• To analyze the optical fiber propagation mechanism within a fiber, Maxwell equations are to solve subject to the cylindrical boundary conditions at core-cladding interface. The core-cladding boundary conditions lead to coupling of electric and magnetic field components resulting in hybrid modes. Hence the analysis of optical waveguide is more complex than metallic hollow waveguide analysis.
• Depending on the large E-field, the hybrid modes are HE or EH The two lowest order does are HE11 and TE01.

### Overview of Modes

• The order states the number of field zeros across the guide. The electric fields are not completely confined within the core i.e. they do not go to zero at core-cladding interface and extends into the cladding. The low order mode confines the electric field near the axis of the fiber core and there is less penetration into the cladding. While the high order mode distribute the field towards the edge of the core fiber and penetrations into the Therefore cladding modes also appear resulting in power loss.
• In leaky modes the fields are confined partially in the fiber core attenuated as they propagate along the fiber length due to radiation and tunnel effect.

• Therefore in order to mode remain guided, the propagation factor β must satisfy the condition

where, n1 = Refractive index of fiber core n2 = Refractive index of cladding k = Propagation constant = 2π / λ

• The cladding is used to prevent scattering loss that results from core material Cladding also improves the mechanical strength of fiber core and reduces surface contamination. Plastic cladding is commonly used. Materials used for fabrication of optical fibers are silicon dioxide (SiO2), boric oxide-silica.

### Summary of Key Modal Concepts

• Normalized frequency variable, V is defined as where,

λ = Free space wavelength … (1.7.1)

Since = NA                                         … (1.7.2)

• The total number of modes in a multimode fiber is given by   ‘d’ is core diameter                                    … (17.3)

Example 1.7.1 : Calculate the number of modes of an optical fiber having diameter of 50 µm, n1 = 1.48, n2 = 1.46 and λ = 0.82 µm.

Solution :             d = 50 µm n1 = 1.48

n2 = 1.46

λ = 0.82 µm NA = (1.482 – 1.462)1/2 NA = 0.243

Number of modes are given by,  M = 1083                                                                       …Ans.

Example 1.7.2 : A fiber has normalized frequency V = 26.6 and the operating wavelength is 1300nm. If the radius of the fiber core is 25 µm. Compute the numerical aperture.

Solution :                    V = 26.6

λ = 1300 nm = 1300 X 10-9 m

a = 25 µm = 25 X 10-6 m  NA = 0.220                                                                     … Ans.

Example 1.7.3 : A multimode step index fiber with a core diameter of 80 µm and a relative index difference of 1.5 % is operating at a wavelength of 0.85 µm. If the core refractive index is 1.48, estimate the normalized frequency for the fiber and number of guided modes.

### [July/Aug.-2008, 6 Marks]

Solution : Given : MM step index fiber, 2 a = 80 µm

\    Core radians a = 40 µm

Relative index difference, Δ = 1.5% = 0.015

Wavelength, λ = 0.85µm

Core refractive index, n1 = 1.48

Normalized frequency, V = ?

Number of modes, M = ?

Numerical aperture = 1.48 (2 X 0.015)1/2

= 0.2563

Normalized frequency is given by, V = 75.78                                                                                             … Ans.

Number of modes is given by,  …Ans.

Example 1.7.4 : A step index multimode fiber with a numerical aperture of a 0.20 supports approximately 1000 modes at an 850 nm wavelength.

1. What is the diameter of its core?
2. How many modes does the fiber support at 1320 nm?
• How many modes does the fiber support at 1550 nm? [Jan./Feb.-2007, 10 Marks]

Solution : i) Number of modes is given by,   a = 60.49 µm                                                                                          … Ans.

### ii) M = (14.39)2 = 207.07                                                                                   … Ans.

### iii)

… Ans. M = 300.63                                                                                                    … Ans.

Wave Propagation

### Maxwell’s Equations

Maxwell’s equation for non-conducting medium:

X E = - ∂B /

X H = - ∂D /

. D = 0

. B    0

where,

E and H are electric and magnetic field vectors.

D and B are corresponding flux densities.

• The relation between flux densities and filed vectors: D = ε0 E + P

B = µ0 H + M

where,

ε0 is vacuum permittivity. µ0 is vacuum permeability.

P is induced electric polarization.

M is induced magnetic polarization (M = 0, for non-magnetic silica glass)

• P and E are related by:

P(r, t) = ε0 Where,

X is linear susceptibility.

• Wave equation: Fourier transform of E (r, t)  where,  n is refractive index.

α is absorption coefficient.   • Both n and α are frequency The frequency dependence of n is called as chromatic dispersion or material dispersion.
• For step index fiber, ### Fiber Modes

Optical mode : An optical mode is a specific solution of the wave equation that satisfies boundary conditions. There are three types of fiber modes.

1. Guided modes
2. Leaky modes
• For fiber optic communication system guided mode is sued for signal Considering a step index fiber with core radius ‘a’.

The cylindrical co-ordinates ρ, f and can be used to represent boundary conditions. • The refractive index ‘n’ has values  • The general solutions for boundary condition of optical field under guided mode is infinite at and decay to zero at . Using Maxwell’s equation in the core    • The cut-off condition is defined as –  It is also called as normalized frequency. Graded Index Fiber Structure

• The Refractive index of graded index fiber decreases continuously towards its radius from the fiber axis and that for cladding is constant.
• The refractive index variation in the core is usually designed by using power law ### … (1.7.4)

Where,                    r = Radial distance from fiber axis a = Core radius

n1 = Refractive index core

n2 Refractive index of cladding and α = The shape of the index profile

• For graded index fiber, the index difference Δ is given by,

###  • In graded index fiber the incident light will propagate when local numerical aperture at distance r from axis, NA® is axial numerical aperture NA(0). The local numerical aperture is given as, • The axial numerical aperture NA(0) is given as,   Hence Na for graded index decreases to zero as it moves from fiber axis to core-cladding boundary.

• The variation of NA for different values of α is shown in 1.7.1. • The number of modes for graded index fiber in given as, … (1.7.6)