Plasma Optics (IV)

To start this section, we need the wave equation expression, which we have derived before (The Wave Equation):

waveEq

where u is wave function and c is the speed of light in medium, defined by

wave func and speed of light

Meanwhile, we know that the relationship between refractive index n has connection with the speed of light in medium by this expression:

n1

We have the speed of light, dielectric function and magnetic permeability in free space, c_0, \epsilon_0, \mu_0, respectively. For nonmagnetic material, the value of \mu=\mu_0, so that we have the simplified expression:

n2

From this equation, we can say that negative dielectric function will give imaginary value of refractive index. In the other way around, positive dielectric function will give positive refractive index. We hope to come back to this section in the next section.

Let us come back to the expression of the wave equation and wave function. We can get the dispersion relation for electromagnetic waves by inserting equation wave function u into wave equation,

disp_rel1

disp_rel2

The last equation is dispersion relation and it tells us two important things:

  • If the dielectric function, \epsilon is real and positive, the wavevector k is real and positive.
  • If the dielectric function, \epsilon is real and negative, the wavevector k is imaginary.

Ok, now we can apply these conditions to the wave function. As a note, absorption is negligible in the following case

First condition is when the dielectric function, \epsilon is real and positive, we will have real and positive value of wavevector k. Let’s say that the value of the k is 1, then

waveEq_pos

 

The result of plotting the wave function with real and positive k as a function of distance is shown below:

Figure 1. This is how light will propagate through the material (grey colored) if the dielectric function is positive
Figure 1. This is how light will propagate through the material (grey colored) if the dielectric function is positive

Figure 1 shows us how light will propagate through the material (grey colored). Since light can propagate through material, it will become transparent. This happens when the dielectric function is positive, i.e. incoming light has wavelength smaller than the plasma wavelength of the respected material (see Plasma Optics (III)), as we expected. Since there is no absorption from the material (from the assumption), the amplitude of light propagating inside the material has the same value of the light propagate outside of the material, and it will look transparent in the infinity. This will be different if we assume that the material has ability to absorb the light, where it will be only reaching certain distance.

The second condition is when the dielectric function, \epsilon is real and negative, we will have imaginary value of wavevector k. Let’s say that the value of the k is \sqrt{-1}, then

waveEq_neg1 waveEq_neg2

The result of plotting the wave function whose the value of k is imaginary as a function of distance is shown below:

Figure 2. This is how light will be reflected through the material (grey colored) if the dielectric function is negative
Figure 2. This is how light will be reflected through the material (grey colored) if the dielectric function is negative

Figure 2 shows us how light will be reflected through the material (grey colored). Physically speaking from the figure 2, the light is not reflected, but rather absorbed. Therefore the material becomes solid. This happens when the dielectric function is negative, i.e. incoming light has wavelength higher than the plasma wavelength of the respected material (see Plasma Optics (III)), as we expected. We can say that with the negative value of dielectric function, the wave of light is damped directly in the surface of the material. This is not the same with the first condition (with the assumption where the material has absorption coefficient), the light can propagate to some certain distance before it is vanished.

How big is the influence of the absorption of the material to the light propagation? We will discuss this in the next section.

Plasma Optics (III)

First, we need to remember that the relation between \omega and f is linear, as described from \omega=2\pi f. We also know that f and \lambda are inversely proportional, which we can see from \lambda=\frac{c}{f}. This means that the behaviour of wavelength and frequency is opposite to each other. When the value of wavelength is high, we will get low value of frequency and vice versa.

The implications of dielectric function, from Plasma Optics (II)

diel func as plasma freq

are described as follows:

  • When incoming light interacting with the material has frequency lower than plasma frequency of the material, such that \omega<\omega_p, we will get negative value of the dielectric function. It implies the light is reflected from the surface of the material when the wavelength of the incident light is higher than the plasma wavelength of the material.
  • When incoming light interacting with the material has frequency higher than plasma frequency of the material, such that \omega>\omega_p, we will get positive value of the dielectric function. It implies the light is propagated through the surface of the material when the wavelength of the incident light is lower than the plasma wavelength of the material.

The above statements agree with the illustration given in the Figure 1 from Plasma Optics (I).

Figure 1. Visible wavelength, with group of alkali metal wavelength (155 - 362 nm)
Figure 1. Visible wavelength, with group of alkali metal wavelength (155 – 362 nm)

 

If the value of \omega_p=1 and the incoming light has \omega varying from 0 to 2, the response of the dielectric function is given in the figure 2.

Figure 2. The response of dielectric function to the omega of incoming light varying from 0 to 2
Figure 2. The response of dielectric function to the omega of incoming light varying from 0 to 2

How can negative dielectric function give rise to the reflected light? How can positive dielectric function give rise to the propagated light? We will try to find out in the next section.

Plasma Optics (II)

Two limits for \epsilon(\omega,K): \epsilon(\omega,0) and \epsilon(0,K). The first one refers to the collective excitations of the Fermi sea, which is related to the plasma, and the latter describes the electrostatic screening of the electron with electron, lattice and impurity interactions in crystals.

To obtain plasma frequency \omega_p, we can proceed with the equation of motion of a free electron in an electric field:

free el motion

x is the position of electron. Assuming electron moves according to the behaviour of simple planar wave, we need to introduce e^{-i\omega t} on both side,

with e

Further, we have polarization: the dipole moment per unit volume,

polarization

Since we are dealing with polarization, we can’t forget electric field, E, which both of them expressed in the electric displacement, D,

almost

That is, we find the characteristic of each material depicted by specific mass, m. Thus, this is the plasma frequency \omega_p,

plasma freq

The dielectric function can be re-written as follows

diel func as plasma freq

In the next section, we will deal with the implication of this equation.

The main concept from Kittel’s book: Introduction to solid state physics, 8th edition.

The Wave Equation

This is what I want to write since a long time ago: The wave equation, derived from Maxwell’s equations in free space. Finally! I can write now :)

To start with, we need to realize that an electromagnetic field is described by two vector fields, both are functions of position and time:

  1. The electric field, \vec{\varepsilon}(\vec{r},t)
  2. The magnetic field, \vec{H}(\vec{r},t)

The famous Maxwell’s equations in free space are defined by

Maxwells equations

the \nabla \times and \nabla \cdot are the curl and divergence. the constants of \epsilon_0 and $latex \mu_0$ represents the electric permitivity and the magnetic permeability in the free space.

Basically, we can express the wave equation based on the described Maxwell’s equation above. So, here we go

calculation1

Next, we need to use the vector identitiy (curl of the curl) of \nabla \times \left( \nabla \times \vec{\varepsilon} \right)=\nabla \left( \nabla \cdot \vec{\varepsilon} \right)-\nabla^2 \vec{\varepsilon}. From (3), we arrive to the equation of \nabla \times \left( \nabla \times \vec{\varepsilon} \right)=\nabla \left( 0 \right)-\nabla^2 \vec{\varepsilon} \longrightarrow \nabla \times \left( \nabla \times \vec{\varepsilon} \right)=-\nabla^2 \vec{\varepsilon}. Therefore, we will have

calculation2

By applying the speed of light equation in free space c_0=\frac{1}{\sqrt{\varepsilon_0 \mu_0}}, we arrive to the wave equation in the free space according to the Maxwell’s equation

wave equation electric field

If we start with (1) i.e \nabla \times \left( \nabla \times \vec{H} = \epsilon_0 \frac{\partial \vec{\varepsilon}}{\partial t} \right) and follow the same step, the final result for the wave equation will be

wave equation magnetic field

When we speak about scalar wavefunction, the electric (\vec{\varepsilon}(\vec{r},t)) and magnetic field (\vec{H}(\vec{r},t)) can be represented in the scalar wavefunction (u(\vec(r),t)), and we will have the wave equation as it is explained below,

wave equation general

 

Great! Now I am satisfied enough :D

Excursion to the Col Dei Rossi

Last week, between 15 – 19 March I went to Italy, attending international conference in Canazei, about 180 km north of Venice. The name of the conference is The 18th Molecular Beam Epitaxy (http://web.nano.cnr.it/eurombe2015/)

Canazei is located in the upper part of the Val di Fassa, and this region is part of the Dolomites, a mountain range located in the northeastern of Italy. This is my first time of my life seeing Alps as it is just standing before my eyes, so huge! According to the wikipedia, The Dolomites form a part of Southern Limestone Alps and extend from the River Adige in the west to the Plave Valley in the east. A huge range of lining mountain in the north of Italy.

Those mountain suppose d to be Ciampac (2100 m)
Those mountain suppose d to be Ciampac (2100 m)

The Dolomites is acknowledged as The Natural World Heritage Sites by UNESCO. Basically this place is heaven for those who can ski, both down hill and cross country. It has almost 260 km range to do ski activity: Ski until you drop :)

It has hundred of squared kilometres of walls, towers, pinnacles and valleys. A well known phenomenon, called Enrosadira, takes place at dawn and dusk when the sun light reflects on the walls of the Dolomites. Unfortunately, I was not able to see this phenomenon. The break I had was unfortunately insufficient to find out. It is supposed look like this:

Dolomiti sunset - Enrosadira Tra Le Odle, copyright: https://www.flickr.com/photos/dancingflowers/
Dolomiti sunset – Enrosadira Tra Le Odle, copyright: https://www.flickr.com/photos/dancingflowers/

In winter, the Enrosadira will appear like this

This is the map for the Val di Fassa:

Val di Fassa, source: http://www.skiforum.it/skimaps/skimaps/displayimage.php?pos=-325
Val di Fassa, source: http://www.skiforum.it/skimaps/skimaps/displayimage.php?pos=-325

And here I was, standing on the Col Dei Rossi, after taking some breathtaking cablecar from Canazei. The transportation was almost made me crazy, the wind blew and shaked our cable car! Who is not afraid of those disturbance?? Anyway, I relieved I could made to the one of the top peak in here.

On top of Col dei Rossi, Canazei (2382 m)
On top of Col Dei Rossi, Canazei (2382 m)

The first two days were cloudy, but the last two days were awesome. Unfortunately, I did not have good schedule on the last two days. I wish I could climbed the mountain on that days :(

Well, eventhough I was not able to see Enrosadira, I was able to see the sun rays were splitted by the mountain behind hotel I stayed and I guess it was Col Dei Rossi.

Sunrays were splitted!
Sunrays were splitted!

That was wonderful 4-day stay in Canazei. The last photo was taken before I left Canazei to Venice, where my flight back to Trondheim departed. I enjoyed and surprised with the great hospitality I received there. I met some of the people who were hardly speaking English, but they tried their best, and the Italian (or Ladin?) language was used eventhough I did not understand what they wanted to say. I wonder why they continue talking in Italian to me? Maybe, they wanted to make me welcomed in Canazei.

Grazie!

Plasma Optics (I)

When we talk about optics, we always relate it with the interaction between light and matters. The interaction will give varying result as it depends on what kind of material are being interacted with. One important properties of material is called dielectric function \epsilon(\omega,K), a function whose frequency and wavevector has impact on the physical interaction between light and matters.

We have two fascinating interaction probabilities in here: the light can be reflected or propagate from/through matter. Before we start with everything, it is better to have understanding of what plasma is. Basically, plasma, one of the fundamental state of matters (others are solid, liquid and gas), is medium with equal concentration of positive and negative charges, of which at least one charge type is mobile. Plasma takes form in gas which composed of free electrons and ions. Plasma has high energy.

Plasma has frequency, called plasma frequency \omega_{p} which becomes a guideline for deciding whether the light will be reflected or propagate. I will try to explain this later. All of us know the relation between energy E and wavelength \lambda: E=h\frac{c}{\lambda}. It is also obvious that metal, in the visible light, is reflecting incoming light. But how does it can be explained by plasma optics?

The answer lies on the value of incoming wavelength light \lambda. When \lambda>\lambda_{p}, incident light will be reflected. In the other hand, as \lambda<\lambda_{p}, incident light will propagate through matter. Illustration in figure \ref{fig:vis} will give a brief explanation of where should the light get reflected or propagate through material. The example is group of alkali metal, with wavelength ranging from 155 – 362 nm.

Visible wavelength, with group of alkali metal wavelength (155 - 362 nm
Figure 1. Visible wavelength, with group of alkali metal wavelength (155 – 362 nm)

We will come back later to this concept in the next section.

Numerical Aperture on The Fiber Optic (3)

3. Fiber optic: application of total internal reflection

The first and second part should be enough to provide illustration of how the fiber optic works: first, it depends heavily on the Snell’s law in governing how the light behaves after passing the interface of two materials with different refractive index number. Second, total internal reflection phenomenon is established owing to the fact that light coming at certain angle, defined as critical angle, is able to have reflection instead of refraction, as the light comes from the denser medium to less denser medium (n_{1}>n_{2}). Mathematical expression for the first and second part are described in here and here, respectively.

Now, we can go back to the figure 1. First, we need to realize that there are three different refractive index, as depicted below:

Figure 5. Fiber optic with three different refractive indexes
Figure 5. Fiber optic with three different refractive indexes

Refractive index for air (where, later, the source of light-laser- comes from), core and cladding of fiber optic are designed as it is so that light with certain range of angle, can be guided (controlled) from one point to another. To get an idea of the value of critical angle need to be fulfilled, we can start first inside the fiber optic, in the interface between core and cladding. We can start describe it as it in the figure 4 in the right picture. To have total internal reflection means that refractive index of core must be larger than refractive index of cladding (n_{2}>n_{3}).

Figure 6. Total internal reflection in the interface between core and cladding
Figure 6. Total internal reflection in the interface between core and cladding

Then we follow the same procedure as in the second part,

tir

Next, we can determine the correlation between air and core of the fiber optic (figure 7). Since total internal reflection in this boundary is not needed (or precisely, avoided), we can proceed with the refractive index of air smaller than core of the fiber optic (n_{1}<n_{2}).

Figure 7. Incident light and its refraction in the interface between air and core of fiber optic
Figure 7. Incident light and its refraction in the interface between air and core of fiber optic

Applying Snell’s law, we can calculate the incident angle required to achieve total internal reflection,

n_1 \sin \theta_1 = n_2 \sin \theta_4

We do not know \sin \theta_4 but we do know the value of \sin \theta_2 or \sin \theta_{c(i)}. Based on the figure 7, we can figure the value of \theta_4 from ordinary trigonometry case,

Figure 8. Trigonometry case to find out the value of Θ4
Figure 8. Trigonometry case to find out the value of Θ4

Mathematically, $\sin \theta_{4}$ is equal to $\sin \left( \frac{\pi}{2}-\theta_2 \right)$. To re-define $\sin \theta_{4}$, trigonometry identity of $\sin (\frac{\pi}{2}-\theta)=\cos \theta$ can be utilized. Then the continuation of the last equation is,

NA

There it is, we find already the incident angle required to conduct total internal reflection in the fiber optic. {\textit{Numerical aperture (NA)} is defined as the qualitative limitation of receiving angle of the fiber optic with respect to the incident angle of light. Therefore, the numerical aperture for fiber optic can be defined as follows:

NAresult