 I will proceed further. So, d omega therefore, I will define differential area d a n is normal to theta phi direction as shown in the figure d a n is normal to the direction of viewing since d a n is viewed from the center of the sphere. So, I draw a line which is going to intersect the center of the sphere point center of the earth to the center of the area outward normal we understand this concept of outward normal where in which direction is this normal if I go to a surface if this is the surface if this is the surface outward normal we do not draw like this this is not an outward normal this is not an outward outward normal is this intuitively we know what an outward normal is that is what we are going to use here the line this is the outward normal in this direction. Now, let us do a little bit of geometry this is a radius r for the sphere this zenith angle is theta. So, this is also r because this is the radius of the sphere. So, this arc is I mean sorry this line here is sub 10 is having a length r sin theta. So, this right angle triangle I will why am I struggling here I will draw it here separately this right angle triangle is going to this is a right angle triangle this is on a sphere of radius r. So, this is my theta imagine in three dimensions. So, this length is r sin theta. So, this length is let me call it a b c whatever. So, length a b is r sin theta. Now, this angle what is that let us go back to the diagram this angle is nothing, but d phi why these are similar triangles here the one on top is similar to this triangle at the bottom what is that this represents roughly the this angle d phi gives you the width of this area here. So, this angle is d phi. So, this sorry sorry sorry wide board. So, this is d phi central angle. So, what is this arc length b c. So, arc length b c is what when d phi is small r times theta is the arc length. So, I will write this as r sin theta which is the radius of this circle for which b c is a part b c is a part of this circle whose radius is r sin theta and the central angle is now d phi. So, I have come from third dimensional to the two dimensional figure which is here. So, r sin theta d phi this is one length also this length is also the same. So, length b e is also the same b e is also the same now I have to get length b c equal to c b. So, length b c is what for that I go to this figure correct this angle is d theta this radius is r. So, this arc length b arc length b e is going to be r correct let me see if I made a mistake. So, r d theta is this arc length r sin theta d phi is the other arc length correct. So, this is r b e is equal to c d equal to r d theta b c is equal to d e is equal to r sin theta d theta what is the area now length into breadth. So, it is going to be product of these two area d a n is r square sin theta d theta this is how I get and solid angle by definition d omega is equal to d a n by r square. So, this divided by r square will give me a solid angle d omega is nothing, but sin theta d theta d phi r square will cancel with r square in the denominator sin sin theta d theta d phi that is what. So, this is the definition of solid angle please keep this in mind and we are just doing a simple match here solid angle of a sphere what should it represent if I am standing what is the solid angle solid angle physically or forget physics common sensically represents the angle subtended by me sitting at the center of the earth to a small area which I am seeing. Now, if for a sphere if I am looking in all directions what am I doing I am going to get the surface area of the sphere. So, answer I should get is 4 pi r square which is what I will get solid angle of a sphere is this d a n full area which I am going to do full sorry what is the total area for the sphere I am going to calculate total area is nothing, but theta going from 0 to 2 pi what is theta again let us remember theta is this what angle zenith angle and phi is going from phi is going from 0 to 2 pi theta is going from 0 to pi why because if I have I will go like this if I go like this I will go to one half and then going from this way up I will get the second half. So, it is hard to show it here let me just draw it how is a sphere obtained let us first understand that how is a sphere obtained sphere is obtained by rotation of what if I take a quadrant if I take a quadrant and I rotate if I take this if I take a strip like this and move it in a plane like this. So, move it in a plane I will get an a quadrant if I move it further I will get a semi circle if I move it all around I will get a circle circle in a two dimensional plane. So, an arc this one is going to give me a circle now if I rotate this circle about this axis I will get a sphere. So, this rotation is going to go complete this is going to be like this. So, what I have told here essentially is solid angle of a sphere theta going from 0 to pi. So, this part is going from 0 to pi. So, 0 to pi when this goes what does it give me this will this is going to give me a semi circle this semi circle if I rotate completely by 0 to 2 pi. So, I have a semi circle strip just a planar semi circle I rotate it in 0 to 2 pi all directions I am going to get a sphere that is what sphere I will get a sphere 0 to 2 pi I will get a sphere and that is what comes out here. So, 0 to minus pi to plus pi with a 2. So, this will give me for a hemisphere and then for a sphere it will be twice sorry I let me just recap this. So, this is a semi circular strip which I have if I rotate this like this how is this semi circular strip obtained when theta goes from 0 to pi. So, this I had I had a small strip here into dimension that strip was going from 0 to pi 0 to pi I got this flat semi circular strip this flat semi circular strip when I rotate about this axis that is phi going from 0 to 2 pi circular rotation. So, phi going from 0 to 2 pi I will get a hemisphere. So, a semi circle planar semi circle when rotated will give me a hemisphere that times 2 will give me a solid sphere. So, instead of going from 0 to 2 pi like this I can go from minus pi to plus pi and multiply by 2 because it is an even function. So, that is what is here. So, this 2 pi r square represents the surface area of a hemisphere that times 2 would give you the surface area of a solid sphere. So, told that in a professor Prabhu wants to emphasize why we take only hemisphere in heat transfer analysis. In real life sun if you take is going to be a sphere sun is going to be a sphere, but as I told you our situations involve what our situations involve practical problems where we are dealing with energy transfer from one side. If you take a furnace wall as I said what is going on the outside will be at a different value of temperature itself. So, I am dealing primarily with interaction between surfaces. So, I deal with the top portion or bottom portion alone therefore, I am it is for me to deal with only hemispherical quantities. So, even if I talk of a light I will talk as if there is a cardboard which is blocking the light which is going down and light is just butting out through a small hole and it is only going to cover a hemisphere that is all I am going to deal with in this course, but yes three dimensional only this is also we are just considering hemisphere. We now have a differential solid angle d omega subtended by area d a when viewed from a point is this is what is it is expressed as d a cos alpha by r square where r is the angle between the normal to the surface and the direction of viewing. What is this? This is essentially we are saying we want to know what is this normal business. What is this normal? Why are we dealing so much with the normal? So, this normal business is nothing but the projected area. How is this coming differential solid angle d omega subtended by an area d a d when viewed from a point at a distance r from d a is expressed as d this is the solid angle definition, but this normal area d a n is nothing but what you see and cos alpha associated with the projected part. Let me just go back to this. We can take this example. This is a problem which we are going to solve. Let us take this area a 2 I want to explain this this way because it is already drawn for us. This one is directly perpendicular a 1 and a 3 are directly facing each other one above the other and a 3 is parallel. So, the normal is this line whereas, if I look at a 2 and a 1 the normal to a 2 is like this, but the line joining a 1 and a 2 is this. So, the area which is seen the part of area the portion of area a 2 seen by a 1 if I am sitting at a 1 and what I see I do not see full area a 2 I see the projected area. So, that projected area is given by this cos alpha business. So, this projected area if I want to draw it here this is my area this is the normal direction to this the projection and this is if this is the angle alpha this projected area is nothing but d a n cos alpha this angle is alpha. So, actual area seen is this one this is my area d a n if this angle is alpha then right angle triangle is formed like this this projected area is nothing but d a n cosine of alpha it is going to be a smaller area that I see this we know from drawing engineering drawing itself we know that projection. So, that we have to keep in mind because the surface does not see this it sees the normal. So, if I am if I have a surface and another surface directly above it that is the best because the it sees the area completely the moment I incline one of this I have to go into the normal business. Before we go here small surfaces when viewed from a relatively large distance can be approximately treated as differential area in solid angle calculation. For example, solid angle subtended by 6 centimeter square plane when viewed from a point the distance 90 centimeter normal to the surface is very it is a point that is why now let us go back why we do this in all these calculations that you did convection conduction where emissivity was given radiation was taken the ball is cooling even if radiation was there in lumped capacitance you did not take into account any of these solid angle business. It is cooling to the environment and the environment is the room is so large compared to the ball you do not have to worry about all these things about how much which direction nothing we worry we take the whole thing because the surface is very small compared to the ambience that is why we will come back to these problems slowly one by one. Now this intensity business I said intensity is related to the direction the same torch light the intensity I am going to the sun white orchard sun intensity of heat received by Malaysia is going to be higher than what you get in Greenland that refers to intensity. Now we are saying intensity we use this definition for all the three quantities which I have defined emission irradiation radiosity what is emission energy it is relating to the energy emitted by a surface by virtue of its finite temperature irradiation energy received by the surface by virtue of interaction with all other surfaces at various other temperatures I do not know what they are radiosity refers to emission plus reflected component of all incident radiation. Now we will go quickly since we have understood all this consider emission of radiation by differential area d a 1 of a surface we are looking at this area which is in pink radiation is emitted in all directions we know that in where into a hemispherical surface and the radiation streaming to the surface area d a n which is here d a n is proportional to the solid angle subtended by the surface. So what this these two lines are telling is the following it is a poorly drawn hemisphere thanks to this pen anyway if yeah this is my d a 1 this is my d a n this area a b c d if I expand this bigger this is a b c d this is d a n and this is subtending an angle solid omega at the surface d a 1 this solid angle is d omega I hope now it is clear. So you are sitting at the center of the earth you are looking at the area d a n. So small area 1 square centimeter will subtend a small solid angle 1 square inch is going to subtend a larger angle 1 square kilometer is going to subtend a larger angle. So essentially what this statement is saying solid angle dictates that is what radiation emitted in all directions and the radiation streaming through a surface area d a n is directly proportional to the solid angle obvious it is now radiation is also proportional to the radiating area of course that is true I have a tube light in your room that tube light is completely covered except for you know a small strip which is left open the amount of energy that is going to come out is less it is tube light is not stop giving out energy what it is coming is what is gotten reduced. So it is proportional to the area which as seen by an observer on d a 1. So what I see exposed which is giving out is what is important not what is not going to give energy out which varies from a maximum of d a 1 when d a n is at top directly we saw that I stand directly below a tube light I feel hotter as compared to when I move away from it you keep your hand directly above a pan when you are making those I you see it is very hot you take it away it is cold this is intuition common sense that is what we are doing now. So when the things are directly normal above 90 degree just parallel to it that is the best when I move away inclined position like this it is going to be different the effective area d a 1 for emission in the direction of theta is projection of d a 1 on a plane normal to theta which is d a 1 cos theta what does this then effective area for emission effective area means what in the direction of theta I want what is going in the direction of theta. So though I am talking of the area d a 1 the actual thing which is of interest because I am talking of normal c is this projected component of this area. So this area is flat does not matter to me because in this direction what is going out depends on the projection that is what is given by d a 1 cosine of theta. So d a 1 is this pink area which is horizontal now if I want to do the projection business this will become like this I can do one of two things I can make the receiving surface do the projection business for it or the sending surface. So when I am dealing with intensity associated with emission because it is based on the surface which is giving out energy I will manipulate the surface. So now we come to the most important set of definition here if we understand this we have understood I think a family of definitions. So I will go very slowly word by word this is just English interpretation only English interpretation there is nothing we have understood the physics we have understood the logic now we are putting it in words that is all radiation intensity what I am doing now I am going to define intensity for emission with relation to emission I can do the same definition relation with relation to radiosity I can do the same definition with relation to irradiation if I understand this those two I will understand in parallel automatically. So every symbol every subscript every bracket has a meaning let us understand there is no memorization here it is just logic. So radiation intensity for emitted radiation so that means what I am talking of intensity with related to emission that is what is this first thing radiation intensity for emitted radiation intensity is given by symbol I as we look at it it is I intensity related to emission is I subscript E this subscript E is for emission. So I will write now without wasting my further time I is for intensity symbol this subscript E refers to emission what is it defined as I subscript E is defined as this theta 5 we will come back to it is defined as the rate at which rate what does rate mean rate means it is energy transfer per unit time so it has to be some kind of watts. So rate at which some watts joules per second at which radiation energy dq so how much energy dq is emitted we are talking with relation to emission so this has to be emitted rate at which radiation energy dq is emitted in theta 5 direction remember we are dealing with intensity. So as I said amount of energy coming to India from the sun is going to be different because of the difference in the angle. So this theta 5 direction becomes very important so I subscript E refers to emission intensity associated with emission now what it is in which direction theta 5 this tells me the relative position of myself with respect to the point. Let us keep this hemisphere in mind always this point here subtends a different theta 5 as opposed to another point here which may be having the same area does not matter it subtends a different angle theta 5. So I subscript E is related to intensity associated with emission this parenthesis theta 5 refers to the angle what angle radiation energy dq emitted in theta 5 direction how much is emitted in that particular direction in that particular direction per unit area. So it is a flux kind of term rate is what watts units per unit area therefore it is a flux units normal to this direction this is very very important. If I have a surface it is going to if I have a surface it is going to emit like this like this like this there is a theta 5 direction point p I will call which is at r theta 5 this gives me the complete coordinates of it intensity at this point is related to what is coming out normally not what is coming out from here we have told normalcy is very important. So normal to the theta 5 direction what is coming so amount of energy per unit time therefore it is watts coming out in theta 5 direction we will put that in per unit area therefore it is meter square what area it is the normal area. So watt per meter square theta 5 direction this area is subtending an angle what d omega solid angle. So about that solid angle am I right so for that unit solid angle which has unit steradian s r. So watt per meter square steradian our definition has to be very general. So I will say if I have red light coming out it is at a particular wavelength blue color light it is at a different wavelength green light it is at a different wavelength there any any object can have a wavelength specific spectrum right. So if I have light of wavelength red light it will have a particular wavelength associated with it so per unit wavelength also has to come why that will come I will tell you in a minute. So radiation energy intensity or for emitted radiation I subscript e theta 5 because in that particular direction is defined as rate at which radiation energy dq is emitted in the theta 5 direction per unit area normal to this direction per unit solid angle about this direction. So this is the general thing I have used micrometer there I will come back to that in a minute. So this is the general definition of intensity what does it mean let us forget all technical words let us go to grandmother's language amount of energy leaving a surface amount of energy leaving a surface per unit time wattage in theta 5 direction. So if I place the computer if I place a paper like this flat I am going to have a very steep angle to be reading it but if I keep the paper like this it is easy for me to read it. So this is the normalcy business keeping the paper like this I will have difficulty reading keeping it like this I will read it better so that is the normal business. So if I have a surface which is emitting in all direction the surface is like this I am going to see from here this ray is going to be very very slant so the normal area of the surface becomes important. So amount of energy per unit area emitted per unit area in theta 5 direction so watts per meter square in the theta 5 direction where what is the infinitesimal angle that I am concerned with d omega solid angle so that is why this thing is there in the theta 5 direction per unit area normal to the direction per unit solid angle. If I have a larger solid angle the intensity is going to be more correct if I concentrate a laser what is the laser high intense beam of light why is it high intense because all the energy is concentrated at a point correct torch light on the other hand is probably brighter than a laser actually but the intensity is going out in all direction that is where the intensity comes in. So per unit solid angle if I take a larger solid angle then intensity is weakened correct I go far away far away from the source of light you do not feel the effect of the brightness or the heat I come closer to the source of light I feel the brightness and the heat more that refers to intensity. So this intensity therefore is d cube per unit area normal d a 1 cos theta is there because of the normal c d omega is there because of the unit solid angle associated now I will substitute for d omega which is nothing but sin theta d theta d 5 which we have derived long ago here sin theta d theta d 5. So that is how I will get intensity of emitted radiation is nothing but watts per meter square steradian comes because of this solid angle definition ok. So this is the general definition. So radiation flux for emitted radiation which we have used so often emissive power is the rate at which radiation energy is emitted per unit area of the emitting surface. So completely per unit area I do not care anymore about all these things this q is the emissive power that is nothing but this intensity times d a 1 cos theta sin theta d theta d 5 this is unit area. So this is your intensity associated with emissive power. Now our fundamental understanding of emissive power is total emissive power we do not when we did emissive power we never bother about angle all these things. So hemisphere above the surface intercepts all direct all radiation emitted from the surface that means what I said the point source is going to form a hemisphere of radius r depending on where you are standing and observing. So this theta going from 0 to 2 pi and sorry phi going from 0 to 2 pi and theta going from please note this 0 to pi by 2 theta going from 0 to pi by 2 gives me a quadrant of a circle. This quadrant of a circle when it is rotated completely in 0 to 2 pi gives me a hemisphere. So that is why hemispherical emissive emissive power we will use this word very often is nothing but this quantity integrated between theta going from 0 to pi by 2 and phi going from 0 to 2 pi. So this is what we have already used all the time emissive power watt per meter square what is it it is obtained from this fundamental definition of intensity. So let us just go back and say intensity of radiation emitted by a surface in general varies with direction this is the most general thing especially with the zenith angle this angle. But many surfaces in practice can be assumed as diffused surfaces what do I mean by diffused surface is already told it is a surface where directional independence does not favor any particular direction. That means this intensity is independent of direction this i e which is a function of theta and phi is no longer a function of theta and phi when something is a function you put it inside the f of x and y means function of x and y when it is independent of x and y you can pull it out as a constant. So this can be pulled out of this integral as a constant it will only be integration of this cosine theta sin theta d theta d phi this can be substituted as sin 2 theta by 2 d theta d phi d phi integration is very straight forward you will go you will get 2 pi sin 2 theta by 2 you can do the integration cos 2 theta so on and so forth and that is going to give me e is equal to pi times i e what it means is for a diffuse emitter when the surface is a diffuse emitter it does not favor any particular direction for such a surface the total the emissive power what per meter square we have not brought in wavelength we have not brought in any wavelength dependency here I am strictly going by what is here wavelength we do not care only the direction we have taken when the intensity of emission is not in any particular preferred direction we call the surface as a diffuse emitter for a diffuse emitter i subscript e bracket theta phi i is intensity subscript e is for emission theta and phi refer to the directionality that comes out as a constant we will just call it as i subscript e the whole function what is there in the bracket gets integrated you would get it as pi so the emissive power associated with a diffuse emitter is going to be pi times i please remember this there is no need to buy had this it comes from fundamentals I am going to go couple of steps forward not do this problem purposely because I want to complete irradiation and radiosity if we have understood emission that means all this subscript brackets business I will be able to extrapolate this to irradiation what is irradiation just I am just going through this slide all surfaces emit radiation but they receive radiation which is emitted or reflected from other surface I do not care what it is but if I have another surface which is giving out energy at a particular temperature this energy that is coming out is not just the emission alone it is also the reflected part of energy received from all other surfaces which with it is interacting. So, for example, you have a bench in your room you are sitting there the bench is going to get energy from all sources of heat and light it is going to reflect some portion to you and also by virtue of it is finite temperature have an emission. So, all those things that is coming out is the irradiation that is what is given intensity of incident radiation. Now, I have to become little bit careful because emission was related to the surface of interest irradiation is not only because of I mean irradiation which is coming to this surface is because of other surface. So, this is the surface of interest and this is another surface to irradiation is because of this energy coming out from here which is nothing but emitted part plus reflected part of all energy that it has received. So, that is let us see this intensity again our word is intensity. So, it is I intensity of incident because it is coming to my surface d a. So, that is why it is incidence I subscript I theta phi that means in that particular direction intensity of energy coming in that particular theta phi direction. How do I get this? I am sitting in Bombay the angle subtended by this with respect to a source outside. So, sun's rays is intensity is changed because of the directionality that is what is given by this theta phi is defined as the rate at which radiation energy d g is incident coming in from the theta phi direction per unit area again we are talking of per unit area what per meter square per unit area now of which surface not the sending surface of receiving surface we are doing everything for us receiving surface. So, emission was with respect to this surface irradiation is also with respect to this surface how much is coming out all those things will be decided by the area associated with the sending surface temperature of the sending surface reflectivity of the sending I do not care about that as long as I know a number associated with it and the relative position of it. We are saying per unit area of the receiving surface in this direction in this direction is given by theta phi rate again. So, it is a what rate at which radiation energy d g incident from theta phi direction per unit area of the receiving surface larger this surface if I make this a size of a notebook page the larger the area it is going to get more energy common sense. So, this is per unit area that is why flux per unit solid angle about this direction again the solid angle business becomes very important larger the solid angle subtended the solid angle larger the solid angle subtended if it becomes a small ice cream cone it is going to bring in a small set of energy coming larger ice cream cone if I subtend here larger is the energy it is going to bring it is like a funnel the funnel is going to take in more and more energy and bring it to d a 1. So, intensity again is I incident radiation because we are talking about irradiation theta phi because of the directionality I will now put this in words it is just I mean I will put it in symbol form it is nothing but d q let me just go back here it is be nothing but d q here similar form it is a flux same units watt per meter square irradiance except that now it is related to incident energy coming to the surface. So, if the irradiation flux incident on the surface from all directions if I know this in this theta phi direction integrating it over all directions is going to give me the net irradiation the total energy coming in from all directions. So, one tube light there another tube light here something behind me all those things is going to be coming into the surface therefore, g is equal to integral dg and this we have not written I will write this is nothing but d q over cos theta sin theta d theta d phi this is I intensity incident theta comma phi in that direction is nothing but d q over this quantity. So, it is nothing but d q divided by d a 1 cos theta sin theta d theta d phi the same thing because we are referring this to the receiving surface. So, it is going to be d a 1 a 1 notice that this is what is referring here to irradiation. So, again the units are watt per meter square steradian and this quantity the total energy that is going to be incident coming in from all directions is going to be given by this I times theta I of theta phi cos theta sin theta d theta d phi area is made. So, this is the radiation flux what we get in problems radiation flux that is what this is if the incident irradiation is diffuse that means, if this I subscript I is not a function of any direction if this is not a function of any direction I can pull this out just as I did for emission I can pull this out here exactly the same way this I subscript e came out and it was just the integration of this same steps I will do I will get g is equal to pi times I I got let me remember e is equal to pi times I e for a diffuse emitter g is equal to pi times I I is it only a difference in the nomenclature it is not just a difference in the nomenclature it is essentially telling incident radiation is diffuse what is coming in is diffuse here in e for e I said what is going out of the surface is not showing any directional preference for g I am saying what is coming into the surface into the surface does not have any directional preference that intensity is not showing any directional preference only in this case I can I can get this g is equal to pi times I subscript I now let us just go and quickly finish radiosity radiosity accounts for all radiation leaving the surface rate at which radiation energy leaves per unit area of the surface. So, this is now just going to be very easy it is by nomenclature just e and r. So, let me go to the whiteboard intensity of emitted part plus reflected this is the emitted part this is the reflected part of all g that came in all energy that came in each of this would have some part would have been reflected. So, this is the intensity of total energy first we saw intensity of emitted energy then we saw intensity of incident irradiation now I am saying some new quantity which is this emitted part plus reflected part of all the incident radiation. So, this part I am just calling as r reflected component. So, exactly parallel definition dq I am not even looking I hope I am right dA 1 watts per unit area what direction normal to that theta phi direction per unit solid angle this came because of the solid angle d omega this came because of the normal to theta phi direction. So, this is again watt per meter square ster area whatever I understood for emission same concept except that this quantity is a new thing reflected part of all incident radiation when I say incident radiation that incident radiation depends on how far I am from that surface which is giving out energy what is the dimensions of the surface which is giving out energy what is the temperature of the surface which is giving out energy all these things are incorporated when I define g, but the thing is I am not concerned with all those things directly at this point I am concerned only with what is coming to this surface I of interest. So, if I go back and complete the definition radiosity rate at which radiation energy leaves completely in all directions. So, if I say this is in all directions I have to integrate with theta and phi. So, double integral going from theta to 0 to pi by 2 will give me a quadrant of a circle that quadrant when it is integrated from phi is equal to 0 to 2 pi will give me a hemisphere this quantity all these radiosity irradiation emissive power e g and j that we define e g and j which we define and which we had units watt per meter square we will call this not just as emissive power irradiation radiosity we will call them total hemispherical emissivity total hemispherical irradiation total hemispherical radiosity why we have lost what have we lost in this we have lost all wavelength dependency. So, there was no there was no lambda in our in our definition wavelength was not involved. So, I did not care about wavelength direction we throughout because we integrated completely in all direction correct. So, and what did we generate we generated a hemisphere. So, this is over all wavelengths all theta phi. So, that is why we will like to call it total and because it covers this hemisphere is something we reiterate only for ourselves we will call it total hemispherical. That means, what is the energy coming out in all directions which is now going to form a hemisphere that is emissive power all wavelength yes all directions all wavelengths all wavelengths that is going to be implied when I use this symbol e if I use this symbol e lambda that means it is at a particular wavelength this means at a given wavelength we will see that later and that we will call as spectral emissive power spectral means at a given wavelength spectral irradiation at a given wavelength. So, and then if I say e lambda theta or phi theta phi that will be sorry next. So, e was total I will do with e everything you can do with g and j again total hemispherical emissive power e subscript lambda will give me a spectral at a given wavelength. So, it will no longer be total, but we have integrated over all directions. So, it will be spectral hemispherical emissive power hemispherical because there is no theta and phi dependency. So, automatically this word has to come this is e therefore, it has to be emissive power and if it is e theta phi suppose just hypothetically sorry theta and phi it is directional emissive power meaning I have not put any wavelength over all wavelengths what is the energy coming out in this particular direction. So, this will become directional emissive power and the most general one e lambda theta phi this is spectral because there is a wavelength dependency directional because directional dependency is there emissive power and I will just finish sign off giving the units for these this came out to me very nicely as watt per meter square. We like this we have used it right from conduction probably even 12 standard we have used this emissive power this is per unit wavelength. So, this is watt per meter square micrometer this is directional. So, this is watt per meter squared only directional. So, it will only be steradian this most general quantity which is the most confusing is per unit wavelength that particular color of light think of a color of light red color light coming out from the sun reaching India that is the direction specific. So, watt per meter square steradian because of the directional dependency micrometer because of the wavelength dependence. So, with this we will sign off.