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  1. Can some one please explain in simple words that what is effective refractive index?

  2. How it is different from the refractive index?

  3. And how we can calculate the effective refractive index?

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Here is nice explanation rp-photonics.com/effective_refractive_index.html – freude Jun 26 '13 at 11:58
    
The term "effective refractive index" is also used in the theory of composite media like photonic crystals in different meaning. – freude Jun 26 '13 at 11:59
    
+1 Welcome to physics SE! Consider marking an answer, that helped you by clicking the green check at the answer. This removes the question from the unanswered questions list. – Stefan Bischof Aug 6 '14 at 6:46

The refractive index $n$ of a homogenous medium can be looked at as a measure of the delay which light propagating through the medium experiences, compared to the propagation of light in vacuum. For example, glass has a refractive index of about 1.3, meaning light waves propagate about 1.3 times slower in water than in vacuum. A refractive index is a property of the material itself.

Now an effective refractive index can be given for an optical component (e.g. a waveguide) as a measure of the overall delay of a light beam in that component. This is not a property of the material the optical fibre is made of, but an effective property of the optical fibre altogether.

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Thanks for your answer but can you please explain little more that what do you mean by overall delay of light beam in that component ??? It is little confusing and do you have any idea that how to calculate the effective refractive index ?? – Rizwan Jun 26 '13 at 11:35
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@Rizwan, the effective index takes into account the different paths light rays can take inside, for example, a fiber. Even in a homogeneous fiber (same material and index everywhere) you have more than one dimension in which light propagates, giving a dispersion in the possible paths and thus in the time necessary to go from one end to the other. Other components may have more complex structures but you also can compare the time between the signal output and the signal input and define an effective index as if it were a homogeneous material where light is propagating. – Andrestand Nov 2 '15 at 13:41

The refractive index of a material, $n$, is the ratio of the speed of light in vacuum and the speed of light in a given bulk media for given frequency. Thus, you could say that $refractive \space index = n(\omega, \epsilon_r, \mu_r) = \frac{c}{v}= \frac{\frac{1}{\sqrt{\epsilon_0 \mu_0}}}{\frac{1}{\sqrt{\epsilon(\omega) \mu(\omega)}}}= \frac{\sqrt{\epsilon(\omega) \mu(\omega)}}{\sqrt{\epsilon_0 \mu_0}} = \sqrt{\epsilon_r(\omega) \mu_r(\omega)}$ in a bulk media with relative permittivity, $\epsilon_r$ and relative permeability, $\mu_r$. But, for most optical media, the permeability in said media is the same as free space, so often $\mu_r = 1$. So, $n \simeq \sqrt{\epsilon_r(\omega)}$

A key point to note about the effective index is that it is intimately tied to the idea of modes in a guiding structure. @theSkinEffect eluded to this by stating that it is defined in an optical component such as a waveguide.

All of these indices are statements of a ratio of the velocity, wavelength, or wavenumber of light in vacuum to that of light in the material. This is because $c=\lambda_0 \nu$ and $v=\lambda\nu$. So, $n = \frac{c}{v} = \frac{\lambda_0 \nu}{\lambda \nu}=\frac{\lambda_0}{\lambda}= \frac{\frac{2 \pi}{k_0}}{\frac{2 \pi}{k}}=\frac{k}{k_0}$. The effective index is no different. The difference is that the effective index tells you the ratio of the velocity of light in vacuum to the velocity of a mode for a given polarization in the direction of propagation in a guiding structure (i.e., along the waveguide in the z direction). The effective index is defined as

$n_{eff_{pm}} = \frac{c}{v_{z_{pm}}} = \frac{\lambda_0 \nu}{\lambda_{z_{pm}} \nu}=\frac{\lambda_0}{\lambda_{z_{pm}}}= \frac{\frac{2 \pi}{k_0}}{\frac{2 \pi}{k_{z_{pm}}}}= \frac{k_{z_{pm}}}{k_0} =\frac{\beta_{pm}}{k_0}$,

where $k_{z_{pm}} \equiv \beta_{pm}$ and $p$ is the polarization (TE or TM) and $m$ is the mth mode of the said polarization.

$k_0$ is the wavenumber in free space for a given frequency of light and is always defined as $k_0=\frac{\omega}{c}=\frac{2 \pi \nu}{c}=\frac{2 \pi}{\lambda_0}$ where $\lambda_0$ is the wavelength in free space and $\nu$ is the frequency in all media (remember that the frequency is constant in all media, but wavelength changes depending on the media).

$\beta_{pm}$ is termed the propagation constant. For a given mode and polarization, it is the component of the wavenumber in the guiding structure which is in the direction of propagation of the light; that is, it is the longitudinal component of the waveguide's wavenumber, $k = n_fk_0$ (where $n_f$ is the refractive index of the guiding material). We can also define a traverse component of the wavenumber, $\kappa$, which is the portion of the wavenumber which is only in the direction perpendicular to the propagation (please see the following diagram). The choice of symbols $\beta_{pm}$ and $\kappa_{pm}$ is a matter of historical convention. But, since these are just the components of $k$ it would be equally correct to designate them as $k_{z_{pm}}$ and $k_{x_{pm}}$ for a system in which the light is propagating in the z direction.

enter image description here

The maximum number of modes supported by the guiding structure in TE polarization, $max(m)_{p=TE}$ may or may not equal the maximum number of modes in the TM polarization, $max(m)_{p=TM}$. This depends on the geometry and refractive index profile of the materials making up the guiding dielectric structure. Additionally, the number of modes supported for either polarization is finite.

You know the wavenumber, $k_0$, of interest and given that you know your structure, you also can easily calculate the working wavenumber, $k$, but to calculate the effective index, $n_{eff_{pm}}$, you also need to know the propagation constants, $\beta_{mp}$ for the modes supported.

The solution? Modes have no meaning unless light is guided. This implies that you must also know the structure of your waveguide and it's refractive index profile. Using that information, you have to use Maxwell's Equations and the boundary conditions for your structure to solve for what are called the $characteristic \space equations$. These equations (one for TE and one for TM) are transcendental equations (meaning that the equal sign is only true for specific eigenvalues). You can solve them by graphing both sides of the equations and looking for the crossings to get the value or you can subtract one side from the other and look for the zeros--it's your choice.

enter image description here

The above characteristic equations are only valid for an ideal planar waveguide (symmetric or asymmetric) which is infinite in extent in the z and y directions, but has a guiding film of a finite height in the x direction. $\gamma_s=(\beta^2 - k_0^2 n_s^2)^2$, $\gamma_s=(\beta^2 - k_0^2 n_c^2)^2$ and $\beta = (n_f^2k_0^2 - \kappa^2)^2$

A short program in Python which I wrote which graphs the characteristic equation for the TE case is as follows. Currently, the height is 0.22$\mu m$ and will only support a single mode. Try a larger height, and you see that changes.

from scipy.constants import pi
from numpy import linspace, arange, tan
from numpy.lib.scimath import sqrt
import matplotlib.pyplot as plt


png_filename = 'beta.png'


nf = 3.48       # refractive index of the guiding film
ns = 1.444      # refractive index of the substrate
nc = 1.444      # refractive index of the cover
h =  0.22e-6     # height of the guiding film

samples = 1000  # number of kappas to sample

wavelength = 1.55*10**-6 
k = 2*pi/wavelength
kappamax = sqrt((k*nf)**2 - (k*ns)**2)
kappa = linspace(1, kappamax, num=samples)

beta = sqrt((k*nf)**2 - kappa**2)
gamma_s = sqrt(beta**2 - (k*ns)**2)
gamma_c = sqrt(beta**2 - (k*nc)**2)


y1 = tan(h*kappa)
y2 = (gamma_c + gamma_s)/(kappa*(1-(gamma_c*gamma_s/kappa**2)))



fig = plt.figure()                                      # calls a variable for the png info

# defines plot's information (more options can be seen on matplotlib website)
plt.title("Characteristic Equation for TE Modes")       # plot name
plt.xlabel('kappa')                                     # x axis label
plt.ylabel('Transcendental Funcs')                      # y axis label
#plt.xticks(arange(0,kappamax, kappamax/20))#5000))     # x axis tick marks
plt.axis([0,kappamax,-10,10])                           # x and y ranges

# defines png size
fig.set_size_inches(10.5,5.5)                           # png size in inches

# plots the characteristic equation for TE
plt.plot(kappa,y1)
plt.plot(kappa,y2)

#saves png with specific resolution and shows plot
fig.savefig(png_filename ,dpi=600, bbox_inches='tight') # saves png
plt.show()                                              # shows plot

plt.close()                                             # closes pylab

One last thing to note. Since the effective index is defined as the propagation constant divided by the free space wavenumber, and since modes will only exist when there is total internal reflection from both interfaces of the film, the effective index is always $n_{max(n_s, n_c)} < n_{eff_{pm}} < n_{f}$. That is, it's value is always between the highest refractive index value of the other materials and the guiding/core/film material.

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In fiber optic measurement the effective refractive index is calculated as follow:

RI eff=(L opt* RI opt)/L eff

where: L eff is the cable or physical distance between two know even on the OTDR L opt is the optical distance between two known events

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You haven't really explained anything useful here. – Brandon Enright Aug 6 '14 at 4:59

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