 Welcome to the next lecture that is on the same transmission characteristics so we have seen the different transmission characteristics in the previous two video lectures. So these are the learning outcomes at end of this session students will be able to calculate the attenuation in the optical signal here. So now in this lecture we will solve one example due to the release scattering here. So in the previous lecture we have seen the losses due to the scattering here. So due to the scattering there are two major problems that is the linear and non-linear scattering. So in the linear scattering we have seen that is the release and the my scattering and in the non-linear we have seen SRS that is the stimulated Raman scattering and the Barolium scattering here. So due to which the data is lost whenever there is a transmission through an optical cable here. So theoretically we have to prove that if you go on increasing your wavelength your release scattering becomes less means we are going to see now the various wavelength at 0.63 micrometer, 1 micrometer and 1.3 micrometers what will be the release scattering. In the previous lecture also we have seen just seen how we are going to calculate the release scattering here that is the gamma R here. So just recall that formula that we have to use to solve this problem here today. So we will switch over for the problem now. So now this is the problem given to the students here. So the silica has an estimated the temperature of 1400 Kelvin and the isothermal compatibility of 7 into 10 raise to minus 11 meter square Newton per Newton and the refractive index and the photoelastic coefficient for the silica are 1.46 and 0.286 respectively. So now we have to determine the theoretically attenuation in d B per kilometer due to the release scattering at the various wavelength that is 0.63, 1 and 1.30 micrometer. And for that solving this we require the Boltzmann constant which is given as 1.381 into 10 raise to minus 21 joules per Kelvin here. So recall that how you are going to solve for the theoretical attenuation in d B per kilometer due to the fundamental release scattering here. So for that we should know that it is being calculated that is gamma R is equal to 8 pi cube n raise to 8 p square B c k T f upon 3 lambda raise to 4 here. So by using this formula so we have to calculate the fundamental release scattering here. So n is nothing but your refractive index here. So the refractive index is given as an 1.46 and the photoelastic coefficient that is p is equal to 0.286 and isothermal compressibility that is given B c is equal to 10 raise to minus 10 meter square per Newton and the temperature is given as 1400 Kelvin and we have to find out the lambda on various lambda that is lambda 1, lambda 2 and lambda 3 which is 0.63, 1.0 and 1.3 micrometer. So by putting all this value into this so we will solve for the release scattering for the wavelength lambda 1 that is 0.63 micrometer here. So now if you substitute all the values into this for the lambda 1 so you will get 8 into pi into cube means what you will get 248.15 into n is 1.46 that is raise to 8 it will be 20.65 into 0.82 into 7 into 10 raise to minus 11 into 1.38 10 raise to minus 3 into 1400 upon 3 into 0.63 micrometer cube. So if you solve this so it will be 1.895 into 10 raise to minus 28 upon lambda 4 so for ease so we are going to keep the lambda 4 only we are not going to substitute the value here right now. So we will solve this that is at wavelength of 0.63 micrometer so your release constant will be 1.895 into 10 raise to minus 28 upon so it is lambda raise to 4 so it will be 158 into 10 raise to minus 24 which will be 1.199 into 10 raise to minus 3 meter per meter. So this is the release scattering so now we will see the transmission loss factor for 1 kilometer of fiber that is in kilometers so it which is given by e raise to minus yr into l so it will be minus 1.199 into 10 raise to minus 3 into 10 raise to 3 so which will be 0.301. So now the attenuation due to release scattering in dB per kilometer is calculated that is attenuation 10 log to the base 10 1 upon in kilometer what we have drawn it is 10 log to the base 3 so which is equals to 5.2 dB per kilometer so for 0.63 micrometer your attenuation due to release scattering in dB per kilometer is 5.2 dB per kilometer here. In the similar fashion so we have to solve for the wavelength at lambda 2 that is 1.0 micrometer so release scattering is we have calculated 1.895 into 10 raise to minus 28 divided by that is 1 into 10 raise to minus 24 that is lambda value is 1 here so it will be same because it is lambda raise to 4 here so which is given by 1.895 into 10 raise to minus 4 per meter so similarly coefficient is calculated exp that is exponential minus yr into l so which is given by exponential minus 1.895 into 10 raise to minus 4 into 10 raise to 3 so which is 0.82 so the attenuation due to release scattering in dB per kilometer is given by attenuation 10 to the log base that is 1.209 which is 0.8 dB per kilometer so this is 0.8 dB per kilometer and due to the lambda 1 it is 5.2 dB per kilometer here so now at wavelength 1.30 micrometer so your railing scattering will be 8.95 into 28 2.856 into 10 raise to minus 24 is equal to 0.64 into 10 raise to minus 4 you will get then calculate coefficient exp into minus r into l so which will be exp 0.4664 into 10 raise to minus 4 into 10 raise to 3 so which is 0.36 and attenuation due to release scattering in dB per kilometer 10 log to the base 10 is 1.069 which is 0.3 dB per kilometer so from clearly we can say that if you go on increasing your lambda your the attenuation will go on less as we have seen in the for lambda 1.63 your attenuation comes to 5.2 dB and if you make use of this 1 micrometer so it will be 0.8 and for lambda 3 so if you make 1.3 micrometer so your attenuation becomes 0.3 dB per kilometer it is clearly that if you go on increasing your wavelength your attenuation goes on becoming less here so these are my references thank you