 Hello everyone, welcome to today's lecture on the course remote sensing principles and applications. In this lecture we are going to continue with the concepts of radiometry what we have started in the last class. Just as a recap in the last class we defined what a plane angle is and what a solid angle is. We have done a small problem in calculating the solid angle certain by sun and moon on earth surface. Then we defined various radiometric quantities. We defined what radiant energy is. We defined what radiant flux is that is energy per unit time. Then we defined radiant flux density that is energy per unit time per unit area. If that radiant flux density is coming towards an object we call it as irradiance or if the object is emitting energy we call it as emittance. And also we defined one important property we often use in remote sensing that is radiance. Radiance is defined as the energy emitted in unit time in a given direction per unit solid angle that is what is the amount of energy per unit time that is radiant flux phi divided by in a unit solid angle. So, divided by the total solid angle omega will come in that is divided by the projected area in a given direction. So, the area what you are going to consider we have to project it in the direction in which we are looking at. Why this projection is occurring because as our viewing angle changes as we look different objects in a 3 dimensional space based on the angle in which we look we will perceive the object in different sizes and different shape. As I said as an example in the last class if you are flying in an aeroplane and looking at like whatever is there on the landscape based on the angle in which we look the area on the landscape be it an agricultural area or like some huge circular surfaces whatever based on our viewing angle it may appear as a circle, it may appear as an ellipse but different area depends on in which angle we are looking at which height we are flying and so on. So, in remote sensing normally a remote sensing sensor which is there in space will perceive objects differently based on the angle in which it is looking and that is why this projected area is really important. And also in a small solid angle because all objects whenever we see or whenever a sensor sees an object on the earth surface it is going to subtend a small solid angle. So, what we are going to continue today is we are going to continue today by looking at the relationship between irradiance and or radiance and radiant flux density. So, before actually seeing this we will make one thing clear that is radiant flux density is actually directionless, directionless in the sense when we define we said if this is the area I am going to put a hemisphere around it and whatever energy that is either coming in or going out within the hemisphere is radiant flux density. So, we are not carrying about in which direction energy is coming whatever energy within the hemisphere we are taking it. Whereas in radiance we are interested in see if this is the area if the energy is going in this direction I am going to calculate like with a given solid angle I am going to calculate in this particular direction I am going to project this project the surface area with respect to the normal and calculated. If the direction changes this is further going to change. So, radiance is directional. One more important property is radiant flux density reduces with the square of distance that is let us assume there is like a small source of energy is here. Like let us say this is point P I have kept like a small source of energy there. Let us say it is emitting P watts of power that is some energy in unit time we call it as P watts. Now that is happening let us imagine I have placed a square of 1 meter square area unit area I place it at a distance of r from this point source. So, this is a source of energy I am placing one meter square area of some object and I am going to calculate what is the power from this particular object is going to fall on this surface area. So, if I see it from the perspective of this particular object see now what I am doing I am going back to the definition of radiant flux density this surface area I am lying it flat on a plane the point source is now here point P it is at a distance of r. So, this is 1 meter square area let us imagine only this is the power source if I put a hemisphere around it if I put like a hemisphere around it only this energy is coming in no energy source from another directions are coming only this is the energy coming and falling over it. So, I am going to calculate what is the irradiance E received by this unit square meter of area. So, what essentially happens E is equal to total power P divided by 4 by r square that is the power emitted by this particular point source P is now spreading uniformly in all directions because it is there in a three dimensional space I said it is like a small point source and it is emitting energy in all directions. So, energy in I am placing an object in one particular small area on the sphere. So, what is happening for this particular point P I put a huge sphere around it of area 4 by r square within this I am placing an object 1 meter square object and I am just going to calculate the irradiance on this particular area. So, this drawing and this drawing are kind of analogy for you to understand. Now, what I am going to do is I am going to move this particular small surface area to a distance of 2 r. So, what essentially is happening now if I move this to a distance of 2 r that is I move it to a further point this point is now move to like a further point something here. So, essentially whatever this point P is emitting whatever power it is emitting now it is distributing itself over a extremely large area like the area is increased 4 folds because of the sphere surrounding this point is going to double in area this what to say the energy emitted by this object is going to now expanded over a area that is 4 times the initial. So, even if I keep the same a square here the same area of 1 unit meter square here the energy I am going to receive is going to reduce by 4 times. Similarly, if I move the distance by 3 times the energy is going to reduce by 9 times and so on. So, conceptually in order to understand as the distance between the source and the object source and the receiver increases the energy received by the receiver is going to decrease by square of a distance that is the radiant flux density. It is very simple analogy is like let us say there is like a big burner that is burning we are standing at some distance from it. Say if we move closer and closer we are going to feel much hotter if we move away and away we are going to feel little bit cooler not that much of heat will be there because the energy emitted by that particular source is going to decrease as the distance between the source and receiver increases. How this thing decreases by a distance square law that is by square of distance as the distance double the radiant flux density received will become 4 times less. As the distance becomes thrice the radiant flux density received is going to reduce by a factor of 9 and so on. So, this is the major thing with radiant flux density the energy received by an object of unit area will keep on decreasing as the object moves away from the source. On the other hand let us take radiance. Now, in the same figure what we are going to do is radiance is defined as like a in concept of like a solid angle. Now, this is again like a point source what I am interested is what is the energy emitted in the solid angle here. Let us call it as like phi or omega or whatever it may be within the solid angle what is the energy emitted. Even as the distance increases even when distance increases we are retaining the same solid angle. As per the definition we have to measure over the entire solid angle itself. So, we are going to increase the size of our receiver in order to collect all the area basically right. That is within the as the distance increases if we want to maintain the same solid angle the surface area is going to increase. So, if I want to place a receiver there at a distance of r I need to have a smaller receiver to collect all the energy at the distance of 2r I need to have a larger surface area and so on. Same thing, but if I keep a larger surface area based on distance I am still going to collect the same amount of energy. So, radiance if you look at the definition of like energy in a given solid angle this will not change with distance. That is just to tell a small analogy say for example like we are standing I said an example of I am near like a small burner burning in front of me. If I go closer to it I will feel hotter if I go back I will feel little bit cooler why the energy from the burner is spreading across in all directions. If I go closer most of my surface area of my body is encountering that energy. If I go back the energy is now spread across a larger area and my surface area is smaller in compared with the total area. So, the energy I receive is only a fraction of what the object image this is radiant flux density. Now, compared with radiance same burner is in front of me the burner if I treat it as collect a small source or something it will have like a small solid angle towards me right basically. As I move back and back I need to maintain the solid angle which the burner had initially then only the radiance definition will be satisfied. Radiance by definition it is the energy in that particular solid angle. So, radiance is here. So, I have to maintain the solid angle. So, what essentially it means as I move back and move back it is akin to myself has to grow bigger and bigger in surface area so that I capture all the energy coming in that particular solid angle. So, essentially what is happening I am maintaining the solid angle and hence I will be collecting the all the energy coming out of the burner which will remain the same. As I move back the surface area in which I am collecting the energy is going to increase because I need to maintain this constraint of solid angle this thing solid angle definition. That is why with respect to radiance energy is not going to change as long as the solid angle is held constant same amount of radiance we will receive. But by definition of radiant flux density surface area has to remain constant. So, even the solid angle is kept same the surface area will become smaller and smaller as the distance between the source and object increases. This is the burner this is me as I move further and further the total proportional area getting this energy is going to decrease based on distance and that is why we say radiant flux density will decrease with distance whereas by definition of radiance it would not decrease with distance. So, you need to understand this difference between them it may appear little confusing in the beginning. But as you try to sit and think deeper about it you will be able to understand it much clearer. Next what we are going to see is the relationship between radiance and radiant flux density. Now, let us say again we have like a small surface area I have like a some small source of energy here. We will take it as an example the source of energy I placed it on a hemisphere surrounding it say rather than placing here I place it here yeah now this is a source of energy. Let us imagine the source of energy is going to emit some energy starting from this particular point and it is going to move all along the hemisphere line and come here and stand and say what to say stop. So, what essentially it means this source of energy is going to see this particular surface area continuously and it is going to move from it. So, essentially what is the energy it is emitting in this particular direction we are interested upon. So, as it is moving here it is going to see like this something of this sort. So, after it has moved one full round I say I want to calculate the total irradiance received by the object okay. So, initially I measured radiant energy from one direction that is radiance now the source has moved in the entire hemisphere now I say I need to calculate the total irradiance. So, what essentially the concept wise is I am going to integrate the energy that came from these different points on the entire hemisphere right because by definition of irradiance it is nothing but the total energy within the hemisphere surrounding the object of interest. So, what I want to do? I want to do integration of the radiance over the entire hemisphere if I integrate it over the entire hemisphere I am going to get the total irradiance. So, the irradiance E is equal to L integrated over the 2 pi solid angle integrated over 2 pi solid angle or to be more precise I write it as 0 to 2 pi solid angle that is hemisphere like for full sphere we had solid angle is 4 pi radiance and for a hemisphere solid angle is 2 pi half of it. But here comes one more tricky issue just go back to the definition of radiance and definition of radiance we spoke about projected area that is whatever be like energy coming in or going out the area which is receiving the energy or emitting the energy should be perpendicular to the direction of motion. So, coming back to this particular example that we had. So, if this area is here when the object is here the area should be rotated like this in a direction perpendicular that is this is the original position by angle of theta I should rotate it in order for the area to receive the energy. Similarly, if this receiver goes somewhere here to another position instead of being horizontal I should now rotate it to another angle to receive the energy this is how we have defined radiance the area should be projected in the direction of radiation and hence there will always be a cos theta term involved in the definition of radiance. If you look at the definition of radiance in this particular slide is defined as the in denominator we have d a cos theta because of this in the integration of radiance rather than just using L I should use L cos theta because of the definition difference in irradiance we did we do not care about directionality but in radiance definition we care about direction and hence due to this projected angle difference I need to introduce a concept of cos theta into it. So, rather than just integrating L over 2 pi radian sorry 2 pi stradian I have to integrate L cos theta I am going to consider the entire direction between the vertical and the horizontal starting from 0 here the vertical all the way up to 90 in this direction similarly 0 here all the way up to 90 in this direction I have to integrate it then only I will be getting the exact irradiance okay. So, just coming back to the derivation so the irradiance E is given by L cos theta into solid angle whatever is a solid angle integrated over 0 to 2 pi solid angles. Now I am not going to go into the detail of this derivation a single solid angle we can divided into two planar angles one is called a zenith angle and azimuth angles and all I am not going to go deeper into it but the final result I am giving out here the final result is E is equal to pi times L that is the irradiance received by an object will be is equal to or will be is equal to pi times the L or if an object is emitting energy in different different directions the total emittance from an object is pi times the L pi times L. So, this is the relationship between radiant flux density and radiance this we will be using repeatedly in various points in the course so you need to remember this relationship always this relationship has a caveat or a like a strict condition what is it? This relationship holds good only if the area is Lambertian so that is given here Lambertian or the source if the area is receiving energy or it is kind of reflecting we call it as Lambertian or if the source is like emitting energy say it is emitting on its own we call it isotropic radiator. So, here we talk about okay we will first concentrate on term Lambertian. So, Lambertian means say I said an object is there now I am going to take an aeroplane fly around an object and look at and measure the energy coming out of an object. So, I have a sensor fitted in the aeroplane I go in like a circle I move in different different directions I move in different angles and so on I fly I fly if I am this angle I then I come closer whatever all possibilities I cover within that hemisphere if that is the case in whatever direction I look if the object is giving out same amount of energy like the particular land surface around which I am flying if it is giving the same amount of energy in whichever direction I look at it that particular surface is called Lambertian it will look exactly the same like same means it will emit the same amount of energy it is not like physically the area will not remain same like as I said the area will project itself but the amount of energy coming out of it will be the same in whichever direction I look that is called like Lambertian. So, Lambertian surfaces it will have like a uniform reflectance properties it will appear more or less equal from all directions. So, only for such Lambertian surfaces E is equal to pi times L the relationship will hold good because the derivate derivation I have not shown you if you want to take the radiance term L out of the derivation we need to treat it as a constant right normally we take only constants out of the integrals you know so out of the integral if you want to take it has to be a constant when it will be like a constant when L should be will when L will be constant is only when the surface is Lambertian that is why I said there is a condition this is like a little bit strict condition because most of the earth surface features are non Lambertian in nature we will see in detail in the coming slides and coming classes but remember this equation as we will be using it often E is equal to L times pi ok I said inverse square law we are going to now see an example of how inverse square law works again we are going to take an example of sun and work around it. So, the question given here is calculate the mean solar irradiance at the earth surface assume equivalent black body temperature of sun is 5770 Kelvin radius of sun 710 power 8 meters and mean earths and distance 1.5 into 10 power 11 meters. So, what is being said it is said sun is emitting some amount of energy calculate what is the energy that will reach the earth surface per unit meter square of area. Let us see how we are going to calculate first step we have to calculate the radiant flux density emitted by sun. We have learnt the law called the Stephen Boltzmann law which says the radiant flux density m is equal to sigma t power 4 where sigma the constant is Stephen Boltzmann constant I gave the value in the initial lectures. So, I am just going to substitute the values. So, m is equal to 5.67 into 10 power minus 8 this is the value of sigma multiplied by temperature of sun 5770 power 4. If you solve this we will get 6.2847 into 10 power 7 and the units is watt per meter square. So, Stephen Boltzmann law will give us radiant flux density what is the amount of energy per unit time per unit area that is radiant flux density. But we all know sun is a sphere big sphere it is not like a one point it has a certain area and the shape of the sun in 3D space is a full sphere. So, this energy that we have just calculated is per unit meter square area of sun. So, what now we have to do we need to calculate the total power emitted by the sun. So, now we have calculated the power emitted by sun per unit meter square of sun. Next we are going to calculate the total power emitted by sun. How are we going to do it the total power emitted by sun is equal to I use the correct technical term the total sorry radiant flux emitted by sun is equal to radiant flux density multiplied by area of the sun. So, here I am sun is like a 3 dimensional figure there in the initial problem of calculation of solid angles in the last lecture we treated sun as like a circle because when we look at sun we are seeing it as a 2 dimensional circle alone but in reality sun is like a sphere. So, for calculating the total power emitted by sun we should treat it as a sphere and take the total value like the total surface area of the sphere. So, here radiant flux density is the m we calculated in the previous slide multiplied by surface area of sphere is 4 pi r square. So, 4 pi into the radius of sun that is given there roughly if you do this the total power radiated by sun will be 3.8698 into 10 power 26 units is watts because now what happened I have converted radiant flux density into radiant flux. So, I have multiplied the power emitted per unit area by I have multiplied the power emitted by unit area with area and hence area terms get cancelled out the unit will remain as watts always have the units written beside your quantities it will help you to cross check whether your calculation is correct or not. So, now I have calculated the total power emitted by sun. Now, let us look at the geometry this is sun as a sphere I have calculated the total power emitted by sun earth is somewhere here at a distance of d. Now, sun is not only emitting energy towards earth it is emitting energy in all direction because sun is there in space nothing there to capture it it is emitting energy in all directions. So, what essentially if I draw a full sphere around it I draw full sphere around it with the radius of d sorry my drawing is not that great but yeah I draw a sphere around it with radius of d and this particular power emitted by sun is actually distributed itself equally in the entire sphere around it because sun is like a sun is what we call as isotropic radiator uniformly radiates energy in all directions. So, if I put like a huge sphere surrounding the sun with radius of d distance between earth and sun it will emit energy equally in all directions within that particular sphere. So, earth 1 meter square on the earth surfaces is actually located somewhere here. So, what I want to do I want to calculate the irradiance at the earth surface that is I have to divide the power within this entire hemisphere for 1 meter square of area that is earth can be anywhere here earth is what is the energy received by 1 meter square area on the earth surface. So, essentially I am calculating what is the power remaining at 1 meter square area on this particular huge sphere that is the total power P emitted by sun divided by the surface area of this big hemisphere will give me the final answer. So, if we can do this if you divide this particular power P divided by 4 pi d square. So, P divided by 4 pi into d is 1.5 into 10 power 11 whole square if you do this we will get 1368.67 watt per meter square again the unit will become watt per meter square because again we are converting the total power into unit area for that particular distance. So, this is known as solar constant roughly this value 1368 watt per meter square value is what we call solar constant that is on an average taken for all season all places together this will be the average amount of energy that we will receive from sun per unit meter square of area per unit time that is solar constant. So, to summarize what we have learned in this lecture in this lecture we have learnt about the relationship between irradiance and or radiant flux density and radiance radiant flux density will vary with decrease with distance whereas radiance will remain constant as long as the solid angle is kept constant. And the irradiance E E is equal to pi times L the relationship we have seen and also we have saw one example of how to use the inverse distance relationship to calculate the amount of energy received on the earth surface. So, with this we in this lecture thank you very much.