Linearly Polarized modes
• Fibers are constructed so that n1-n2 << 1. The field components are called linearly polarized (LP) modes
and are labeled LPjm where j and m designate mode solutions.
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
• LP01 mode alone exists
• 0 <V<2.405
43. Total number of guided modes in graded index fiber is
given by
M = (α / α + 2 ) ( n1 k a)2 Δ
Normalized frequency
V = n1 k a (2 Δ) 1/2
M = (α / α + 2 ) (V2 / 2)
For a parabolic refractive index profile α = 2
M = V2 / 4
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 = n0 sin θa = (n12 − n2 2 )1/2
For Air medium n0 = 1
NA = θa = 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:
θas = 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%
Solution
Given n1 = 1.46, a = 4.5 and Δ = 0.25%
We know that V = 2 𝝅 a(n1
2 – n2
2)1/2 ; where V = 2.405 (Single Mode Fiber)
λ
λ = 2πan1(2Δ)1/2
V
λ = 2 x πx4.5x1.46x(2x.25/100)1/2
2.405
= 1.214𝝁m.
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 n1 =
1.54,n2 = 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 n1 = 1.5 and n2 = 1.47 .Determine the solid acceptance angle
