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{\fnArial\fs36}The fundamental entry to all laser materials processing is the question

{\fnArial\fs36}What happens if electromagnetic radiation laser radiation hits the surface.

{\fnArial\fs36}Basically, there are three things that can happen.

{\fnArial\fs36}It can be reflected, it can be transmitted or it can be absorbed.

{\fnArial\fs36}So consequently, this session is structured according to these three characteristics.

{\fnArial\fs36}We would start with some conditions on the surfaces, on real surfaces.

{\fnArial\fs36}Then we look at plane waves at interfaces

{\fnArial\fs36}so the condition of what happens at the interface to the electromagnetic field.

{\fnArial\fs36}Then we will look at the amplitudes of reflection and transmission.

{\fnArial\fs36}The reflectance and transmittance.

{\fnArial\fs36}Then we will discuss what is called the Brewster effect.

{\fnArial\fs36}A very special absorption at a specific angle of incidents

{\fnArial\fs36}and last not least in the outlook we'll take a look at the absorption the definition of the complex index of refraction.

{\fnArial\fs36}Looking at ideal surfaces, usually, we have the case that we have in perfectly straight, levelled interface

{\fnArial\fs36}between the incoming laser radiation and the medium,

{\fnArial\fs36}to which, we focus the radiation and we will have very clear geometries

{\fnArial\fs36}in terms of reflectivity and absorptivity and transmission.

{\fnArial\fs36}In reality surfaces look more like this, so we will have a direct part of reflection

{\fnArial\fs36}so, yes, a lot of the radiation will be reflected according to the ideal situation

{\fnArial\fs36}however, we'll have scattering due to the roughness of the surface.

{\fnArial\fs36}Sometimes we have periodically structured surfaces.

{\fnArial\fs36}Then of course we get periodically modulated reflectivity.

{\fnArial\fs36} In terms of application like welding, deep penetration welding,

{\fnArial\fs36}when we form keyholes, we will have multiple reflection in the laser induced surface geometry.

{\fnArial\fs36}Sometimes we use coatings and in these coatings we will have multiple reflection

{\fnArial\fs36}like in a Fabry-Perot interferometers in which we then will have interference of the reflected beams

{\fnArial\fs36}and thus a very specific wavelengths filtering in terms of the reflected light.

{\fnArial\fs36}And sometimes we have high intensities, so we don't only get the,

{\fnArial\fs36}what we call cold absorption processes or reflection processes

{\fnArial\fs36}but we get the formation of a plasma and in this plasma also we can have reflection, absorption or transmission.

{\fnArial\fs36}All in all, this equation actually represents the energy balance

{\fnArial\fs36}so whatever comes in, imagine the intensity plotted here will be either reflected to a certain fraction or it will be absorb

{\fnArial\fs36}and you already see here a exponential behavior of the absorptivity

{\fnArial\fs36}and the last part of the radiation of the incident radiation will be transmitted.

{\fnArial\fs36}So, absorption plus reflection plus transmission equals one.

{\fnArial\fs36}To define a few things.

{\fnArial\fs36}We start with what we call the plane of incidents

{\fnArial\fs36}which is formed by the vector which represents the normal of the surface and the incoming laser beam.

{\fnArial\fs36}This angle defines the area in which we use all our applications and this is called the plane of incidents.

{\fnArial\fs36}At the interfaces we can look at what happens now to the electromagnetic waves.

{\fnArial\fs36}Here you see the formulation of a simple propagating linear wave.

{\fnArial\fs36}E equals the incident not times the exponential of a Omega T minus K X.

{\fnArial\fs36}The same thing holds for the reflected electromagnetic wave here.

{\fnArial\fs36}So we denote that with ER for reflected and ET, consequently, for the transmitted part of the wave.

{\fnArial\fs36}We use complex amplitude vectors which we see later

{\fnArial\fs36}because we need to take care of the phase changes at the interface

{\fnArial\fs36}depending on as we will see the orientation of the electro-magnetic field.

{\fnArial\fs36}We will have phase jumps in the reflected beams.

{\fnArial\fs36}In general we can say that the frequency is unchanged among all three parts of the beam

{\fnArial\fs36}so the incoming laser frequency will be exactly the same to the reflected one and the transmitted one.

{\fnArial\fs36}However, which is not true for the wavelengths that's where the index of refraction will come into play

{\fnArial\fs36}and all three vectors will remain in the same plane in the plane of incidents.

{\fnArial\fs36}Now as you know or may know the electrical field has one more property compared to the direction of propagation

{\fnArial\fs36}and that is the orientation of the electromagnetic field vector E, in terms of its polarization property.

{\fnArial\fs36}So, if you take this into account the following situation appears.

{\fnArial\fs36}We have a general orientation of the electric incident field EI

{\fnArial\fs36}and we split that up into a fraction in the direction of the plane of incidents

{\fnArial\fs36}and one perpendicular to the plane of incidents.

{\fnArial\fs36}So, we call these IP for parallel and IS.

This comes from the German “senkrechtes” meaning the perpendicular part of the electromagnetic field.

{\fnArial\fs36}So, if we know the properties of these two components we can construct any orientation of the electric field

{\fnArial\fs36}by just super-posing the two components parallel and vertically perpendicular to the incident plane.

{\fnArial\fs34}If you look at the boundary condition that is exactly at the interface between the medium number one and the medium number two.

{\fnArial\fs36}Then we learn from Maxwell's equations that we have two conditions for the magnetic and the electric field respectively.

{\fnArial\fs36}By looking at the magnetic field we find that the dielectric displacement and the magnetic field are continuous

{\fnArial\fs36}and that means that the vertical components the components perpendicular to the surface

{\fnArial\fs36}so in other words in the direction of the surface vector N are the same before and after and during the medium.

{\fnArial\fs36}So, this is denoted here by N times B2 is the same as N times B1.

{\fnArial\fs36}So, these vector components are exactly equal

{\fnArial\fs36}because over the change of the medium they are continuous and the same thing holds for the electric field

{\fnArial\fs36}but no not for the vertical parts of the electrical field

{\fnArial\fs35}but for the components of the electric field parallel to the surface at which we have the change from medium 1 into medium 2.

{\fnArial\fs36}Taking into account these dependencies

{\fnArial\fs36}We now can formulate the ratio of the reflected electromagnetic field with respect to the incident electric field

{\fnArial\fs36}and that is what we call the reflection coefficient R.

{\fnArial\fs36}And correspondingly to the transmission

{\fnArial\fs33} the transmission coefficient is given by the ratio of the electric field transmitted with respect to the incident part of the electric field.

{\fnArial\fs34}For the reflection coefficient and the transmission coefficient we now can calculate the dependence of the angle of incidents theta

{\fnArial\fs36}and we end up with four formulas for the transmission coefficient

{\fnArial\fs36}and the reflection coefficient in the case of perpendicular orientation of the electric field

{\fnArial\fs36}and the same thing for the transmission coefficient and the reflection coefficient for the case of parallel polarized radiation.

{\fnArial\fs36}In this special case when we neglect absorption we are just left with reflectance and transmittance

{\fnArial\fs36}and if we look at the reflectance which is just the square of the index of reflection

{\fnArial\fs36}and if you look at the transmittance which is defined again as the index of transmission squared

{\fnArial\fs36}however here we see that we have two more influence factors

{\fnArial\fs36}number one is the ratio of the indices of refraction and the angle of incidents.

{\fnArial\fs33}Of course, ultimately energy conservation if you have no absorption then reads that the reflectance plus the transmittance equals 1,

{\fnArial\fs36}so everything which is reflected and transmitted needs to be the same as the incoming beam.

{\fnArial\fs36}Now if you look at the graph here's one more thing.

{\fnArial\fs34}If you look at a certain part of the beam which has a certain cross-section then the reflected beam will look exactly the same.

{\fnArial\fs36}It will have the same cross-section but the beam which is transmitted as you can imagine

{\fnArial\fs36} especially for larger angles of incidents

{\fnArial\fs36}then the projection will be such that the diameter of the beam will be expanded

{\fnArial\fs36}and the fact is that this expansion depends on the cosine of the angle of incidents

{\fnArial\fs36}though this area grows with the cosine of the angle of incidents

{\fnArial\fs36}and thus the energy density.

{\fnArial\fs36}So depending on the index of refraction ratio from incident area to transmitted area

{\fnArial\fs36}we get this influence on the transmittance of the radiation.

{\fnArial\fs36}Now let's recall what we have so far

{\fnArial\fs36}If you look at a surface and you come in with an electromagnetic wave

{\fnArial\fs36}the electrical field will be oriented somehow perpendicular to the direction of propagation.

{\fnArial\fs36}Somehow meaning that parts of the electrical field will be perpendicular to the area of incidents,

{\fnArial\fs36}the plane of incidents others will be parallel

{\fnArial\fs36}it will be in the plane.

{\fnArial\fs36}And now if you imagine the two limiting cases what happens when the incident beam hits the surface.

{\fnArial\fs36}In the case perpendicular polarized radiation of perpendicular polarized radiation

{\fnArial\fs36}we will have an intuitively you can see that probably there won't be a big dependence on what happens

{\fnArial\fs36} because the electrical field is always so to say parallel in the surface.

{\fnArial\fs36}However, if the polarization is parallel to the incident beam you can see that depending on the angle

{\fnArial\fs36}the field vector will kind of stick into the surface and come out again

{\fnArial\fs32}so, there will be a pretty strong dependence expected for the angle of incidents for the polarization direction parallel to the incident plane.

{\fnArial\fs36}So, what I just described by gesture, now you can see in a plot of the friendly equation

{\fnArial\fs36}which describe the dependence of reflectance and transmittance.

{\fnArial\fs36}Again absorption is not included in this model and you can see that for polarization perpendicular to the plane of incidents

{\fnArial\fs36}which is as polarized radiation we have very little dependence on this angle of incidents

{\fnArial\fs36}so it just changes from pretty high values metal reflect very good at least for visible light

{\fnArial\fs36}and that's what is assumed here.

{\fnArial\fs36}As an example you see that this changes very, very little.

{\fnArial\fs36}On the other hand for the polarized light in the direction of the plane of incidents

{\fnArial\fs36}you see a strong decline in reflectors up to very high angles

{\fnArial\fs36}and then we come back to the value of one at an angle of incidents of 90 degrees.

{\fnArial\fs36}So again, if you approach 90 degrees by increasing the angle of incidents

{\fnArial\fs36}then of course at 90 degrees we have a limiting case in which the radiation is simply on the surface

{\fnArial\fs36}so reflectance is one and there is no transmittance.

{\fnArial\fs36}So, all curves have to finally end in this point here and depending on the index of refraction ratio at the interface

{\fnArial\fs36}we get a starting point for the reflections as I said for metals

{\fnArial\fs36}now we can discuss an insulator, think of glass.

{\fnArial\fs36}We usually have for visible light a value of about 4% of reflectants for vertical incidents.

{\fnArial\fs36}That is the value here 4% now and if you look at the polarised light then

{\fnArial\fs34}of course you see not very thrilling changed but somehow the curve needs to go up to 1 and that is the behavior of this part.

{\fnArial\fs36}If you look at the parallel polarized light then we observe the same effect as for metals

{\fnArial\fs36}we have an increase in transmission a decrease in reflection at a certain angle here the value even goes to zero

{\fnArial\fs36}and it is exactly zero and then it increases back up to the value of the limit of grazing incidents with the reflections of one.

{\fnArial\fs36}A little addition if you look at the complex values of these parameters

{\fnArial\fs36}then you see for the reflection coefficient that depending on the polarization

{\fnArial\fs36}we have a negative and a positive value

{\fnArial\fs36}which simply represents for the complex notation that for the polarization parallel to the plane of incidents

{\fnArial\fs36}we have no phase change but for the S polarized beam we get a phase change of 180 degrees.

{\fnArial\fs36}At the interface we get a phase jump as you may know it from mechanical waves which are reflected at surfaces.

{\fnArial\fs36}In order to make this a little more practical we here listed a few values for typical index of refraction values

{\fnArial\fs36}and absorption values of different materials in the combination of different light wavelengths.

{\fnArial\fs36}So, for instance if you look at a diamond at visible light we see it has a very large index of refraction.

{\fnArial\fs36}Water is about one point five for visible light and for diamonds we observe a value of three.

{\fnArial\fs36}The absorption is negligible as we know from diamonds they are very nice and transparent.

{\fnArial\fs36}And if you look at the consequence for vertical incidents you look at the reflectivity

{\fnArial\fs36}we can calculate it and it comes to a value of about 25%.

{\fnArial\fs36}Remember glass was something like 4%.

{\fnArial\fs36}Now you know why diamonds are so desirable because they reflect so intense

{\fnArial\fs36}and so, they shine and they twinkle, that's why we like them so much.

{\fnArial\fs36}So, if we look at other examples like glass I already said that the value for the reflections is about something like 4%.

{\fnArial\fs36}For a CO2 LASER however, in the far infrared at 10 micron wavelengths

{\fnArial\fs36}we have strong absorption of the radiation and consequently the reflection changes at zero incidents to about 25%.

{\fnArial\fs36}And as we are very familiar with for metals in this case

{\fnArial\fs36}here at the example of aluminium we have values of reflectants of the order of 90-95 %.

{\fnArial\fs36}They're very shiny they reflect the visible light and if you look at values for solid or for liquid aluminium

{\fnArial\fs36}we see that the values of index of refraction and index of absorption are in between 25 and 40

{\fnArial\fs36}so relatively very high values.

{\fnArial\fs36}Coming back to the picture you have seen already I now want to try to get a little more light into this effect here.

{\fnArial\fs36}So, you see for insulators at a certain angle which we will call the Brewster angle we will have practically no reflection.

{\fnArial\fs36}So, all light is transmitted or absorbed but nothing is reflected.

{\fnArial\fs36}In metals it's a little different.

{\fnArial\fs36}In metals we don't go really to zero we go to low values still a significant effect for the parallel polarized light

{\fnArial\fs36}and the values for the angle at which this occurs is much higher 85, 87, 88 degrees depending on the metal.

{\fnArial\fs36}Now how does that come about?

{\fnArial\fs36}There's a straightforward explanation to this and this is sketched here in this picture on the left

{\fnArial\fs36}if you look at the situation where the incident beam

{\fnArial\fs36} and the reflected beam of course are under the same angle with respect to the normal of the surface

{\fnArial\fs36}however the transmitted beam the transmitted and absorbed beam is in this case,

{\fnArial\fs36}in this special case, sketched here exactly perpendicular to the reflected beam.

{\fnArial\fs36}So if the transmitted beam and the reflected beam share an angle of 90 degrees

{\fnArial\fs36}something very special happens and this can be explained by the dipole radiation.

{\fnArial\fs34}Imagine the electrical field being perpendicular to the direction of propagation of course also in the transmitted part of the beam

{\fnArial\fs36}then the electrons or the ions of the matter

{\fnArial\fs36}with which this electrical field interacts will oscillate in the direction of the stimulating electric field.

{\fnArial\fs36}So, the electrons will move in the direction perpendicular to the direction of propagation of the light wave

{\fnArial\fs36}so, in this direction here.

{\fnArial\fs36}And now if you look at the characteristic how accelerated electrons radiate you may know from antennas that

{\fnArial\fs36}when electrons are moved in this direction or in this orientation here, up and down,

{\fnArial\fs36}they will radiate perpendicular to this direction of motion and only perpendicular to this direction of motion.

{\fnArial\fs36}There will be an intensity displayed here in these curves in this, it's like a doughnut,

{\fnArial\fs36}if you think of a spherical geometric characteristic of a mission.

{\fnArial\fs36}So, if you look at the formula you see it depends on the square of the angle under which the radiation is emitted.

{\fnArial\fs36}So, here's a sine square of the angle so at 90 degrees we have a maximum emission sine of 90 is 1.

{\fnArial\fs36}So perpendicular to the direction of motion of the electrons

{\fnArial\fs36}we will have the maximum emission and sine of zero is zero.

{\fnArial\fs36}So, in the direction of motion we will have no emission at all.

{\fnArial\fs36}So as a consequence, this explains why we have these minima in reflection at specific angles

{\fnArial\fs34}which represent the case where the reflected beam is exactly perpendicular to the transmitted or absorbed beam in the medium.

{\fnArial\fs36}Some examples of how we can make use of that Brewster effect in the left part of the picture

{\fnArial\fs36}you see a typical situation of a bore, a drilled bore,

{\fnArial\fs36}and in some industrial applications we want to harden the inner parts of that bore

{\fnArial\fs36}so we somehow need to get into this bore under very, very small angles.

{\fnArial\fs36}And now if you remember what we just learned, if we choose the Brewster angle

{\fnArial\fs36}which is a very, very small angle compared to the direction of the bore or a large angle compared to the angle of incidents

{\fnArial\fs36}then we get the Brewster conditions

{\fnArial\fs36}so all the light will be absorbed very efficiently even in otherwise very high reflective metals.

{\fnArial\fs36}Under perpendicular incidents we would have extreme reflection

{\fnArial\fs36}but if we use the Brewster angle everything will be absorbed.

{\fnArial\fs36}So here the condition of how to get into a bore and the vision or the condition of high absorptivity really match

{\fnArial\fs36}and get two coherent goals.

{\fnArial\fs36}And a similar application in all laser applications and cutting and welding immediately shows you

{\fnArial\fs36}it will depend quite a bit on the orientation of polarization

{\fnArial\fs36}how this cutting curve or welding boundary layer between the cut and evaporated material

{\fnArial\fs36}and the base material will effect the absorption.

{\fnArial\fs35}Finally, we need to come back to the understanding that we discussed everything for the ideal situation of very plain surfaces.

{\fnArial\fs36}Remember as I said in the very beginning that usually we have rough surfaces not ideal surfaces

{\fnArial\fs36}so we will have parts of diffuse reflection.

{\fnArial\fs36}Usually we will have a mix of all we will have some direct reflection and some diffuse reflection

{\fnArial\fs36}so we have to be careful with these formulas for real surfaces.

{\fnArial\fs36}So as an outlook we here see displayed an electric field entering some medium

{\fnArial\fs36}and if the index of refraction in the medium is larger than in the vacuum

{\fnArial\fs36}then we get a smaller wavelengths and the wavelength scales with one over the index of refraction.

{\fnArial\fs36}Then when the wave again comes out of the medium as sketched here

{\fnArial\fs36}the original wavelengths again can be seen on the way through this part of the medium

{\fnArial\fs34}we see there is an absorption which took place and this absorption scales with the exponential factor of an absorption index.

{\fnArial\fs36}This index is defined with respect to the electric field as you can see here.

{\fnArial\fs36}And so the characteristics for what happens can be expressed by exact two properties of the material

{\fnArial\fs36}which is the index of refraction and the absorption index.

{\fnArial\fs36}And that is why we combine those to a complex index of refraction.

{\fnArial\fs36}This N Tilde is N minus I times kappa, kappa the absorption index.

{\fnArial\fs32}So, for the intensity we measure an exponential behavior of the loss of intensity along the transmission through the absorbing medium.

{\fnArial\fs36}And this is represented by Alpha, Alpha being the absorption coefficient

{\fnArial\fs36}and it is connected to the absorption index by an exponential factor of 2.

{\fnArial\fs36}And this comes just from the square of the electric field to the intensity.

{\fnArial\fs36}Finally, the penetration depths of the radiation into a media then simply is 1 over the absorption coefficient alpha.

{\fnArial\fs36}So as a final aspect we now see that we take only

{\fnArial\fs36}it takes only two parameters to describe the full optical properties of the material in its reaction to radiation

{\fnArial\fs36}and that is in this case the index of refraction and the index of absorption.

{\fnArial\fs36}You can define five of such parameter pairs in total

{\fnArial\fs36}with which we can describe the nature of what happens if electromagnetic radiation hits the surface of a specific medium

{\fnArial\fs30}and we only need two parameters of the medium to describe everything that happens with respect to reflection, absorption and transmission.

{\fnArial\fs36}You already know these two the index of refraction and the absorption index N and kappa.

{\fnArial\fs36}In the last slide we also learned that the change of the wave number in the medium

{\fnArial\fs36}and the absorption coefficient can be taken very similar to the ones above.

{\fnArial\fs36}However, then you may know from maximal equations

{\fnArial\fs36}the parameters of permittivity and electrical conductivity from atomic absorption processes in plasmas or in metals

{\fnArial\fs36}we know the plasma frequency and a collision frequency

{\fnArial\fs36}which represents the dissipated, the energy lost, the absorption part of the process.

{\fnArial\fs32}And using ellipsometry we can just by analyzing the reflected part of a polarized incident beam to identify the parameters of the material

{\fnArial\fs36}defining what happens in terms of absorption transmission and reflectants.

{\fnArial\fs36}So I think this is really fascinating that if you ask the question.

{\fnArial\fs36}What happens if electromagnetic radiation hits the surface?

{\fnArial\fs36}It takes only two parameters of the material to define everything

{\fnArial\fs34}which is happening in terms of fractions of absorption, fractions of reflectivity, fraction of transmittance, change of polarization

{\fnArial\fs36}all these things which can happen

{\fnArial\fs36}if we shine light on a surface are described but only two parameters

{\fnArial\fs36}and we learn that there are five of such parameter pairs which we can use.

{\fnArial\fs36}And once you have two have such parameters you can directly calculate the other four pairs immediately.