 So, how do you not begin with the question, but anyway how do you how do you introduce these concepts of emissivity meaning in fact I do not remember studying irradiation, radiosity, intensity all these terms I do not remember them as distinctly as how I remember now because probably I have not studied it the way it has to be studied, but do you introduce these words irradiation, radiosity anybody who does not introduce these terms to students. No, but I saw in many of the curriculum there are no definitions it is starting from black body in many of the curriculum what I saw in their syllabi I saw there is there is no definition of blanks distribution and all I mean directly we are getting into black body no I would say half half probably it is there it is not put in the syllabus no, but Mangesh was for example Mangesh's case he was saying that we assume that everything they know in plus 2, but I would suggest that even if they know in plus 2 it is a good idea you see professor Arun brought that red light thing so maybe we might have studied in what is that plus 2, but it might have not got registered. Another thing is plus 2 to 3rd year 3 years yeah maturity levels have gone up how plus 2 students study we all know it is only from an examination point of view ok 98.6 or 98.8 is what people care about so whether you have understood or know I do not think that is ever taken into consideration so actually if you are thinking of introducing I do not know it is some it is like a wall which has to be broken with students actually so I think the best way to start these things especially these definitions is by I think illustrating with some kind of an example I do not know how I did it at engineering level when I studied I did not do it this way I did it when I first started teaching so what I think of is always I think of an orange ok so the center of an orange is a point ok and instead of a think of a light bulb which is placed at the center of the orange ok. So the light bulb is going to emit in all directions so it will it will form a sphere ok now through out here from now on instead of a full sphere we will deal with only the hemisphere ok that is the first thing so you will see what is like hemispherical in irradiation total hemispherical etcetera what it means is we are looking at something which is going to occupy only the top portion of the sphere which we will call as the hemisphere is that acceptable so I have I have an orange which is cut into half nothing is going down the center of the orange is where the bulb is the bulb is turned on it is going to emit in all all the directions such that this angle goes from 0 to pi and this angle goes from 0 to 2 pi ok so this this picture if you remember I think life becomes very very easy I do not say it becomes very easy it becomes easier so now we have to look at the definitions from this picture point of view so if in Cartesian in two dimensions especially we always deal with arc of a circle now arc is a difficult concept in terms of a hemisphere so in fact this picture in single is very nice actually this he says you think of a plane angle two dimensionally we will call it arc length and arc length is nothing but central angle is equal to arc length by the radius when the central angle is very very small now this watermelon which has shown here is exactly orange or watermelon does not matter it is the same thing in terms of concept I am looking only at the top half ok and a portion which is shaded here surface area s ok this part is what I will be dealing with so everything we like to start with a differential control volume differential area why because unfortunately radiation is direction sensitive ok so if I can apply my logic of the definition to an infinitesimal surface area on the hemisphere and so I am imagine the area of the hemisphere imagining that I am going to sit at this point ok I am sitting here and the central center of the hemisphere is going to see me how is it seeing me what is the angle subtended whether it is of radius r or radius 2 r or 3 r the central angle is the same so the radius of the hemisphere does not matter is that appreciated so all that matters is the angle subtended which is a function of my area so instead of calling it as a plane angle we will call it as a solid angle and the solid angle by definition is nothing but the area d a n we use the normal area ok that is one thing which we have to keep in mind always we are talking of the normal or projected area if the area is normal area means this is the direction to reach my surface is normal area means I always use my hands it should be normal this is normal what professor is saying is that if my area is like this on the top then I have to make it later on we will see that some cause alpha will come to make it normal but I always use my hands to this is my direction and this is my surface I have to make it so this is very normal so this I have drawn is the normal area now if the area was oriented like this actually throughout today both of us will I think interchangeably talk so because the subject is such that whatever input the other person gives is definitely going to be useful so honest actually what we have realized we have to be thankful to all of you because you have propelled us to listen to each other we have been teaching in independently in different classes at the same time same material but we never get an opportunity to sit in each other's class because of our whatever mundane regular stuff but this is given as an opportunity to listen from each other and see the different perspectives so we are learning from each other simultaneously because of this course that is that is one perceptible advantage what we have got so normal essentially refers to something which is perpendicular to the line which is drawn so this area here is normal to the line joining the center to the surface everybody appreciates that okay this solid angle is nothing but dA by R square the units is going to be steradian S T E R here S T E R A d I A N steradian S T E okay so this picture essentially is a part of the area which forms the hemisphere okay so we will call this angle in the horizontal plane I get confused always with azimuthal and zenith angle so zenith I think is the one upright and this is azimuthal yeah so that is how I remember but sometimes I get confused so this y is the azimuthal angle this theta will be called as the zenith angle so now just look at this picture or look at this watermelon and we will sit there at this area and see so all this while we have dealt with heat flux Q double prime okay now in radiation we are just changing the nomenclature we call it intensity that is the only change the conceptually it is the same thing heat coming per unit area heat going per unit area water heat receive per unit area this per unit area instead of calling it as flux heat flux heat flux we will call it intensity of radiation okay so it is a terminology which is changing concept is not changing it is per unit area so what is the energy that is absorbed whatever we will not we will fill in the appropriate words per unit area that is this shaded area of this surface okay so the word intensity whenever we see the word intensity it is like your heat flux energy per unit area watt per meter square is that okay so yeah so I will come back to all this a little later intensity of emitted radiation why is radiation felt to be so complicated I am sitting here by virtue of a finite temperature I will give out energy right I will also receive energy from everything surrounding me I am standing here this table generates radiation it comes to me you all of you each one at a different location generate some amount of energy by virtue of your temperature your surface area etcetera some Q is coming depending on the relative position a part of it comes to me this light here gives me different energy this light gives me much different energy so for that infinitesimal area d a all the surrounding parts give some kind of energy okay so one thing the body does is by virtue of its finite absolute temperature everybody above absolute zero is going to have emission what is emission giving out of energy emit that English is clear so by virtue of its finite temperature it will give out some energy we will refer to that word as emission anything with the word emissive emission emissive will be something related to out going emitting emitting now what we are saying is okay per unit area we want so we will call it intensity related to emission okay emission is what how much energy is emitted how much heat is emitted what intensity of emission would mean how much energy is emitted per unit area of that surface I am not showing any figure just listen carefully okay so this surface by virtue of its finite temperature is going to emit a certain amount of energy per unit area we will call it as intensity associated with emission okay intensity is given by symbol I capital I okay now books will say I subscript E will be intensity associated with emission so I is intensity watt per meter square in corporate I think events angle follows this this E subscript refers to emission related quantity okay so intensity emission is double M right single M okay whatever is the spelling okay single M so if it had been this simple life would have been easy unfortunately we have just seen sir has told us that there is a wavelength dependency and an angular dependency correct directional dependency as well as wavelength dependency now I am saying I am looking here I am standing on the road I am counting the number of vehicles which are passing by outside IIT gate that is easy I do not have to deal with too much of concentration I just have to sit and count now I say count the number of red color cars going red color cars okay I have to be little bit careful okay because I am giving it an attribute okay life is a little bit difficult but not so difficult now I say count the number of red color cars going only from right to left so I should ignore what is coming from left to right okay so when I put in these conditions of directionality that is too trivial an example but what I am saying is at this point when I am sitting intensity of emission of watt okay it is going to emit over all the wavelengths or a range of wavelengths what I am saying is okay I do not care about all these wavelengths I look at a particular wavelength lambda how much energy is emitted per unit area of that surface at that wavelength lambda okay now that is the red color car now I say okay that emission will happen in this direction in this direction always think of this bulb bulb is going to emit in all directions right now I am saying I do not want all directions this particular direction what is that particular direction this direction associated with the position of the infinitesimal area is that reasonably clear I think if this part we appreciate life becomes a lot easier okay so this subscript theta is there right theta comma 5 only theta I will put because it makes life or only theta right so what started off as I count all the vehicles going so left right you do not care you will count everything that is heat flux intensity so this is analogous to q double prime okay we relate to q double prime very well because we have seen it so many times now the problem is this is heat per unit area okay very nice first thing is how much energy is what is this this is intensity associated with emission therefore the subscript e has come okay emission of what emission of energy at that wavelength lambda therefore another subscript has come that wavelength lambda may like only one particular direction or two directions or four five directions it may not emit uniformly in all directions or it might emit uniformly in all directions I do not care all I care is this direction what is the emission so that is why what is this direction this direction represents the relative position of my surface area dA with respect to the source of okay so my area dA I am the area dA I am emitting let us say some amount of energy what is the energy of wavelength lambda which is reaching Mr. Vivek Kortay that is the directional dependency what is reaching Vivek Kortay will be different from what is reaching Parshura right so that is what is the directionality okay I am looking only at a particular wavelength if I say all wavelengths then I will all I will have to do is remove the wavelength dependency integrate over all the wavelengths is that right now if I say I want yeah Mangesh theta phi r becomes 2 regular 2 most of the text books but now here if you see that this one that is right here this is the observational angle and this one is the theta it is the vertical angle theta now it is radiating in all directions correct now our surface is standing here correct okay so this I know it is the angle which is on the other side how will I place my area the other way you need 2 angles no how will I locate my area this how will I locate my point for example in the space r theta and phi only we need we need we cannot locate only with r and theta it is correct correct correct what is the question I am not understanding the question see what is your question will you please elaborate it is going in the theta direction correct correct and the radiations are also in this in all directions there yeah but we have taken we have preferred only one direction one particular direction theta we have preferred later on I have to later on I have to take all directions but at this instant just in fact no no my question is that whether the surface is standing at theta right surface cannot whatever we are considering that infinitesimal small surface okay it is at an angle of theta my how will I locate that angle how will I locate that area it cannot be 2 coordinates r theta phi r is not needed because we are we do not we do not care about the size of the right but my position on the orange depends on r n depends on theta and phi okay in fact we did if you go strict by nomenclature this is how it is I e the subscript I will it will become like this I yes sir I got it no no what was your question it is that this is the phi and you have to locate this area with respect to the theta as well as this phi correct correct correct that is otherwise I cannot locate no otherwise yes we are taking direction theta but I have to just locate my area as well no it is a ring when you peel off the orange each ring is at a different angle correct okay any any questions confusion now we will go to the blah blah blah text book radiation intensity means heat flux associated or for emitted radiation is given by I subscript e bracket theta comma phi that means the function of the direction is defined as rate at which radiation energy dq watts is emitted in that particular direction per unit area that is why it is heat flux normal to that direction because if I am oriented in a non normal direction the surface area as seen is going to be smaller which will be the projected area so it has to be the normal area per unit solid angle about this direction what is this per unit solid angle current what it means is see here this I have understood this way maybe sir can have a different or better explanation I do not know as my slice of the orange changes okay this sector shaded area sector is subtending a solid angle d omega at the center now if I take a larger sector larger piece it will subtend a larger angle at the center ice cream cone ice cream cone smaller the sorry smaller the surface I mean area of the cylinder of the circle on the top your cone solid angle is smaller same thing okay so per unit everything is per unit basis per unit area of the surface per unit solid angle okay and it is in that particular theta phi theta phi theta and phi again are married it goes together okay is that clear so when I apply this concept to emission it is called intensity of emitted radiation so it is a function of theta phi applicable to e related to emission and what is this missing in fact this one concept I like incorporate a little bit better because he starts with the most general you will say intensity of emitted radiation d q by d a cos theta d omega d lambda also that means my units will be what per meter square steradian micrometer all the dependencies have come this equation which is projected there is integrated over all the wavelengths so that d lambda has vanished so my definition actually would be so this is radiation intensity that would be called as spectral intensity because it is that at that given wavelength so again it is to be very honest with you radiation is only understanding the terminology when the word spectral comes it is at a given wavelength okay I write it down for this even for the sake of being repetitive this means at given lambda okay when it comes total hemispherical these two words come together that means it has taken all wavelengths all directions because hemispherical means you are covering the entire hemisphere so total hemispherical would mean all directions all wavelength total spectral what would that mean total spectral a particular wavelength how much of that is emitted all directions is that clear total hemispherical hemispherical and spectral when this spectral word comes it is a wavelength specific term specific wavelength specific specific quantity okay so this would probably be written as e i e lambda theta phi which would be dq by d a cos theta d omega d lambda correct correct yeah so what meter square one minute what meter square steradian micrometer yeah everybody will emit energy at a particular wavelength and it can emit over range of wavelengths different wavelength you have different wavelength you for every temperature for example you see hundred Kelvin what is the wavelength range which is having emissive power the lowest one near about eight micrometer to thousand micrometer so now you go to the highest wavelength sorry highest temperature five thousand eight hundred Kelvin yeah so it is who is going to contribute that contribution is going to be dependent on temperature interdependency both temperature and wavelength here of course temperature is not there that will come later on but here only worry about yes how do I know which wavelength I should take that will be decided by temperature yeah hey month has a question sir why are we taking this as we see that the radiation is multi omnirection phenomena and it is all to get a creating a spear to us so why are we studying at land the heat transfer this between the surface is in hemispherical concepts yeah yeah I do not know this is why it is hemisphere I will this is how I have understood we all to get a cut we all to get a cut the parts and we have counseling mainly on the hemisphere understood why let me tell may be professor Arun can correct so the way I understand is how much my plate is going to see around me if this is all I am emitting let us say behind I am I may not be emitting if I am not emitting behind why should I be worried about the other hemisphere I am emitting only in front of me if I am emitting in both directions then I have to take the whole whole sphere I am emitting only that is a very good question and anyway I intended to tell that later on but this I always my students always ask this question who decides a hemisphere hemisphere this is how so in fact the bulb example I bulb is a sphere it is giving you a sphere of field of interest I think I think to my mind we merely concern being in heat transfer engineering merely concern the heat flux of the wall at boundaries so as I see not really not really I disagree with you not really not really one see no not really it cannot be only on one direction it can be both if you look at flames for example if I look at flame if I put a body in a flame if I put a body in a flame I am I am discussing this thing in context with the say for the example I am having the radius in a cuboid ok so if I see the one surface I am having the temperature of this one surface if this is emitting the radius into other surface I think yes yes yes in your application what you said is yes we usually insulating electrical furnaces or any furnaces what you said is right other side we are insulating only one side is we are forcing it to emit that is their hemisphere is no no it is three-dimensional only when I am taking r3 to 5 where am I not taking three-dimensionality just for the case in which the emission is only in wonder in one what to say one side that is the right word one side not the other side if it depends on application to application as you said in application of furnace it is only on one side their hemisphere is enough but if I have let us say a plate which is emitting on both sides then I have to take complete sphere this is just because we are not considering the participating video no no nothing to do it is probably for first of all I think it is for simplicity if you take see once you do for hemisphere is it going to be difficult to cover the other sphere other hemisphere in fact see this angle 0 it is not going to be right it is only integration you can integrate it so it is nothing to do with why did you bring the word participating medium because see when when the when the intensity of radiation comes out from the surface and it travels to the path suppose some particles are there which is which is really interacting so instead of having the hemisphere hemispherical surface at the wall we have to consider this sphere when it is exactly in the medium they are also same thing there is no no we have to understand this clearly see it is irrespective of whether it is participating or non-participating first thing is we are we are linking something which is not to be linked no it is a very specific issue because hemispherical means what hemispherical means we are merely worried about participating has nothing to do with surface it is it is representing the property of that medium here I have not brought any medium when I started the you are very much true but my concern is we are simply taking this hemispherical concept just for the wall of the surface the just for the wall but as soon as it leaves the wall we have to consider the not hemispherical complex sphere now let us say now let us say if I take that door this door I take this door I take is emitting it is sitting at a higher temperature and emitting if I take r theta phi hemisphere will I am going to am I going to miss any of this am I going to miss any of the point in this room am I going to miss any of the point in this room if the answer is no then I am not making any mistake but if the same wall same door is emitting on both both the both the sides someone helped me with the word side both the sides then whatever I did the for this hemisphere I have to do the same thing for the other hemisphere if the two surfaces are different temperature different coating etc that side that is the hemisphere the intensity quantity can be same or different depends on that is true day and night correct so but then but then it is all radiation is going to be hemispherical is also not a right thing see I think his question was when you go to participating media it is the gas which is which is like a source which is going to go in all directions we should not get confused with participating medium why did I say let me complete see is again the same example if I take what is that which is which will happen in the participating media what is participating medium so far we have taken non participating there is vacuum but what am I saying in a participating medium what will happen in a participating medium with the basics whatever we have learnt so far my electromagnetic wave cannot move at the same speed it is going to be so and the medium which is there like for example in a flame I said if suit is there the suit also starts emitting so I again take packets of those suits all over this room and then start interacting with my door again so where is the question of in participating medium full sphere has to be taken there also I am taking I am covering full room we are not taking two we are covering this half we are covering this half I will not spend more time on this I think we have over coffee again we will discuss but I think we have covered every point let me summarize for the other guys if you get confused because of this question so in fact then extrapolating is why are we not doing it for quarter of a sphere why are we doing it for half the sphere yeah why do not I do that is a good way of convincing you why do I take hemispheres why cannot take quarter of a sphere or one eighth of a sphere that is what same thing we are saying no same thing we are saying but you brought up with this question top and bottom we are team are getting it into two top and bottom what this is emitting all this hemisphere I can cover with r theta 5 if this fellow is also emitting then I have to be worried about this r theta 5 but if it is indeed emitting we have to do it we will do it we do it in this analysis to make it simple we have taken right now why because if this is a solid surface then you will have to take conduction effect through this are these at the same temperature or no then is it emitting at the same rate or no what is the wavelength of this what is the wavelength of this what is the boundary condition on this surface what is the boundary condition on this so to forget all that and especially when you are learning it at the initial stage let us take it as one surface the other surface we can treat it the same way as we treat the surface provided we understand the concept associated with the emission both are not talking to each other we can feel this light cannot go it cannot it cannot go that that is one if that if we understand that point the light coming from this cannot go into the any of the ray which is coming from the top hemisphere and not going to the bottom hemisphere if I cut a circle here and keep that bulb that bulb is though it will it has the property to emit in all of entire sphere by virtue of this solid surface it will emit it will the light will come out only in the top part that is what we are focusing on by virtue of a solid surface you have a question for sure we are considering the radiation only one one one part of the one direction one side one side one side we just showed the temperature wavelength okay okay no problem no problem what does this figure tell what does this figure tell different for a given temperature I can have several wavelengths okay it is okay because it is okay it is okay we will be as patient as possible for radiation for a while and before we get into surface now I think you realize why conduction we breeze through okay we want this conscious decision why we have put it conduction we consciously I think even if we skip conduction it does not matter you guys are all teaching great there but convection radiation involved so much of thinking so much of thinking okay one second so surface to surface radiation that is why it is called okay probably because one surface will see only okay if you are talking of two surfaces only the one portion is what is going to be seen okay so I will come to the solid angle definition etc little later see now we said intensity associated with emission at a given wavelength at a particular in a particular direction theta phi okay where is that this one so if I want to get q that q would be obtained this q would be obtained by dq is nothing but I e theta phi if I want I can even put lambda d a cos theta d omega this is just bringing the numerator and the denominator multiplying it so q essentially would be integration over all wavelengths over all theta over all phi is that right so q that is that what the amount of heat if you want to understand in layman's language the amount of energy which is emitted by the entire area irrespective of the wavelength irrespective of the directionality that is obtained by integration of this so we define something like a heat flux because it is not a constant we have to go this way and then integrate it back to give you the total heat transfer and when this quantity is associated with emission we will give that symbol e okay is this okay now we will just this is called as emissive power by the way I forgot to mention it is the radiation heat flux for emitted radiation is called as the emissive power which is defined as the rate at which radiation energy is emitted per unit area of the emitting surface okay so you have taken the definition just put it on the other side and you will get the amount of heat transfer now what I am saying is keeping the same idea same concept if I do not talk about emission and I talk about reception of energy okay that surface is getting energy I am the surface or let me go back to my watermelon I am this sitting here I am getting energy in all possible directions all possible wavelengths all possible directions per unit area so this is my unit area that is very nice I am saying what is the energy that is coming at a particular wavelength to me in that particular direction so from Vivec Cordet to me what is the amount of energy which is coming at a wavelength lambda I do not care about lambda minus delta or lambda plus delta this wavelength lambda which is specify in this direction what is the amount of energy coming that when I apply it to incoming radiation I will call it as irradiation okay irradiation refers to terms associated with energy coming to received by that surface exactly the same concept nothing different only thing will change is all surfaces emit radiation also receive radiation emitted or reflected by other surfaces I do not care what is the nature how is it coming is it coming directly from him or reflected portion of this tube light energy which is coming to me I do not care as long as an energy of wavelength lambda is coming from that direction to me that I will call as irradiation associated with that wavelength and that particular direction so this is given by symbol G okay so G essentially is just a nomenclature again and what happens is this I E instead of E the subscript will become small I because it is for irradiation that is the only change conceptually it is exactly the same when it was for emission it was associated with energy leaving the surface by virtue of its temperature I am stressing this again and again because we will introduce radiosity which is a slightly different concept emission is associated with the temperature of that body so energy leaving that surface per unit area in that direction theta phi per unit wavelength lambda through that solid angle d omega and reaching the area normal d a n all those things remain the same now instead of emission I am talking of reception incoming so it is called as irradiation I think is connect has been made I think now last thing okay every surface gets energy also emits energy amount of energy coming from there to me I will reflect also a portion of it I will reflect so a portion of that reflected energy plus the emitted energy so E is emission this part you see here R is the reflected part exactly the same definition so amount of energy blah blah blah blah everything emitted plus reflected portion of incident radiation again it is at that specific wavelength at that specific direction so when I have the direction wavelength dependency I will call as spectral when I do it for the whole area and over all wavelength it will be called as a total hemispherical are these terms clear I think this terms once they become clear it is very very easy to communicate so it is we are teaching English so what is spectral at a given wavelength what is total hemispherical all wavelengths all directions okay irradiation is referring to quantities which are related to incoming radiation radiosity given by symbol j refers to emitted portion which is a virtue of its finite temperature plus the reflected portion of incoming radiation so incoming radiation can be from him from him from him from all the places to me out of the incoming radiation what is the reflected portion in that theta phi direction okay I do not care from where the incident incoming radiation has come so this tubelite this wooden shelf all of you are going to emit radiation which I will receive out of the incoming radiation what is reflected rho times that quantity in that particular direction at that particular wavelength plus the corresponding quantity which I am emitting by virtue of my finite temperature in that direction at that wavelength is what is radiosity I do not think I can make it any clearer okay so this is the g which is the irradiation this is the emissive power this is the radiosity the mathematics is just same thing dq is blah blah blah so I will just go through the mathematics quickly for complete yeah I will go to solid angle first time okay now having understood hopefully this terms will come to a easier part which is solid angle so we have defined solid angle as dA n by r square so differential area dA n which is normal to theta phi direction as in the figure dA n this one is we are going to give it a form in the r theta phi that way so we are calling this angle theta this is r sin theta this length radius of this arc is r sin theta so r sin theta times this angle plane angle which is d phi because this triangle and this triangle are similar triangles the triangle at the base and this triangle are identical in shape similar triangles the central angle here is d phi the central angle here is also d phi so the arc length of this surface and the corresponding surface is essentially r sin theta d phi this surface is nothing but r d theta this arc length also is not moving this arc length is r d theta the move over so the area essentially is rectangle which is given by r sin theta d phi times r d theta which will come out to be r square sin theta d theta d phi all these I think is just geometry all of us know it so d omega is essentially dA n by r square which comes out to be sin theta d theta d phi so wherever in the next page when we start to do the intensity etc d omega is replaced by sin theta d theta d phi so for the sphere solid angle is essentially obtained by integration over all theta and all phi so it comes out to be 4 pi r square which is the total area how can we get coordinate area and coordinate orbit angle because it is connected by radius what do you mean area or angle I took dA n what area I took dA n says to which area it is going ice cream cone ice cream cone it depends how big is my ice cream if my ice cream cone is a big ice cream cone I need lesser ice cream cone to complete my sphere if I have smaller than this so this is how much it is covering so how much direction it is covering that is dA n that is dA n this is what so this one is smaller this one is much larger dA what will be like this when I say that meter square is source. Irradiation is referring to collector irradiation is reception emission and radiosity are linked because both of them have to be for the same surface which is emitting by virtue of a temperature it is emitting also it is reflecting so it is for the surface which is giving out this surface which is which this is talking to it is also receiving so when I talk of irradiation it is for the receptor so at that time this becomes a receptor taking in energy from all other surfaces surrounding. Correct. Correct. Exactly. Exactly. See I I think a visual picture if it is there I think life becomes very easy to teach this. So differential solid angle dA n by r square so intensity of emitted radiation has covered this total emissive power so hemisphere above the surface intercepts all the radiation rays emitted by the surface the total hemispherical emissive power from the surface surrounding it is given by integral from theta going from 0 to pi by 2 and so I do this and go on because I am taking only half the hemisphere now. Let me cover I have to cover the hemisphere how will I cover the hemisphere this is 2 but any point if I have to cover 190 is sufficient no. So I do this and I rotate so I form a quadrant on a plane a rotation of a quadrant gives you hemisphere right apart from dancing we have done everything it is it is hard it is hard. It is a triangle triangle and rotating. Yeah. Come back again minus minus 4 pi how is it possible yeah theta will be minus pi by 2 to plus pi by 2. Phi is this angle here phi is this angle phi is only 2 pi 2 pi is 360 degrees the rotation is only for 360 degrees which is 2 pi the theta will theta will go from minus pi by 2 to plus pi by 2 that 4 pi is actually 2 times 2 pi this 2 pi is because of this and the second is because it is an even function. No, no, no do not confuse do not confuse. I am not going to discuss that I am not serious I am not going to discuss that. Hey human this one this theta is going from 0 to pi by 2 so theta going from minus pi by 2 to plus pi by 2 show me. Convince me. Convince me. One second what is this theta. Rotating you said rotating so how much is the rotating angle that is it that is it. So you do not take a quarter anymore you take a semicircle and rotate. The semicircle is you take a semicircle and rotate a semicircle you will get it. So semicircle is that is 2 you are still rotating by 2 pi the semicircle quadrant has become a semicircle because I have changed it from 0 to pi by 2 it has become minus pi by 2 to plus pi by 2. So this is my semicircle minus pi by 2 to plus pi by 2 when I rotate which will become 2 times integral 0 to pi by 2. I think it is fine. A month you go back and visualize we will discuss. I think too much in discussion. So this is intensity of it is flipping intensity of emitted radiation we said this already. The intensity of radiation emitted by a surface in general varies with direction but many surfaces in practice can be approximated as diffuse what is diffuse surface independent of direction. For diffuse emitting surface intensity of emitted radiation is independent of direction therefore I E is a constant so I can just pull that out I will say E is equal to pi times I this is just maths you can do the maths and there is a small problem with that. Radiation incident which we call as irradiation intensity of incident radiation is defined as the rate at which radiation energy DG is incident from theta phi direction per unit area of receiving surface again the same blah blah blah normal to this direction per unit solid angle about this direction per if I was using lambda subscript per unit wavelength lambda d lambda blah blah between the direction of incident radiation and the normal to the surface. So once the concept is clear you can put it in words. So G is nothing but BG integrated for the hemisphere the total radiation flux incident on a surface from all directions. So I am sitting at the center of the hemisphere what is the total flux which is coming on to me that is what I am calculating here which is nothing but integration over theta going from 0 to pi by 2 phi from 0 to 2 pi which comes out to be again pi times I subscript I this subscript is for irradiation. Radiosity accounts for all the energy leaving the surface by radiation it is the sum of reflected portion of incident energy plus the emitted energy that is why E plus R. So the rate at which radiation energy leaves unit area of surface in all direction and we have seen that here we come spectral quantities are wavelength dependent. So I will define only one of this everything else will follow. Spectral radiation intensity for emission I intensity lambda spectral E emission this is at that wavelength theta phi is nothing but dq divided by dA1 cos theta which is the normal area d omega is that per unit solid angle business d lambda is that per unit wavelength. Essentially because this is the definition actually dq if I have to calculate is just the product of the left hand side and this one. So I lambda E which is a function of lambda theta phi times dA1 cos theta sin theta d theta d phi which is a solid angle times d lambda. So a triple integration that is what will be there on the next next page triple integration over theta over phi over all lambda will give me whatever I got total hemispherical emissive power that is over all wavelengths over all directions. You had a question sorry we have not brought any temperature into picture here. Then its characteristics of emission will be different at different points of the surface correct sub divide and this is applicable to a surface irrespective of the size of the surface of course it is applicable to a surface at a finite temperature the surface can be a small surface at a finite fixed temperature then only it gets range of the wavelength gets fixed otherwise as you said if it is multiple temperature then I will have multiple wavelength. So this definition if I take the denominator up and integrate with respect to theta and phi that means I am keeping the wavelength dependency still. So spectral hemispherical emissive power that means at that wavelength lambda over the hemisphere what is the emissive power that is obtained by E subscript lambda. So all of us are familiar with E when these subscripts come that is where the problem occurs. So this lambda indicates it is emission at a particular wavelength that concept should become clear. So it is just obtained by integration with respect to theta and phi again it is watt per meter square at a particular wavelength. So when the variation of spectral it should be watt per meter square steradian actually yeah watt per meter square steradian. When the variation of spectral intensity radiation micrometer sorry not steradian micrometer when the variation of spectral radiation intensity with I lambda wavelength the total radiation intensity for emitted incident or emitted and reflected whatever will just be integrated over the entire spectrum. So I lambda E would be integrated from 0 to infinity over all wavelengths. So let me just ask a question just to check whether you are we have reached you or not. What does this 0 to infinity mean? What does this 0 to infinity mean? I just answered yeah we just answered but let us see 0 to infinity means what literally should I be taking 0 to infinity? No, the band of the spectrum which is really contributing the thermal radiation for a given temperature for a given temperature we have to thank you you reminded that this is for a fixed temperature that was not stated explicitly it is fixed temperature when I fix the temperature my wavelength range gets fixed. So once I get the wavelength range my this 0 to infinity gets fixed. What is the total? So it is not 0 to infinity it is some lambda 1 to lambda 2. That is what I wanted to emphasize it is not it is just stated for the sake of. Generality. Generality but case by case that is case is getting fixed by temperature. If I want to make life difficult for a student I will give some multiple peaks just nonsense but what we are saying is it is integrated over several wavelengths. So this same definition holds good for E, G and J. So this is something which we present. So this is first one is you tell me total. What is? No, no, no you cannot. Full, full big word. You have to qualify it Hemanth Kumar from Sardar Vallabhavi Patel that is what qualifies you. You have to qualify what is this emissive power, total emissive power that is at that temperature. G is J total hemispherical radiosity I think the message is gone. So now we have got the definitions clear actually there are two ways of teaching see to not to scare the student even I was just in the morning also I was dilly-dallying should we be starting radiation with intensity. In fact we can go Chungal's way also this is the way incorpore approaches thermal radiation. So what perhaps to make to ease I mean or the to keep the students at ease with radiation what Chungal does is actually what he does is first he gives this introduction all this introduction and he immediately after this he will immediately go to black body. Why black body because in emissive power when I say emissive power emissive power is something which is independent of I use the word emissive power only when it is independent of direction. It is used when directionality comes I use intensity that is what his approach is. So it is not a it is not a bad idea to come first come from black body radiation here you do not need intensity anything you have to just state maybe it is a good idea maybe it is a good idea otherwise if they do not get those definitions they are thrown out of gear. So my suggestion would be after going through this this time we are going to teach this way only that is let us not first give them the definitions. So we can come to black body they would be knowing Planck's distribution we will be stating that I mean I do not know what professor Arun would opine on this this approach but only thing is that otherwise we have to spend so much time as much as professor Arun spent maybe one class two classes only going through those definitions you see we had enough fights in understanding each other. So we should go through those fights in the class so that we have reached them clearly then perhaps we can go this otherwise we can make little easier by coming till here but there is a problem when we come to band emission because I have to introduce intensity again. So so that is that is the problem so I think we can we can go the way we went then I think that is the best way I think we will calibrate we will go best way I think we will calibrate ourselves to go to incorporation, incorporation, okay. One word of caution you will give notes for this for how many of you give notes for students NITs I think they will not give other places so you write on the board no the thing is because these definitions all these things I think they have to just listen first and appreciate it okay for a student to read this and understand is next to impossible first time. So I think you have to insist that they hear you out first they listen to you first and then write it down so maybe I will suggest that do not spend time writing on the board and then just hand out maybe yeah something like that you ask them do not do not the time can be better utilized to okay no dictation right that is something because if you are writing the students are just busy copying them actually to come back to the same question we now that this course is over and you have complete material given PDF file you can make it as a book or a handout or chapter by chapter you can give they can bring that I was thinking that if they only print it on one side on the left side whatever extra you tell and write it on the board over and above the material what is there in this notes they can write that then perhaps even if they do not read a textbook perhaps we have reached them of course we should encourage them to read a textbook in fact I tell my students to bring textbook to class whichever textbook you like then I keep saying I like you because you have brought the textbook just to inculcate the habit of having a book with them okay so I think this way if you do we can reach them in a faster fashion and more effectively perhaps perhaps okay.