 ச mall அர்ஞாகchmal் இவர் சரிதாக மன்னமாகத் கவனில்ாளம் நான் அனையுங்கள் பழக்குவிட்ட நியாளரத்தான். அதிக Abu கத்திப்புதில் மின்றுபிக்கு வைத்து உடுன திருமுன்றது make the theory agree with the experiment, it was necessary to move away from classical physics and then propose a D equal to h nu. So you have to go away from the classical physics and propose that energy transfer takes place in multiples of h nu, where h is not infinitesimally small 10 to 0, h is a finite constant, it's a fundamental constant of nature, which is the Planck's constant which is 6.627, ஐயா, Yongtian to the power of minus 34� median of, then we looked at the distribution, this is the Planck distribution, the monochromatic spectral radiation intensity, black body. In the engineering radiation, we would like, we call it as, so that pi is taken care of, but seraidin will not come, so that is e, takes care of that solid angle. நான் நல்லேய் அதனால் இடத்துகொள்ளலாமல் எங்களைத் தெரிகி ஆகிவிட்டு, பண்ணகள் சென்று மைக்கிறத்லு நாள் வீட்டு வட்டு வேண்டும் என்று கொடுக்கக் கொள்ளும். சரி,தான்போதுмதான் என் கசையில் 1.45ன Посெலும ஊர்வேசி 3 மாடிப்பாற்று என் storyline நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்சியில் நிகழ்சியில் நிகழ்ச்சியில் நிகழ்ச்சியில் நிகழ்ச ஆகிவில் கடக்கமியைத்தந்த விடங்க முன்மன்று திடீர் சின்ன நிலையில் புர்வவுக்கு பெரிந்துவிரை மாகத்திரி அதாக்காமுும் ஒருiotதிரராகி வழம்பு Osaka மாகத்திரி சின்ன அதிகலனுல் வலூகால மருவி says வெளிசார்க்ஞதால பரிடி சொல்ங்கள்часு ஹர்ந்த்துபோமில் � கில்வின், நாகோஹக்கை வேண்டுர் தெரிப்புக்குப் ப isso அந்த அறைந்ததில் க வாணிகில், நான் இரணு இருங்க நிடம் போகிரு அழாதம் இரைந்தால் மட்டா விளைக்கில் நின்று கடிக்கடும் நாங்கள் உறுக்கு வketμ கடிக்க மனம் அன்று நிற்க செய்ய மற்றது தருப்பட்டை அல்லாது வாணில்,முறை美味 நலையில் நிறக்குத்தால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால்ஹால் நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங்கள் சரியா? நீங ஆனால் சிருந்திருந்து சிம்சன் சிருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந்திருந ஆகாகத்திருள்வேண்டுமொல்ல, அவ்வு செல்கிறாய் என்பது இரynamுகிறேன் சரி தலைப்பட், María, வந்து இர்க столько சோக்கிறாய், இதை என் அவரை செல்கிறாய்? of four so this is called the Staphan befestman's law this law was known much before planh headlights figured out the Planck's distribution because this has come from thermodynamics from thermodynamics we can prove that the radiation the emissive power of a black body is proportional to the power of four over for getting to the sigma we cannot get it purely from thermodynamics you required experiments. so if you may t to the power of four it is mass to the experience with the experiment ராஜாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவின் ஐயாவ 56.7 வாட்சி, நான் வாட்சி, 5.6 and 10 to the power of minus 8 into 1000 is 10 to the power of 3, 10 to the power of 12, so there will be 10 to the power of 4, so 5.6 and 10 to the power of 4, 56.7 and 10 to the power of 3, 56.7 into A is 1 meter square, 56.7 kilo வாட்சி, so 1 meter by 1 meter plate, which is at 1000 kelvin has a emissive power of 56,700 kilo vats, the solar constant, the solar radiation which is entering the earth that has got a power of 1353 vats per meter square on the equator, on a clear day at 12 o'clock in the noon that is the value 1353 vats per meter square, therefore if you have a 1 meter square area, the maximum you can get is only 1.35 kilo vat, if it is cloudy and this thing you correct for altitude latitude blah blah blah all that, so it will be much slower than that, then take care of these rainy days and you take care of the evening and this thing and all that and fine it will be a non-zero value, but still okay, but you will have to take into account all this fine. So, we have derived both the laws from first principles, now it is time to solve some problems involving these alright. Problem number 45, very good, so we will hit half century very soon, yes, problem 45, calculate the equivalent black body temperature, any doubt, do you think it is clear, anyway for those people who are mathematically inclined, so this is the algorithm for what is called the successive substitution method, we will start with the value of x, let us say x equal to 3, we will say e to the power of 3 minus 1 by e to the power of 3 into 5 will be x i plus 1 that would not be 3, then the new value again substitute, find the new value of x i plus 1 and keep on doing it till the left side and right side are very close to each other, so this algorithm, so this is called the trial and error, this is called the successive substitution method, if you apply the method, you will get a value which is equal to 4.965, we already helped it by saying that e to the power of x is much greater than 1 and therefore e to the power of x minus 1 and this can get cancelled, x is almost equal to 5 right, for those people who are very mathematically inclined, you asked this question yesterday, I am trying to answer this, you can solve it using this, it will turn out that the answer is 4.965, which will lead to 2898 micrometrical rather than 2900 micrometrical, is it okay, so this is called the successive substitution method, for the sake of completeness, somebody take the value of 4.965 and let us get reassured that right side and left side are 4.965, when x i is 4.965, x i pack it okay, 4.965, very good, so doubting Thomas' put your doubts to rest, 4.965 is indeed the correct answer alright, if you want it to be more threateningly formal, please go ahead and use the Newton-Raphson method, put it as f of x, then x of i plus 1 is x of i minus f of x divided by f dash of x, keep on iterating okay, you will get the same answer 4.965 okay, I got distracted, we are supposed to do problem 45 right, problem 45, calculate the equivalent black body, calculate the equivalent black body temperature Te okay, calculate the equivalent black body temperature Te of the solar photosphere, of the solar photosphere bracket, outermost visible layer of the sun, solar photosphere is the outermost visible layer of the sun, based on the following information, the flux density of solar radiation, the flux density of solar radiation, the flux density of solar radiation reaching the earth fs, just now I told you, I told you 1353, I think it's 1368, not a big difference, the flux density of solar radiation reaching the earth fs is 1368 watts per meter square, the earth sun distance d and the radius of the solar photosphere, the radius of the sun, the radius of the sun is 7 into 10 to the power of 8 meters. If you have the flux density on the outer layer of the photosphere, you are done right, if you have the flux density on the outer layer of the photosphere, then that flux density is equal to your sigma t to the power of 4, how do you have the flux density on the outer layer of the photosphere, you know the flux density at a far off place and you know there is something called the inverse square law which is at work, now use it, you know the flux density at a far off distance, so find the flux density at a lesser radius, that's it, you understood, what was the distance, you know rs, d, what did you call it, d, what did you say earth, fs, what did you call that, fs, rs is it, this is correct, why you messed it up, it's correct right, the Vishree, what is the problem, yeah, yeah, no, no, wait, wait, what, this is correct, what is d, that I called it as d, then this is correct, then is it okay, yeah, yeah, what is troubling you Anusha, what is it, so it is going like this, now we know the flux density here, we want to find out the flux density here, that's it, Sneha is it okay, so now tell me what is the photo sphere, son, okay, see at this radius, if it is having 1,368, at this radius, if it is having 1,368, at this radius it will be much more, what is that much more, it is equal to sigma t e to the power of 4, that's it, yeah, so what is this, 1,368 into rs is what, divided by 10 power 11, so f photo sphere, how much is that, 6 point watts per meter square, no, no, no, no, t, ts, yeah, what is it, too low, why, what made you think so, no, no, you may think about many things, but according to the values I gave, is it correct or not, okay, just hang on, we can check it, if I get 5,800 Kelvin, then it is correct, okay, so let us see what the girls, whether what the girls are saying is correct, 6.28, 6.28, 10 to the power of 7 divided by sigma, 4 to the power of, yes, how much is it, then it is fine, what is your problem, why, no, no, I am also getting watt per meter square, what is your problem, what, just check, just check, relax, it should be fine, so this is 5, okay, so we approximate usually to 5,800, okay, alright, so 1,368 watts per meter square, it is possible for you to measure, so if you are able to measure the radiation falling on the earth, and from astronomy, you get the sun earth distance, but somebody has to give you the measurement of the radius of the photo sphere, that somebody, some physics people, some cosmology, somebody is giving you, based on this data, it is possible for you to extrapolate, or I can directly get from emission, from the emission spectra, I can take sigma t to the power of 4 and figure out that it is 5,700, so if inversely we can work out, if 5,770 is correct and this temperature is correct, and then you can actually work out the, you can work out astronomy from radiation, you can actually work out the distance if you know the radius and all that, right, okay, anyway let's not get into all that, actually they are moving, the stars are all moving, they will find out how it is moving, they will measure the radiation spectra, then it can be related to the solid angle, okay. Problem number 46, to complete the discussion, let us do the Ulta, let us calculate the earth's temperature, okay, so that your understanding is complete, but now one more thing is required for calculating the earth's temperature, that is the reflectivity of the earth, that is called the albedo, okay, the reflectivity is called the planetary albedo, albedo means how much it reflects back, I will give you the data, calculate the equivalent black body temperature of the earth, problem number 46, calculate the equivalent black body temperature of the earth, calculate the equivalent black body temperature of the earth, assuming a planetary albedo, assuming a planetary albedo of point 3, open the brackets, fraction of the incident solar radiation, I am defining albedo, open brackets, fraction of the incident solar radiation that is reflected back into space, fraction of the incident solar radiation that is reflected back into space without absorption, fraction of the incident solar radiation that is reflected back into space without absorption, closed brackets, assume that the earth is in radiative equilibrium, assume that the earth is in radiative equilibrium, it is called planetary albedo, people are working on those presentations, the planetary albedo can change, if the ice keeps on melting, the snow cover changes and that will lead to a feedback effect and all that, snowball earth will be the opposite of that, albedo is a very critical factor in the earth, how much it reflects, now what is radiative equilibrium, radiative equilibrium means you are considering the whole, let us say the whole earth is at one temperature, T, now under equilibrium the rate of change of enthalpy of the earth is incoming minus outgoing, the incoming is from the solar and the outgoing is basically emission, under equilibrium temperature does not change with time, now this 1,368 watts per meter square that is based on the projected area of the earth, Lakshmikanth that 1,368 is based on the projected area of the earth, what is that, do you have any doubt, so we will have to be careful, so if you want to solve problem 46, so the radiation which is coming from the sun, okay, this 1,368 this is earth is a sphere, that 1,368 is based, if you cut like this, if you cut like this, what happens, if you cut like this, what do you get, a circle, pi r squared, okay, so the 1,368 is based on pi r squared, but the emission back is based on 4 pi r squared, yes please use this, okay, so this is incoming, this is outgoing, emission is based on 4 pi r squared and please take factor in the albedo, okay, but what will be the absorption, absorptivity, 0.7, that you have learnt somewhere no, I think I should that teach that first, you already learnt it somewhere, okay, so that funda is required now, right, okay, hang on, so if radiation is falling on the surface, what is also falling on the, not only radiation which is falling, okay, so g incident, what can happen to this radiation if it falls on the surface, this can be reflected, okay, this can be absorbed or this can be transmitted, there is no chance of transmission from the earth, if it is glass, it is possible, okay, so using energy balance, g incident, g is the commonly used nomenclature for incident radiation, okay, g is irradiation, then we qualified with subscripts, reflection, transmission and absorption, so this is equal to, I will divide by g throughout, make this 1, so 1 equal to reflectivity plus absorptivity plus transmissivity, okay, transmissivity is 0 for the earth, but for the earth's atmosphere, it is not 0, okay, for an opaque surface, when tau equal to 0, therefore the absorption of the, the absorptivity of the earth, planetary absorptivity will be 1 minus 0.3, in simple English, 1 minus 0.3, 0.7 into 1,368 watts per meter square will be absorbed for every meter square of the area and what is the total meter square, what is the total area which is absorbing this pi re square, but this must be equal to the emission, the emission is 4 pi re square into sigma Te to the power of 4, that is the energy balance, please do it, we will complete it, we will get the temperature of the earth, it is 255, 255 is the temperature of the earth, okay, you are happy, you are at equilibrium now, you figured it out, okay, let us finish this, சைத்தன்யா, for this also the radius of the earth is not required, correct, 4 pi re square equal to 1,368, so the temperature of the earth is about minus 18 Kelvin, okay, do not worry about the temperature in Chennai, that is something that is okay, so overall it is about, you also have the poles, you have got சைத்தன்யா, you have got all these places, the average temperature is assuming the earth's temperature to be 1, the average is 255, assuming the earth's temperature is 1 is also a bad idea, but if you have to do that, it is 255 and there are other assumptions, there is something, when you do climate predictions, they assume something called aqua planet, the aqua planet means the earth contains 100% water, so there are simulations with aqua planet simulations, it is easy to do, but otherwise, you have to take 3 fourths, 71% water, 29% land, then you would have emissivity for land, emissivity for ocean, problem gets more complicated, so first level modeling, they will say aqua planet, that means everything is water, 100% water and they will do the simulations, okay, now this power point I want to show, just 3 minutes, it is already there, very good, so mouse, okay, so these are the black body radiation curves, so if you see, this is from the book, the great book Wallace and Hobbes, much of my material is taken from that book, atmospheric science and introductory survey, so you see the Planck's black body distribution drawn side by side, so the left curve, basically the units have been normalized, because 255 Kelvin, it cannot be so much, that should be a big mountain, this should be a hillock, right, but we have adjusted that, okay, now if you see, just look at the nature of the curves, so the peak is around 0.5, that is what we saw in yesterday's class, 0.475 for the solar, it is about, where is this, for the solar incoming radiation, I mean radiation from the sun, so it is about 0.5 Kelvin, so this is the spectrum, black body spectrum for the sun, this is the black body spectrum for the earth, therefore the incoming radiation is largely in the visible part or the near infrared part which is less than 4 micrometre, so actually for the radiation from the earth, it is always in the infrared part of the spectrum, okay, so this is the first major difference and all these greenhouse gases do not absorb much in this part of the spectrum and they absorb a lot in the other part of the spectrum, till out if you don't believe me, so this is the absorption spectrum, so if you see, this is the absorption spectrum as measured, so at 11 kilometers at the top of the atmosphere and at ground level, at the top of the atmosphere is what is of interest to us, so same wavelength if you see, 0.1, 1, this is 1, 5 and this is 100, so remember this, so in the incoming radiation you see the absorption is very less, but whereas in the outgoing part, the absorption is very high, so because of this the greenhouse gases, so the top is basically carbon dioxide, the bottom, I mean the top is at 11 kilometers, the bottom is ground level and the various gases absorption are given here, oxygen and ozone and so on, so you can see that there is very little absorption in this part of the spectrum whereas a lot of absorption, therefore these gases, mostly diatomic gases because of the electric dipole, because of the particular property, the nature of the properties, they allow their incoming radiation, incident radiation to come through, whereas they do not allow the radiation which is going back from the earth, this is responsible for a continuous build up of this radiation imbalance, the budget is affected, the radiation budget is affected, this leads to global warming and so on, always water vapor is there, water will evaporate and the water vapor is not under over control, it is a by-product of the natural hydrological cycle, but this carbon dioxide and methane are under control, so the whole idea is to reduce this carbon dioxide so that all these stay within limits, okay. So in the next class we will go a little deeper and then look at the greenhouse effect and then we will have to look at the equation of transfer, how radiation gets transferred in an absorbing and emitting atmosphere.