 So, let us get started with may be one question, Amal Jyothi college Kerala any questions. Ask you earlier, now suppose you are considering a person, a man who is standing inside a room and suppose the walls of the room are having a temperature which is greater than the temperature of the human body, suppose a room is having a greater temperature. Now the room will emit some radiations and it will be absorbed by the man, now after that the, suppose instead of a man suppose there is an object there, now after absorption it will transmit some amount of radiation, now it will go it is going, so absorbed radiation it is going to be transmitted, it is going to be transmitted and it is going to be hitting on the same surface itself. So, in that concept do we want to take the transmitted amount of radiation. See the question is actually this is the question is if I put a body inside a furnace and there is an interaction of the body and the wall and do we have to take the transmissivity also into account. I think you will have to, we will be answering this question, we will be in fact solving a problem on this, so you will have to wait until I solve, the point is there is view factor which is taken into account, I have to take into account the irradiation and also the emission, so this is the point, but anyway you will be able to appreciate this when we get to the, when we complete the basics of radiation, so I am going to get started where we stopped, see where we said that the black body radiation is the radiation depends on the temperature of the material, wavelength of the emission, surface roughness and surface roughness, but not considering the wavelength, temperature and the surface roughness, if I define an ideal body that is the black body which absorbs what is a black body, it has to satisfy three conditions, it should absorb all the energy which is incident on it and it should emit maximum amount of energy, no one should be able to emit more than the black body and black body has no preference for direction, that is it is a diffuse emitter that is black body is this, this is not black body, this is black body, so it is a diffuse emitter under these conditions, so we can say that that is the black body and all real bodies are going to have a emissive power lower than that of the black body, so now we have Stefan Boltzmann law where in which we said that E b equal to sigma t to the power of 4, where sigma is 5.67 into 10 to the power of minus 8 watts per meter square Kelvin to the power of 4 and then this was contributed by Stefan and Boltzmann and Stefan never left the university, he was always in his own place, he never left the university, in fact he never went home also, but Boltzmann was a peripatitic traveler that is he used to travel all over the place and unfortunate the bad news is that Boltzmann died of suicide that is he committed suicide perhaps he thought that he did not get the recognition as much he deserved during his lifetime, so it is a bad news maybe Boltzmann would have been alive for few more years few more laws would have come out anyway, so there is we said that a cavity can be imagined as a perfect black body because it is a multiple reflections because of multiple reflections it is going to absorb everything and at the same time it is going to emit everything. Now the spectral emissive power of a black body is given by Planck's distribution as you can see that it is a function of lambda and T and rest are all constants, so and I know the black body is a diffuse emitter, so if I calculate emissive power instead of intensity of radiation pi I lambda, B lambda , T is I if I substitute the Planck's distribution I am going to end up with C1 upon lambda to the power of 5 exponential of C2 upon lambda T minus 1, so C1 and C2 are constants. So now if I plot this spectral emissive power I find that this is the spectral emissive power and this is the wavelength plotted for different temperatures, so what I can see here is that the emissive power increases and decreases and there is always a maxima at a particular wavelength and that maxima at any particular temperature can be computed by Wien's displacement law that is lambda maximum into T equal to 2897.8, so that is Wien's displacement law and also it states that any wavelength below 0.08 micrometer and near about 2000 to 3000 micrometer there is no contribution of any other waves other than these wavelengths contribution towards thermal energy that is the important point we need to recognize or realize or emphasize and one important point professor Arun emphasized is that area under this curve represents the total emissive power of any body at of a black body at that temperature at that temperature covering complete thermal radiation wavelengths. So is professor Wien and this is the Stefan Boltzmann law if I integrate this with respect to lambda covering all lambdas I am going to get sigma T to the power of 4 and that is how I end up getting sigma equal to 5.67 into 10 to the power of minus 8 Watt square meter squared Kelvin to the power of 4, so intensity of the black body is equal to EB by pi that is another way of looking at it. Now the question is that is where we stopped in the law just before signing off for lunch, so now the question is now I want to find an emissive power in a particular band of wavelength it is not from 0 to infinity let us say I want it from lambda 1 to lambda 2 that is 3 micrometer to 5 micrometer or 7 micrometer to 12 micrometer if I am interested in only band emission that is what is this band emission what is the band emission means I am interested in a particular band how do I get that band emission, so that is defined as a fraction fraction of the total emission of from a black body that is in a certain wavelength interval that is fraction so what is that fraction so that let us say I define it as 0 to lambda E lambda B D lambda upon 0 to infinity E lambda B D lambda what is this denominator what is this denominator total emissivity that is total emissivity complete spectrum it is covering the complete spectrum whose spectrum it is covering black bodies so it is black bodies of course even in the numerator it is black body only for a black body if I cover complete spectrum what is this E lambda comma B D lambda just now we have integrated this and found that that is nothing but sigma T to the power of 4 so the denominator turns out to be sigma T to the power of 4 but then what is the numerator that is between that is the emissive power in a particular wavelength what is that particular wavelength band 0 to lambda so this fraction between 0 to lambda and 0 to infinity is the fraction that fraction is called 0 F 0 to lambda you see capital F means fraction 0 to lambda means I am interested in the emissive power only between 0 to lambda compared to the emissive power of complete spectrum which spectrum thermal spectrum that is the spectrum over which the thermal radiation contribution is inherently there now I will do some algebra so what I am doing here is lambda T equal to m let me call this as lambda T equal to m so if I differentiate this I get D lambda into T this remember all this I am doing at a particular temperature temperature is fixed D lambda into T equal to D m so D lambda equal to D m upon T so when lambda equal to 0 m is 0 when lambda equal to lambda m is lambda T so if I substitute that E lambda comma B D lambda T what will happen D lambda becomes now D lambda T upon T so T is there in the denominator and you have now the limits have changed from 0 to lambda to 0 to lambda T so 0 to lambda T E lambda comma B D lambda of T upon sigma T to the power of 5 what is E lambda comma B lambda comma T that is pi I lambda comma B lambda comma what is pi what is this I intensity of radiation that is Planck's distribution that is C1 upon lambda to the power of 5 exponential of C2 upon lambda T minus 1 is that okay you see 1 by pi is there that pi as vanish okay so I get 0 to lambda T C1 D of lambda T upon sigma upon sigma where did the sigma come from where did the sigma come from yes it is there in see what am I doing I am substituting this in the above equation so if I substitute this in this equation what am I getting see E lambda comma B is C1 all this blah blah blah and in the denominator I have sigma T to the power of 5 but here lambda to the power of 5 is sitting so lambda and T I will combine and make it as lambda T to the power of 5 so what is this whole of this which is there in the integral it is a function of what lambda T so fraction fraction of the total emission from a black body we have shown that it can be represented as a function of lambda T why are we doing this because every time we do not want to calculate this fraction once we want to calculate this fraction for different lambda T and keep it and use that table again and again that is what we are going to do so again the same thing I am not done this font is big that is all F0 lambda is 0 lambda T C1 D lambda T upon sigma lambda T to the power of 4 exponential of C2 upon lambda T minus 1 F of lambda T if I have to understand what is this fraction this is the fraction this is the fraction of the emissive power compared to the total emissive power I am looking for is that ok and that fraction happens to be a function of lambda T that is F lambda T ok so now this has been integrated because I know C1 C2 if I choose a lambda T I should be able to get the I should be able to integrate this I can use either Simpson's one third rule or trapezoidal rule and integrate this you can also do this integration using excel sheet simple excel sheet ok so that is for 0 to lambda T now what if I have to find a fraction from lambda 1 to lambda 2 that is let me show that pictorially that means earlier I was trying to find this fraction what if if I want to find the fraction of emission between lambda 1 and lambda 2 that is the area under the curve which has been shown that is it is emissive power from 0 to lambda 2 minus of emissive power 0 to lambda 1 upon sigma T to the power of 4 same thing whatever we did for 0 to lambda I can do the same thing and show that fraction of lambda fraction of the total emission between lambda 1 to lambda 2 is equal to fraction between 0 to lambda 2 minus fraction of 0 to lambda 1 ok is that ok similarly if I have to find the intensity of radiation I lambda B lambda T upon sigma T to the power of 4 this is given by this so I can compute this ratio also by computing this ok so all these are plugged in here for a given lambda T I get the fraction f of 0 f of 0 comma lambda and this is the ratio that is this ratio I lambda B lambda comma T sigma T to the power of 5 otherwise every time I have to compute instead of computing that intensity of radiation has been given and this is given as a ratio this intensity whatever I get upon I lambda comma B at lambda maximum comma that is intensity of the radiation of the black body at a given temperature at maximum at a wavelength at which the emissive power is maximum so that is what is this ratio. Now let us just quickly solve a problem so that we understand this concept of band emission ok consider a large large isothermal enclosure that is maintained at a uniform temperature of 2000 Kelvin. Calculate the emissive power of the radiation that emerges from a small aperture on the enclosure surface if you see the figure this is the enclosure and this is the small object and the enclosure is sitting at 2000 Kelvin. Now what is the calculate the emissive power of the radiation that emerges from the small aperture on the enclosure surface how will I calculate this what is the emissive power there is a small object there is a small object ok so it has to undergo through multiple reflections so the it does not matter what is the emissive emissivity of the small object if the small object inside a large cavity can be considered as a black body. So if I do that so the emissive power is sigma T to the power of 4 sorry sorry what is that question as calculate the emissive power of the radiation that emerges from a small aperture on the enclosure surface who is emitting this the enclosure surface is emitting so enclosure surface is at what temperature 2000 whatever is emitting from the surface only has to get out through the aperture so the emitting thing what is the emitting thing that is sigma T to the power of 4 so that is what has been substituted and we are getting 900 kilo watts per meter square ok. So now what is the next part this is the easiest part what is the next part what is the wavelength lambda 1 below which 10 percent of the emission is concentrated that means what what is the fraction or what is the wavelength lambda 1 below which 10 percent of the emission is concentrated what is this 10 percent that means f 0 lambda 1 is 0.1 what what is the wavelength at which f of 0 lambda 1 is 0.1 that is the question is that ok ma'am is that ok fine. So you get f of 0 lambda 1 is 0.1 so now how do I get that lambda 1 how do I get lambda 1 f of 0 lambda I have to just go back I have to just go back and see this table what is this table showing what is this table showing lambda t and f 0 lambda for this problem for now these two columns are not important for this what is given f 0 lambda has been given to be 0.1 at what lambda t it is equal to 0.1 somewhere between 2000 and 2200 little below 2200 if I am right it is 2198 if I interpolate that if I am right no 2195 2198 is that maximum Wien's displacement sorry this problem I get lambda 1 t equal to 2195 micrometer kelvin. So how do I get this lambda 1 now what is temperature temperature is given to be 2000 kelvin so if I substitute lambda 1 t equal in lambda 1 t for t 2000 kelvin if I substitute I am going to get lambda 1 as 1.1 micrometer what was the question answered now the wavelength at which 10% of the emission this figure is wrong this complete thing is sorry this green colored thing should have displaced little bit here and this is lambda 1 so this has got distorted that is what I am trying to say is this fraction is found to be 0.1 and this is lambda 1 so that is what is been shown here that is what that is the figure so this is the emissive power now what is the next part what is the wavelength lambda 2 above which 10% of the emission is concentrated that means what that is what is the wavelength lambda 2 above which only 10% of the energy that is this is 10% of the total emissive power is concentrated that means what is F0 lambda 2 F0 lambda 2 should be 0.9 so now life is again F0 lambda 2 equal to 0.9 that is how did I get this area under the curve is 1 fraction area under the curve is 1 so 1 minus 0.1 gives me 0.9 gives me 0.9 that is what I have done here in this problem so F lambda 2 to infinity lambda 2 to infinity what is that infinity this is infinity lambda 2 to infinity is 1 minus F0 to lambda 2 that is 1 minus 0.1 you get 0.9 so okay so now lambda 2 I T I should be looking for in the table where at which lambda at what lambda 2 T this is fraction is 0.9 I know the font is very small I am sorry about that that is it is somewhere between 9000 and 9500 okay so you cannot see it 9000 and 9500 okay so lambda 2 T lambda 2 T lambda 2 T lambda 2 T lambda 2 I get 9382 micrometer kelvin if I interpolate so I get lambda 2 please add this one lambda 2 equal to 4.69 micrometer 4.69 micrometer is the wave length. Now what is the next question asked the question is not over determine the maximum spectral emissive power and the wave length at which this emission occurs is that right that is where does the maximum emission what is the what is the condition for which or what is the wave length at which the maximum emissive power occurs have we studied any law for this. Wien's displacement law that is lambda maximum if you see the Wien's displacement law we see that if you go to Wien's displacement law what is Wien's displacement law saying if you see this Planck's distribution it is saying that this peaks are curve fitted by an equation lambda max into T equal to 2897.8 micrometer for our problem I know the T what is T 2000 kelvin if I substitute T equal to 2000 I will get my lambda max so that is this lambda max that lambda max equal to 1.45 micrometer at 1.45 micrometer I can get the what is that the emissive power at which it is maximum see because we defined this fraction because that table is there we are able to solve this problem very easily otherwise you have to sit down and do the integration every time that is not easy so that is the reason this table helps us in easing out our problem very very fast. Now what is the next portion determine the maximum spectrum emissive power we found just the wavelength what is the spectrum emissive power how will you find that spectral emissive power how will you find that again from the table now you will realize you have I lambda comma B lambda T upon lambda comma B lambda max comma T now let me just see what exactly is going on here yeah yeah so that is in the third column I will get that and then I will multiply fine that is you see the third column what is the third column it is if you are not able to see let me write that that is I have the third column as I just help me out I lambda comma B yes I mean upon lambda lambda comma B upon sigma T to the power of sigma T to the power of 5 I lambda comma B sigma T to the power of 5 of course I lambda comma B is function of lambda and T this ratio is there so I know that black body is a diffuse emitter if I get the intensity of radiation I can get the emissive power so I know the temperature I know sigma all that I need to do is at lambda max T equal to whatever I had got that is the maximum wavelength at which that is at maximum when does the maximum spectral emissive power be there at lambda max T equal to 2198.5 for 2198.5 what is this ratio I am sure that at least should be there okay I can take 2200 thing that is it is 5.89 649 into 10 to the power of minus 5 let me see is that what is the maximum what I have taken yeah is that is that right no I have taken something else oh no 2898 I saw wrong 2898 means I should be seeing I should be seeing 2900 so that is something around 7.23 into 10 to the power of minus 5 so if I put that 7.23 into 10 to the power of minus 5 I think above values are something wrong please be careful with this table there is something wrong with this table in the top these three values are wrong and I guess these two values are also wrong 10 to the power of minuses are missing okay we will check that and correct it back so if I put that ratio that is 5. that is 0.722 into 10 to the power of minus 4 and sigma T to the power of 5 sigma is 5.67 into 10 to the power of minus 8 and T is 2000 to the power of 5 I get 1.31 into 10 to the power of 5 I am not interested in spectral intensity I am interested in emissive power so to get the emissive power because I know the black it is black body all that I need to do is multiply by 5 I will be getting that per meter square micrometer this is the emissive power at the maximum spectral emissive power at the wavelength at which it can be maximum fine so now what is the next part of the question what is the irradiation incident on the small object placed inside the enclosure that should also be equal to what sigma T to the power of 4 whatever is being emitted by the enclosure has to be absorbed by my black body because it is very small so irradiation of any small object inside the enclosure may be approximated as being equal to the emission of from a black body at the enclosure surface temperature so because of which I can compute that so this is same as what we had written here so that is how we have solved this problem that is band emission so this concept what we have elaborated in the last one half an hour is that we have introduced the concept of band emission that is what is the fraction of emissive power between two wavelengths that band emission concept we have elaborated and for one time these integrations these integrals are integrated and kept as in the form of a table and now I can using this table I can calculate the intensity of radiation and the fraction at any particular wavelength so that is the that is what we have learnt from this what is this band emission so there is another problem I am not going to solve this problem I would request you to solve it yourself because it is not so difficult problem now let me introduce for next one half an hour I am going to spend on some properties which have been postponing since morning that is we are going to introduce three properties or four properties emissivity absorptivity reflectivity transmissivity first is emissivity okay so if you see here I am giving definitions why I am flashing all these transparencies is because morning professor Arun said four things one is things can vary spectrally I can have things can vary direction on the basis of directionality and I can have total and I can also have hemispherical so based on these four four definitions we will define emissivity also as four different definitions so if I define emissivity similarly I would have defined for absorptivity transmissivity and reflectivity I am not going to define for absorptivity transmissivity and reflectivity whatever I am defining for emissivity is valid for all others but everything is there in the notes I would request you to see the notes but I am going to define you only the emissivity okay so what is black body is an ideal emitter no surface can emit more than a black body but all real surfaces are going to emit less than the black body so that how much it is less is characterized by emissivity emissivity is the radiation emitted by a real surface to the radiation emitted by a black body remember the pointer at the same temperature both numerator and denominator has to be written that is the radiation emitted by the surface at the same temperature radiation emitted by a real surface at a given temperature upon radiation emitted by a black body at the same temperature that is what we mean okay so that is emissivity okay now emissivity is going to vary as I said spectrally and this is spectral distribution if I have this is the black body and this is real surface okay the real surface cannot be above this it can only be below this that is why the emissive power of a real surface is going to be lower than that of black body this ratio if I have to calculate at every wavelength at every wavelength if I calculate the ratio that is what is called as spectral emissivity now emissive power can also change with the direction okay can also change with the direction so if I have to take the directionality in token black body is has no preference to the direction but real surface has preference to direction okay so if I have to take into account then I have to have directional dependency also in my emissivity definition so real surface will have directional distribution it need not be diffused unlike black body and real surfaces can have spectral distribution also of course black body also add even black body also is having spectral distribution so is the real surface okay there is nothing new there but directionality wise real surfaces and black surfaces are different okay now what is the definition what is this telling spectral directional emissivity first is let me write let me restate spectral directional emissivity all these definitions four definitions I am writing let me write these headings so that we do not lose track of what we are doing spectral directional emissivity spectral I want all of you to write along with me spectral directional emissivity that is epsilon lambda theta, 5 this is what we mean by spectral directional emissivity and all these definitions I am going to state that at a particular temperature at a given temperature okay so that is comma this subscript only says that it is varying with lambda it is varying with direction theta comma 5 at a given temperature now next definition what is that this is spectral directional emissivity next is total directional emissivity total directional emissivity okay so what does the total directional emissivity total means what it is spectrally averaged spectrally averaged so that is why I am seeing here only epsilon theta theta comma 5 comma t epsilon theta comma 5 comma t so it is usually called it is a function of that it is usually written as epsilon theta that is what you are written here epsilon theta it is a function of theta comma 5 now next one is spectral hemispherical emissivity spectral it is all English there is nothing very different or difficult spectral hemispherical emissivity that is epsilon lambda lambda comma t note here epsilon is not a function of theta and 5 because it is averaged over complete hemisphere so epsilon lambda lambda comma t here it is only a function of lambda and t the last definition is total hemispherical emissivity total hemispherical emissivity it should be a function of what this epsilon should be a function of what it is spectrally averaged that is why it is called total and it is in all directions it is averaged that is for complete hemisphere it is averaged that is why it is hemispherical so it is neither dependent on lambda nor dependent on theta comma 5 then it should be dependent on what only temperature so epsilon t okay so spectral directional total directional spectral hemispherical total hemispherical that is all you see directional is common hemispherical is common otherwise you have spectral total spectral total is that okay so now having understood this let us define each one on a fast note epsilon lambda comma theta lambda comma theta 5 comma t equal to i lambda comma e lambda comma theta comma theta comma 5 comma t upon i lambda comma b lambda comma what does this mean what does the numerator mean it means that this is the radiation intensity at a particular wavelength in the direction theta comma 5 at a given temperature upon black body will black body have theta comma 5 no because it is diffused but black body intensity is it dependent on wavelength yes it is dependent on wavelength and it is dependent on temperature temperature will not keep stating every time because both in the numerator and the denominator in all the definitions temperature is going to be same so temperature is not a variable at all for now so this is the intensity of radiation emitted at a particular wavelength in a particular direction at a given temperature upon intensity of radiation emitted by a black body at the same wavelength and temperature is that okay I am not writing this again I hope you all have notes and you are seeing this now total directional emissivity so epsilon theta theta comma 5 comma t equal to i e intensity of radiation emitted in the direction theta comma 5 t upon i b t why here lambda is missing because I have averaged for all wavelengths okay but this averaging I have done in only one direction that is theta comma 5 direction in only one direction I have done in the numerator in the denominator there is nothing like direction because in all directions it is same it is uniform so total directional emissivity is given by i e theta comma 5 comma t now what is spectral hemispherical emissivity now I am going to ask you a question spectral hemispherical emissivity I am not writing intensity I am writing emissive power why you notice please notice this spectral hemispherical emissivity I am not writing intensity I am writing emissive power why intensity is the only one which is having directionality but the emissive power it is it is integrated over all directions theta comma 5 that is what we realized when we defined intensity and emissive power so that is the reason why we have defined spectral hemispherical emissivity as e lambda lambda comma t upon e lambda comma b lambda t please note this please note this it does not occur to us that easily if you are not understood also go back and think you will understand go back to the definition of spectral intensity and spectral emissive power you will understand this this is a subtle thing because student is going to ask you how did I write this you need to know that now what is that I am going to do relation between spectral hemispherical emissivity and directional emissivity how do I do that let me start off with spectral hemispherical emissivity that is spectral hemispherical emissivity is epsilon lambda lambda t is given by e lambda lambda t what is e lambda lambda t given by is i lambda e lambda comma theta comma 5 comma t note this why am I writing always subscript e emitted if you are doing absorptivity you will have to take irradiation okay so so lambda comma e is lambda comma theta comma 5 comma t and then for directionality you have to take sin theta cos theta d theta d phi this is the where did I get this from this is the definition of emissive power okay okay so if I take that and integrate that and on the denominator in the denominator I have i lambda comma b lambda comma t sin theta cos theta d theta d phi but i lambda comma b is independent of direction I can pull this fellow out of the integration and integrate only sin theta cos theta d theta d phi that is what I am doing here so I am pulling this out I am pulling this i lambda comma b and pulling this out and getting this into the below the denominator I can take this i lambda comma b lambda comma t anywhere why because if I pull it in it only means that it is it is a constant and the integration we I am doing with respect to direction when I say constant not with respect to wavelength or with respect to temperature but in terms of it is uniform in all directions this integration is with respect to direction so I can push it in that integral so i lambda e lambda comma theta comma 5 comma t upon lambda comma b lambda comma t what is this ratio we had just defined that ratio is this so it is spectral directional emissivity so if I substitute that as spectral directional emissivity I have sin theta cos theta d theta d phi and if I do this integration which I am not going to do if I do that integration you are going to get epsilon lambda lambda comma t equal to 2 into 0 to pi by 2 epsilon lambda theta sin theta cos theta d theta here in this I have taken epsilon lambda comma theta as independent of pi as you with the angle I have taken it as independent of pi so that is why I am able to pull out that phi and say that as 2 phi if I integrate d phi I get 2 pi so that is what I have done you can ask me and say how did you do that you cannot do that yes I cannot do that that is an assumption I have made okay so I have no justification in making that assumption there is no justification in making that assumption I have just to make it look little better I would made that assumption if you are a purist your definition of connection between connection between spectral hemispherical emissivity and spectral directional emissivity ends here okay so now what is the last definition left out total hemispherical emissivity here also I am not using intensity I am using emissive power because it is hemispherical the moment you see hemispherical you are going to get into emissive power why you please sit down and think okay so epsilon equal to emissive power at a given temperature upon total hemispherical emissivity equal to emissive power at temperature T upon emissive power of the black body at temperature at the same temperature so what is this E of T just now we have derived this what is this this is nothing but what is this how did I write this this is E lambda T I can write that as epsilon lambda lambda T into E lambda comma B lambda T that is what I have done here now that is it so these are the four definitions which you need we have defined that is spectral directional emissivity total directional emissivity spectral hemispherical emissivity and total hemispherical emissivity these are the four definitions we have introduced so now having introduced them before I go to various band emissions and things like that similarly one can define what is called as okay I have not let me formally define before I go to the band emissivity is I will define what is called as absorptivity reflectivity and transmissivity it is quite straight forward I think every one of us are very comfortable with this if there is an irradiation G all of that cannot pass through that is some amount of it is transmitted that is tau G is transmitted rho G is reflected and alpha G is absorbed if I combine these three I should be getting G so G equal to G absorbed plus G reflected plus G transmitted which is in which G absorbed is equal to alpha G rho here G reflected equal to rho G G transmitted equal to tau G G G gets cancelled out everywhere I get alpha plus rho plus tau equal to 1 so what is alpha alpha is absorptivity what is rho rho is reflectivity what is tau tau is transmissivity compared to incident radiation how much is absorbed is absorptivity compared to incident radiation how much is reflected is reflectivity compared to incident radiation how much is transmitted through that body is transmissivity so for opaque surfaces generally for opaque surfaces transmissivity is 0 opaque means solid to take any solid plate transmissivity is 0 okay so I will have absorptivity plus reflectivity later on we will show that absorptivity is equal to reflectivity under restricted condition but that we will do little later on but that is why we first teach emissivity and then come to absorptivity transmissivity and reflectivity so I have absorptivity plus reflectivity equal to 1 for opaque surfaces now similar definitions the way we define spectral directional emissivity we can define spectral directional absorptivity spectral directional reflectivity spectral hemispherical absorptivity and reflectivity all definitions roll in the same manner so I do not have to go through all that I need to make differentiation is that for absorptivity I have to consider intensity that is absorbed for reflectivity I have to consider intensity that is reflected that is all the definition or instead of emissive power I need to take for absorption irradiation similarly for reflection also that is all the differences otherwise all definitions roll to the come to the same way or turn out to be looking similar complete derivation I would look I would request you to look into this in the sit down and re derive or rework what all derive all the definitions in the same way I did for emissivities four definitions okay I think if we do that I think if we do that now we can come to what is called as emissivity function like band emission emissivity is also not going to be spectacularly same everywhere we said that there is going to be spectral emissivity if you take emissivity usually this emissivity is measured by what is called as FTIR spectrometer Fourier transform infrared spectrometer so if you put this if you put your surface under FTIR spectrometer let us not get into how it works and things like that we will be digressing. So if we get into FTIR spectrometer if you put my surface under FTIR spectrometer I typically get an emissivity at a given temperature typically FTIR does at room temperature so if you do that you are going to get the emissivity something like this okay so that means what emissivity is a continuous function of lambda now I need to break this so this is the function what am I assuming now between 0 to lambda 1 I will make a reasonable assumption and say that that is constant and that constant is epsilon 1. Now between lambda 1 to lambda 2 instead of this sinusoidal wave looking thing I will convert and imagine that it is a square wave it is looking something like a square okay it is looking like a square and here the peak emissivity is epsilon 2 and then it is dropping to epsilon 3 at lambda equal to lambda 2 and subsequently it is constant whose emissivity epsilon 3 now the question is what is the average emissivity if this is the question if this is the total thing this is the emissivity function as lambda with spectrally that is how it is if it is varying what is the average emissivity how do I compute this average emissivity so that is what is given here how do I do this I will be taking the recourse of band emission how did I write this equation what is this equation what is this equation emissivity by definition is what let me just take the range 0 to lambda 1 emissivity is what emissive power of a real surface upon emissive power of a black body emissive power of a black body is E b but what is emissive power of a real surface emissivity into emissive power of the black body but emissivity epsilon 1 I can take out from this integral only within the range of 0 to lambda 1. So, epsilon 1 into E b lambda d lambda within the limit 0 to lambda 1 similarly epsilon 2 is epsilon 2 into E b lambda d lambda between the limits lambda 1 to lambda t upon E b plus epsilon 2 into E b lambda d lambda 2 to infinity upon E b. So, epsilon 1 what is this what is this by the way this fraction we have just defined that is the band emission fraction f of 0 to lambda 1 what is this fraction f of lambda 1 to lambda 2 what is this fraction lambda 2 to infinity what is this fraction f of lambda 2 to infinity. So, that fraction band emission idea is helping us even here to get the average emissivity. Let us quickly solve a problem what is the problem given is emissivity is given to be epsilon 1 equal to 0.3 from for lambda 0 to 3 micrometer let me see if there is any figure no epsilon 1 equal to 0.3 for lambda 0 to 3 micrometer and between 3 micrometer to 7 micrometer epsilon 2 is 0.8 and between 7 micrometer to infinity it is 0.1 why am I looking for the figure this is the figure epsilon 1 is 0.3 lambda 1 is lambda 1 is 3 micrometer and epsilon 2 is 0.8 lambda 2 is 7 micrometer and epsilon 3 is 0.1. So, that is what is this so all that I need to do is apply this if I do the application and I need to get the band emission what is that I need I need to know the temperature that is the temperature of my surface is 800 Kelvin. So, if I have to find that lambda t so there are only two wavelengths that is 3 into 800 is 2500 micrometer if you see that table I am not going to go to that table if you see that table you get F03 as 0.140256 and lambda 2 t is 5600 micrometer Kelvin and if you see that fraction let us go and see because my students here are seeing so that means everyone is tempted to see 5600 is 0.701046. So, if I that fraction if I substitute that here that is 0.701046 so what is my relation epsilon 1 F0 lambda 1 epsilon 1 is 0.3 F0 lambda 1 is 0.14 plus epsilon 2 is 0.8 what is F lambda 1 to lambda 2 what is this yes F lambda 0 to lambda 2 minus of F0 to lambda 1 that is what we did. So, 0.701046 minus 0.140256 plus 0.1 into this is what this is 1 minus that is whole area under the curve minus that fraction. So, 1 minus of 0.701046 you get a net emissivity of 0.521. Now if I have to find out what is the emissive power net emissive power from my body epsilon sigma t to the power of 4 what is my net epsilon epsilon is 0.521 if I put that this is the surface my surface is going to emit 12.1 kilo watts per meter square of radiation that is what we understand from emissivity I think there is another problem I am not going to solve this problem you please go through this problem yourself. So, because essentially it is dealing with directional distribution and spectral normal distribution and spectral emissivity. So, please go through this problem every step is there here please go through this problem you will be able to understand it yourself. I think I have loaded too much in 45 minutes. So, I will not I will I will not take up this emissivity variation with theta and things like that tomorrow morning we will start off the day with how do we get the emissivity how do how do we measure emissivity for example and how do we get the absorptivity and how does it vary why is emissivity dependent on surface temperature and absorptivity dependent only on the source temperature we will try to understand all of this tomorrow morning 9 o'clock.