 lecture on the topic the radiance reaching the sensor. Last lecture we started discussing about what are the different paths in which radiance can reach the sensor like we saw 5 different paths as given here in this slide. We just went through what are these 5 paths in which energy can reach the sensor. Then we neglected to what of it and we started to look at 3 major components of energy that will reach the sensor especially in the shorter wavelength domain that is lambda less than 3 micrometers. Then we started calculating each and every path separately we started with path 1 that is the energy from the sun interacting with the terrain reaching the sensor. There we calculated the energy coming in from the sun as 3 different steps like first we calculated using Planck's law what is the energy emitted by sun within that particular wavelength. Then we calculated how much of the energy will reach the top of atmosphere using inverse square law. After that we apply that we apply the atmospheric transmissivity to the radiance top of atmosphere for calculating the irradiance at the earth surface. We saw what fraction of energy will reach the earth surface after it we also saw why this cos theta term is there if the solar radiation is not normal to the surface if it is coming in at an angle we need to multiply with the theta cos theta where theta is the angle of incidence at what angle solar radiation is coming in with respect to normal to the surface. So, that is the total energy that reaches the sensor sorry surface from sun. Then we multiply that particular energy from sun with respect to sorry with reflectance from the surface rho and then divided by pi in order to correct it for radiance. Assuming a lambertian surface we have now converted the radiant flux density from the sun into radiance reaching the earth surface. When it reaches the sensor it has to again go through the atmosphere there is one more transmissivity that came in. So, this is the final equation for path 1 that reaches the sensor. Now we are going to proceed and look at path 2 and path 3. Let us first talk about reflectance of the surface from diffuse skylight component. So, diffuse skylight it is just a scattered light that remains in the atmosphere which travels towards the sensor and also towards the earth surface. So, essentially there is we can measure diffuse radiance using what is known as instrument called shaded pyronometer. We can measure diffuse radiance that is diffuse energy coming in or we can model it using various atmospheric and radiative transfer models. So, here we are not going to derive it because there are not actual derivations to account for how to do it. But we are just going to assume we have some measurement or some model output of how much diffuse radiation is present ok. Because direct solar radiation it is easy for us to calculate everything is fixed. But diffuse radiation it is not fixed it will vary with time with cloudiness, where sun is located etc etc. So, we assume some measurement is available to us right now or some model output is available assuming this. So, this is the radiant flux density from diffuse component the diffuse skylight component units of watt per meter square per micrometer ok. So, this is what is emanating effectively from the atmosphere itself that is reaching the surface. So, assume we have a instrument called shaded pyronometer here and we are measuring it. There is we are not going to go in detail and derive what factors are using it we are just going to assume we have that measurement. So, once we have that measurement so skylight is coming in getting reflected back. So, again we are using the reflectance term here. So, again we are using the reflectance of the surface. So, this is the radiant flux density multiplied by reflectance of the surface. Again we are dividing it by pi assuming a Lambertian surface in order to convert radiant flux density to radians with a unit of watt per meter square per stradian per micrometer. This pi is to account for that. So, this diffuse skylight component the incoming energy we know by some measurements let us assume. Multiply it with reflectance of surface divided by pi which will give us the radians leaving the surface due to diffuse skylight component. Now, this radians leaving the surface has to travel entire atmosphere again. So, when it reach the earth surface that is fine because the radians can start from any point in the atmosphere. So, we have not taken into account atmospheric transmissivity because it is that is the atmospheric components only is what is creating this diffuse skylight. But once it reflects with the object it has to travel back to other sensor. So, this energy will be again absorbed will beginning to be absorbed or scattered by atmospheric components. There we have to use this transmissivity term between earth and sensor. So, here we are using tau v where tau v is the transmissivity towards earth and sensor. So, this is effectively the radians reaching the sensor due to diffuse skylight component. What is this f part? f part is to account for the fraction of sky. So, maybe I am not going to explain it in detail. Let us assume the surface is completely free from any obstacles whatever the surface we are talking about is not covered by any feature surrounding it. So, f will be 1 I am not for the sake of simplicity I am just going to neglect this. Assume f as 1. So, E lambda D is the radiant flux density due to diffuse skylight component multiplied with rho divided by tau to calculate the radians leaving the surface multiplied with tau v to get radians reaching the sensor. So, now we have calculated two parts. One is reflectance from the surface due to direct sunlight component, parts 2 reflectance from the surface due to diffuse skylight component. Third component is the path radians what energy scattered energy that directly reaches the sensor. Again we have to model the path radians somehow like we have to use what is known as radiative transfer models to calculate it how much is the path radians at a given time with the given atmospheric conditions temperature pressure cloud fraction and everything what is the diffuse radiation that is going towards the sensor that is path radians we have we cannot measure it actually we have to model it. Let us assume we also have that model there is no defined equations for us to derive here. If we assume that value is available to us so we get what is known as L lambda sp. So, this is the radians due to path or diffuse component. So, this is the total lambda or sorry total radians that will reach the sensor from earth surface. If you look at this this signal about the earth surface is present in two components. One is the direct sunlight component this is the direct sunlight component this is the diffuse skylight component in this term and this is the path radians component. So, this will effectively tell us what is the radians that reaches the sensor in visible NIR and SWIR wavelengths. Next we are going to see what is the radians that reaches the sensor in thermal infrared wavelengths or mid wave infrared wavelengths. First we will talk about thermal infrared wavelength. Okay what will happen in thermal infrared wavelength that is lambda between 8 to 14 micrometers what we call it as thermal infrared region. Already we discussed that in this particular wavelengths solar incoming radiation is almost zero and whatever the energy that reaches the sensor is purely because of energy emitted by features on the earth surface and atmosphere. So, it is effectively due to earths own temperature that is what is going to reach the sensor. So, again here we have three paths path one is direct surface emitted component that is earth certain temperature let us say this is like 300 Kelvin. Due to this temperature this will emit a radiation again using Planck's law we can calculate. So, this is thing. So, these sensors will essentially operate at this wavelength not at shorter wavelength most likely let us say 10.2 to 11.2 micrometer and so on. This will be like the wavelength range at which thermal infrared sensors will operate. So, similar to calculating the energy from sun apply the same principle here use Planck's law substitute earth's temperature of 300 Kelvin if it is so most likely substitute the wavelength range 10.2 to 11.2 example for example say if you integrate Planck's law we will be calculating what is the radiant flux density from earth. So, this is the direct emitted component. Second thing atmosphere also has certain temperature that will also emit some energy and when it emits it will emit in upward direction and also in downward direction. So, whatever is coming towards the ground a fraction of it will be reflected by the surface that will reach the sensor. So, this is the surface reflected down emitted radiance. So, this is very similar to the if we compare it with shorter wavelengths shorter wavelengths what we discussed we discussed that diffuse radiation will be there that will come towards the surface will be reflected back same thing. But instead of diffuse radiation here we have emission due to atmosphere atmosphere also has certain temperature air will have certain temperature it will emit energy in these wavelengths. So, that particular energy will come down towards the surface and a fraction of it will be reflected back that is it. And third thing is the direct atmospheric emission reaching the sensor we call it as path emitted component. So, in thermal infrared wavelengths the energy reaching the sensor will have again 3 paths. First path is the direct emitted component. Second thing is surface reflected atmospheric emitted energy and third thing is path emitted energy. These 2 paths carry some signals about the surface and this is entirely unwanted energy which reaches the sensor. If it is thermal infrared wavelength only emitted components will be there. If it is not thermal infrared if it is MWIR that is between 3 micrometers to 5 micrometers wavelength or 3 up to 5 micrometers wavelength especially during daytime the energy containing or energy reaching the sensor will have both reflection components and also emission components. Because at 3 to 5 micrometer wavelength the energy coming in the sun coming from the sun cannot be neglected. Similarly, energy emitted by the earth also cannot be neglected both will be there during daytime. So, that is why remote sensing in mid-wave infrared band between 3 to 5 micrometers especially during daytime is a bit complex process. It has 6 different paths now. 3 paths for surface reflected component, 3 paths for surface emitted component. But if it is purely thermal only like between 8 to 14 micrometers only emitted part will be there. If it is less than 3 micrometers only solar reflected part will be there. In this portion 3 to 5 both will be there especially during daytime. At night time reflected part will not be there because sun is absent also only emitted part is there. This please make it clear and based on time availability we will see microwave infrared portion also how remote sensing can be done in mid-wave infrared portion later in the course. Right now we will focus on calculating the emitted components. So, how to calculate the emitted energy? I already said in the last slide that this is calculating the energy emitted by the sunness very similar to how we did it for sun like calculating energy from earth is very similar to how we did it for sun. Say let us say our sensor is sensing between 10.2 to 11.2 micrometers. This is Planck's law. So, let us assume the earth is at 300 Kelvin temperature. So, if you substitute this 300 Kelvin integrate Planck's law between whatever wavelength the sensor is observing we will get the energy emitted by earth this part. Here there is a new term called epsilon. What is this? This is called surface emissivity. What surface emissivity is whenever we discussed about Planck's law Stephen Boltzmann law and everything not all objects will obey those laws. That is those laws are defined for objects what are known as black bodies. What black bodies are? Black bodies are can radiate energy can emit energy with its full efficiency that is. So, Planck's law Stephen Boltzmann law are kind of like the peak energy that can be emitted maximum limit that is the maximum energy an object can emit at a given wavelength at a given temperature. But not all objects will obey that law or will emit radiation with that maximum efficiency I will say. Like if I say like maximum there can be 100 units of energy emission that is the limit. But certain objects cannot emit at that particular efficiency. Say if I can like let us assume like I am having 100 rupees. If I have the capacity to spend all the 100 rupees that is fine. I am having 100 spending 100. But I have 100 rupees there is some limitation on me that I can spend only 90 rupees I cannot spend the 10 rupees somehow something is stopping me from doing it. So, my efficiency of spending there is 0.9 same thing here there is some energy content within an object. It can radiate energy as prescribed by Planck's law. But there is something that is stopping it is it cannot radiate with its maximum efficiency something that is stopping it from radiating with maximum efficiency. And hence it is emitting energy only a fraction of what is described by Planck's law. Say Planck's law tells this object at a given temperature can emit 100 units of energy. But if the object has an efficiency of let us say only 0.9 it can emit only 90% of what is prescribed by Planck's law. This term what I am calling it as efficiency is actually known as emissivity. We will describe in detail about emissivity when we discuss about thermal infrared remote sensing. But here you just know that emissivity is kind of like a efficiency with what efficiency objects can emit radiation. Certain objects like water bodies deep water bodies have an emissivity of 0.98, 0.99 and so on. For vegetation emissivity will be 0.9 to 0.99. For some man-made objects concrete and all efficiency is somewhat lower. Highly polished metallic surfaces steel highly polished steel surfaces or highly polished mirrors which can reflect a lot for them efficiency will be much lower say 0.8, 0.85 and so on. So we have to account for it how much with what efficiency an object can emit that fraction is known as emissivity. That is what we are using here. So here it is to account for say the earth surface features cannot radiate energy as prescribed by Planck's law. It can radiate only a fraction of energy and that fraction is emissivity. See this is the object some vegetation is there. Let us say it can emit only 90% of Planck's law what Planck's law says. So it is emissivity is 0.9 multiplied with what is given by Planck's law that will give you the energy emitted by it. So again this is in radiant flux density you are dividing it by pi in order to convert it into radiance. So this is the energy emitted by the surface or radiance emitted by the surface. Now this radiance from the surface that is emitted by this feature has to travel through sensor and hence there is an atmosphere in between. Similarly we have an atmospheric transmissivity term tau. So if you multiply the energy emitted by the object with atmospheric transmissivity we will get the radiance reaching the sensor from the object of interest. So this is how we have to calculate it. And this law one just one more information reflectance will be is equal to 1- emissivity for most of the objects. This is known as Kirchhoff's law just for your information I am posting it here. I will explain this in detail in later classes when we discuss thermal remote. Now the second term down emitted surface reflected component. So atmosphere has some temperature that is also emitting some energy. So this is essentially emissivity of atmosphere multiplied by Planck's law at whatever wavelengths we need at atmospheric temperature. So again we are applying Planck's law multiplying with emissivity of atmosphere and getting this that will reach the earth surface. Here we are using reflectance of surface okay that is at this particular wavelength. So let us say like 10.2 to 11.2 this is the example we are seeing at this particular wavelength what is the reflectance for surface. Reflectance is equal to 1- emissivity that also I told you. If you do this that is the fraction of energy that will be reflected back to the space multiply with tau v transmissivity divided by lambda sorry divided by pi you will get the radiance reaching the sensor due to down emitted surface reflected component. The third component is path emitted component that is what fraction of energy that is emitted by atmosphere reaching the sensor. But again we have to model it using radiative transfer models we cannot have a measure for it or measure it from the ground we have to model it from the sensor. So the final equation of the radiance reaching the sensor in thermal infrared band comprises again three components where this first component is essentially the surface emitted. This is of real interest to us what is energy emitted by surface. The second component is surface reflected atmosphere emitted component this is kind of like noise we do not need this basically we have to correct for this effect. And third thing is path radiance again we have to correct for it. So why we discussed all these things in detail the major reason is the signal received by remote sensing sensor is not simple or direct lot of energy components come in and have a kind of a mixed energy only will be reaching the sensor. What we need what we do not need for what things we need to correct all these things we should be very clear when we do remote sensing say for example when we do remote sensing either in wavelength less than 3 micrometers or in 8 to 14 micrometers path radiance is unwanted term it is kind of like a noise that is getting added into the system we have to remove it. Similarly if you look at thermal infrared band there is a small surface reflected component which we do not need because whatever is emitted by the surface is of real interest to us we do not need this surface reflection in those wavelengths we have to remove that effect. So all these things we should know why this energy component is coming in due to which it is there whether we should remove it or not we have to clearly understand it and work according to it and I will give a very good example for what will happen especially in shorter wavelengths in this particular slide. This is the spectral reflectance curve of vegetation this dotted line. So I described what is meant by spectral reflectance curve in the last class so that is at different wavelengths objects will have different different reflectance values if we plot them together that is known as a spectral reflectance curve and this spectral reflectance curve is actually will be helpful for us to identify the object if the curve is having a certain pattern we call it okay this is vegetation if the curve has a certain pattern we call it okay this is a water body like this the spectral reflectance curve will help us to identify features that is why sometimes we call this spectral reflectance curve as spectral signatures but this is true only when the signal we have is pure without any contamination. We are taking all the measurements in controlled lab environment in a ground means it is fine directly we can measure spectral reflectance use it for object identification in remote sensing images but as I just described in these two lectures the remote sensing signals that reaches the sensor not only has signals about the target of our interest it also has signals from various other objects like atmosphere, neighboring pixels lot of those things. So finally if you look at the if you process the signals you got from the sensor and try to calculate reflectance from it we may have something like this what is given in the slide. What we need is these dotted lines what we may get is this dark black line actually so this black line is the accents are radiance this is how the shape will look whereas the shape essentially should look like how this reflectance features is unless we get this proper reflectance curve we would not be able to classify this. So what this accents are radiance has this accents are radiance has atmospheric effect effect from neighboring pixels and so on. So the accents are radiance must be corrected for atmospheric solar like solar radiation can vary I said like if the solar radiation is not perpendicular to the surface we need to multiply with some cross-seater terms if the surface is not flat if it is slope there may be other errors and all those things. So the accents are radiance whatever energy that reach the sensor must be corrected for atmospheric solar and topographic effects before we compare them with spectral libraries that is before we compare them with spectral reflectance curve. See unless we correct this dark black curve the radiance curve for all these effects we would not get this proper spectral reflectance curve for us to do object matching. So in the next set of lectures we will be discussing some simple strategies to correct these effects and calculating surface reflectance especially in the wavelengths less than 3 micrometers. Thank you very much.