Beer-Lambert’s Law Experiment

Beer-Lambert’s Law Experiment Design Project on  Beer- Lambert’s Law. Saswati Rakshit Aim: To learn and understand the basics and mathematical calculations of the following problem and write programs accordingly. Problem 1: Suppose an outdoor multispectral image is captured by a camera with a path length of 1-3 microns. A part of the intensity is attenuated by the absorption of small particles in the atmosphere for that spectral range and let the scattering by the small particles for that spectral range is 0. Find the total attenuation in the spectral range using Simpson 1/3  ­, trapezoidal and Euler integration methods and comment in your findings. Scope/Application: Beer Lambert’s law relates the attenuation of light to the properties of the material through which the light is passing. When light passes through a medium some amount of light is absorbed by the medium. For this absorption intensity of light reduces. Beer Lambert’s law states that the quantity of light absorbed is directly proportional to the concentration of the substance and the path length of the light through the transmission medium Beer Lambert’s law is used to find total attenuation of light when light passes through a medium(considering scattering is zero).It is also used to find the concentration of medium in chemical analysis, medium length in some application and absorbance of medium when necessary. Introduction of Beer Lambert Law: Lambert’s law is (Related to thickness/path length of medium) When light passes through an absorbing medium its intensity decreases exponentially as the path length of the absorbing medium increases. I = I0 e-k1 L †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.(i) (where L is the length of medium and k1 is molar extinction/absorption co-efficient for the absorbing material) Beer’s law is (Related to concentration of absorbing medium) When light passes through medium(absorbing) its intensity decreases exponentially as the concentration of the absorbing medium increases. I = I0e-k2C †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦..(ii) (C concentration of medium and k2 is absorption co-efficient for the absorbing material) Combining both Beer’s and Lambert’s law we get I = I0 e-k3CL combining eqn (i) and (ii) Where I0 = Incident light intensity I = Transmitted Light intensity C=concentration / volume L= path length of medium We consider an outdoor multispectral image is captured by a camera with a spectral range of 1-3 microns. A part of the intensity is attenuated by the absorption of small particles in the medium.so image will be attenuated. Beer Lambert’s law find the attenuation caused by absorption for that spectral range and let the scattering by the small particles for that spectral range is 0. Objectives: In a given path length 1 to 3 micron, we have considered a constant value of molar extinction/absorption co-efficient.and now we need to find the absorbance total attenuation of the incident light using Beer’s Lambert Law. And then applying Simpson’s 1/3, Trapezoidal and Euler Integration in it compare the result. System flow: Implementation of the Beer Lambert’s law needs a proper mathematical understanding of the Beer Lambert’s law. Here I am showing how to set the equation using its mathematical basics First we apply Beer Lambert’s Law for a medium which absorbs light in spectral range 1-3 micron. Considering no scattering we assume I0 is the incident light to the medium Air of attenuation coefficient 1.64at temperature 20oC. Here path length travelled by light is 1 to 3 micron. According to Beer Lambert’s law light intensity is decreased if concentration path length increase. So we get the equation I=Io e-kcl = Io e- µl Now as we know the path length l and attenuation coefficient  µ, we calculate absorbance of the medium using eqn Log10 = kcl (Where l is constant) = e kcl But now for a spectral range 1 to 3 micron path length we need to formulate a new equation by integrating ranged from .001mm to .003 mm [Here x is path length and c is attenuation coefficient] Solving the above eqn we get total absorbance. Thus Beer Lambert’s law is successfully implemented in our problem. Now we apply Simpson 1/3 rd, trapezoidal rule on Beer’s Lambert Law to find total attenuation. Finally compare the result of Simpson’s 1/3 rd and Trapezoidal rule with the actual integration. Flow Diagram: Math: Beer-Lambert Law Concept Consider a light incident on a medium with area A and thickness dx and concentration of molecules C. Number of molecules illuminated by light of incident intensity Ix is CAdx. Total effective area ÏÆ'CAdx. Probability of light being absorbed in thickness dx is = dx [where dIx is the change in intensity across dx and ÏÆ' is scattering coefficient] So we can write, = dx †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦ (i) Now we integrate both sides of (i) ln (I) – ln(I0) = ln = ÏÆ'Cx I = I0e-ÏÆ'Cx = I0 e µx †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.(ii) The co-efficient  µ=ÏÆ' C is the linear attenuation co-efficient. Here C=Absorbing co-efficient ÏÆ' = Scattering co-efficient. The ibrightness of light decreases exponentially with depth in the medium. So we can tell Beer-Lambert Law is also a function of( ÃŽ »), i.e. I (ÃŽ ») = I0 (ÃŽ ») e- µ(ÃŽ »)x†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦(iii) Calculation: (Here we considered linear attenuation) Given spectral range 0.001 mm to 0.003 mm Let we consider Absorbing coefficient(C) of Transmission Medium (Air) = 1.64 at 20o C. So total absorption (A) of light is calculated by integrating in the spectral range, [from eqn ii] = (1) = = 0.002006571 Now we can easily calculate attenuated intensity of light (I). Attenuation is the loss of light intensity over distance. The greater the distance, the lower is the intensity Where I=I0 -Attenuation Trapezoidal rule:- We know, in case of multiple application of Trapezoidal rule, the formula is: = [f(x0) + 2) + f(xn)] So, here applying the above formula for equation (1) we get: = 0.0005[[ 0.001 + 2 + [] 0.003] = 0.0005[1.001641346 + 2.00657077 + 1.004932123] = 0.0005[4.013144239] =0.002006572 (Ans.) Here, x0 = x0.001, xn = x0.003, b = 0.003, a = 0.001, n = 2, = 0.0005. Simpson’s rule: We know, composite Simpson’s rule formula is written as: = [f(x0) + 4) +2) + f(xn)] So, here applying the above formula for equation (1) we get: = 0.0003[[ 0.001 + 4 + [] 0.003] = 0.0003[1.001641346 + 4.01314154 + 1.004932123] = 0.001805914 (Ans.) Here, = 0.0003. Euler’s formula: Here, = y(0.001) = = 0 (assumption) y(0.003) = ≈ and we have to find the value of equation (1) using Euler’s formula which is, = + f (, ) h let us choose h = 0.001 Step-1 i=0, = 0.001, = 0, h = 0.001 = + f (, ) h = 0 + f (0.001, 0) 0.001 = 0.001001641 Step-2 i=1, = 0.002, = 0.001001641, h = 0.001 = + f (, ) h = 0.001001641 + f (0.001, 0.001001641) 0.001 = 0.0020049426 This is actually the value of the function at i.e. at (+h) or (0.002+0.001) or 0.003. So, = = 0.0020049426-0 = 0.0020049426 (Ans.) We find that the result of all above technique is almost same if we take approximation i.e.0.002. CODES and OUTPUT: Beer Lambert’s Law: #include #include #include int main() {float absorbtion,m,l,u; printf(nEnter spectral range); scanf(%f%f,l,u); printf(nnenter the value of absorption cofficient); scanf(%f,m); absorbtion=(1/m)*(pow(2.718,(m*u))-pow(2.718,(m*l))); printf(nnTotal absorption is %f: ,absorbtion); getch(); } Output: Simpson 1/3rd rule: #include #include #include void main() {float x[10],y[10],Total=0,h,t; int i,n,j,k=0; printf(nhow many values you will enter: ); scanf(%d,n); for(i=0; i { printf(nn x%d: ,i); scanf(%f,x[i]); printf(nn f(x%d): ,i); scanf(%f,y[i]); } h=x[1]-x[0]; n=n-1; Total = Total + y[0]; for(i=1;i { if(k==0) { Total = Total + 4 * y[i]; k=1 } else { Total = Total + 2 * y[i]; k=0; } } Total = Total + y[i]; Total = Total * (h/3); printf(nn I = %f , Total); getch();} Trapezoidal rule: #include #include #include int main(){ float x[10],y[10], Total =0,h; int i,n,j,k=0; float fact(int); printf(nhow many values of ranges you will be enter: ); scanf(%d,n); for(i=0; i {printf(nn x%d: ,i); scanf(%f,x[i]); printf(nn f(x%d): ,i); scanf(%f,y[i]); } h=x[1]-x[0]; n=n-1; for(i=0;i if(k==0) { Total = Total + y[i]; k=1; }else Total = Total + 2 * y[i];} Total = Total + y[i]; Total = Total * (h/2); printf(nn I = %f , Total); getch();} Future Work Scope: This Beer Lambert’s law can be used in image processing application where atmospheric condition is poor to find the attenuation of light and image by absorption of light. Implementing Euler Method. References: Weisstein, Eric W. Simpsons Rule. From MathWorldA Wolfram Web Resource. http://mathworld.wolfram.com/SimpsonsRule.html. (Accessed on 26.04.2015) Basics of Trapezoidal and Simpson Rules, www.math.umd.edu/~jmr/141/Simpson.pdf. Lal, A. K., Simpsons Rule, 2007, http://numericalmethods.eng.usf.edu.(Accessed on 20.04.2015) http://numericalmethods.eng.usf.edu. (Accessed on 19.03.2015) Garrett, P., Absorption and Transmission of light and the Beer-Lambert Law, Lecture 21, 2006, www.physics.uoguelph.ca/~pgarrett/Teaching.html. (Accessed on 26.04.2015) Mudakavi, J. R., Modern Instrumental Methods of Analysis, Lecture – 07, Ultraviolet and Visible Spectrophotometry – 3 Theoretical Aspects, http://nptel.ac.in/courses/103108100/7  (Accessed on 26.04.2015). www.chemwiki.ucdavis.edu. (Accessed on 19.03.2015)