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
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

Numerical Aperture on The Fiber Optic (2)

2. Total Internal Reflection

As it has been discussed in the first part that, the incident light coming from one medium to another as it strikes the boundary with certain angle, will be refracted in definite manner according to the n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2}. Let’s just start this section by increasing incident angle of light in figure 2, first part. This is what will happen if the incident angle is increased:

Figure 3. Refraction phenomenon, with n1 is smaller than n2
Figure 3. Refraction phenomenon, with n1 is smaller than n2

As we know from the behavior described in the first part, we observed the light being refracted in increasing manner when the angle of incident light is increased as well. It is less likely that we can notice something intriguing in here.

How about if medium one has larger value than medium two (n_{1}>n_{2})? We should have light will have lower velocity in medium 1 than in medium 2 (v_{1} < v_{2}) and thus angle of light with respect to the normal in medium 1 is smaller than in medium 2 (\theta_{1}<\theta_{2}). As we increase the angle of incident in medium 1, we will this result:

Figure 4. Refraction phenomenon evolves to total internal reflection, with n1 larger than n2
Figure 4. Refraction phenomenon evolves to total internal reflection, with n1 larger than n2

Total internal reflection is depicted in the right illustration of figure 2. In this phenomenon, the medium boundary acts as a reflector, instead of refractor under two conditions:

  • Light propagates from medium with higher refractive index to the medium with lower refractive index (n_{1}>n_{2}).
  • The angle of incident light has to satisfy minimum of certain angle, i.e. critical angle \theta_{c}.

Total internal reflection can be considered when the refracted light is in parallel direction with the interface (figure 2, middle illustration), i.e. \theta_{2}=90^{\circ}. Using Snell’s law equation described earlier in the first part and two conditions stated above, we can quantify the required incident angle so that it can trigger total internal reflection:

n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2}

n_{1} \sin \theta_{c}=n_{2} \sin 90^{\circ}

n_{1} \sin \theta_{c}=n_{2}

\sin \theta_c=\frac{n_2}{n_1}

\theta_c=\sin^{-1} \frac{n_2}{n_1}

Total internal reflection is occured when the incident angle fulfills the critical angle whose value is determined by the medium 1 and 2, as it is expressed in the last formula of the above equations, called the critical angle equation:

\theta_c=\sin^{-1} \frac{n_2}{n_1}

This equation establishes the requirement for total internal reflection to take place: medium 2 must have refractive index smaller than medium 1. If it happened in opposite way, there will be no solution for this equation, meaning that total internal reflection does not take place.

Let the density and its respective radius being known, then a critical beam angle shall be found!

It was happened when I read a paper from V. Consonni et al published in Physical Review B (or you can find it in here) about the shadowing effect part, in the page 4. V. Consonni et al gave a mathematical description (modelling) for the growth rate of self-induced GaN nanowires, and extracted several important parameters (for instance effective diffusion length on the sidewalls and substrate surface, desorption rate, driving forces for the diffusion of gallium adatoms to the nanowire top) based on the finding of equations from the fitting with the experimental results, which are as a function of growth time, gallium rate and growth temperature.

So, I found this “magic” at page 4, on the sub-part of “Theoretical modelling of the NW axial growth rate for the self-induced approach”, namely “shadowing effects”. Beforehand, I agreed with them regarding the geometrical considerations in the molecular beam epitaxy chamber determining the final form of nanowires which are heavily influenced by the incident angle of the gallium effusion cell and nitrogen plasma source with respect to the substrate (read here). It is important also to realize that the incident angle of these sources will be having an impact in the shadowing of the grown nanowires after they reach certain height.

To my surprise, V. Consonni et al provided three examples, just straight answer, of how shadowing effects took a role in affecting structural morphology of the nanowire (density, radius, spacing, height) and the respective angle of gallium beam. In all cases, they assumed nanowire to have regular square area. The first case is nanowire having density, radius, spacing of 100 {\mu}m^{-2}, 30 nm and 51 nm, respectively. They found out that with an angle of 21 degree, the gallium impinged on a nanowire with height of 133 nm. The second example is nanowire with density and radius of 100 {\mu}m^{-2} and 30 nm, a critical gallium beam angle of 49 degree was found. What… The last example, nanowire with radius and gallium beam angle of 30 nm and 21 degree, the density was deduced to be 200 {\mu}m^{-2}. I just don’t understand that straight answer.. It is just BOOM! Just let the radius and density to be known, then gallium beam angle will be discovered by some calculations which I do not understand.

To elucidate this matter, I make simple self-explanation. I used plain figure, based on the first case, which I think the most reasonable example I want to approach due to the more data in it. The objective is simple, before shadowing plays role, I want to find the maximum gallium beam angle and density, while radius, spacing and height are known. The figure has same scale measurement as with nm:

Nanowires configuration based on the first example
Nanowires configuration based on the first example

Let’s consider that gallium beam angle comes to the nanowires in a single line (of course in reality, the nanowires are received bunch of gallium flux on them). The consideration of “just before” shadowing effect means that this effect does not occur, i.e. maximum height of nanowires are found to be around 133 nm. The figure below gives an illustrative idea of how the maximum gallium beam angle “just before” the shadowing effect.

Maximum gallium beam angle with respect to the nanowire before shadowing effect
Maximum gallium beam angle with respect to the nanowire before shadowing effect

If I made a larger angle, than shadowing effect would take place. This is the maximum angle before shadowing. We can calculate that angle by doing a little hack. We can consider those figure with this:

The alternative approach for angle calculation
The alternative approach for angle calculation

I doubt the angle will be 21 degree. With \tan \theta = \frac{111 nm}{133 nm}, the maximum gallium beam angle is 39.84 degree.

The density of the nanowire itself is lower with my calculation. Since 1 {\mu}m^{-2} is equal with 10^{-6} nm^{-2}. We have 1 nanowire every 111 nm, meaning that in 2000 x 500 nm^{-2}, we have only 81 nanowire for 1 {\mu}m^{-2}.

See! The protractor also shows around 39 degree
See! The protractor also shows around 39 degree

It might be that V. Consonni et al have lower maximum limit to ensure that shadowing effect really will not be occurred, while my illustration really push the maximum angle of gallium beam angle. Using my approximation, with gallium beam angle of 21 degree, the maximum height of nanowire can be around 192 nm.

Maximum height of nanowire with gallium incident angle of 21 degree
Maximum height of nanowire with gallium incident angle of 21 degree

Ok, I will leave the shadowing part here.