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Imaginary Part of Dielectric function 
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#1
Feb711, 04:39 PM

P: 3

Can someone please explain the concept of optical losses and its correlation with the imaginary part of the dielectric function in elementary terms. I am confused.



#2
Feb811, 02:33 AM

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P: 3,593

An imaginary part of the dielectric constant can only be due to the polarization which, in frequency space, is [tex]P=j(\omega)/(i \omega)[/tex] (please don't rely on my signs and factors) and therefore, [tex] \epsilon=\epsilon_0+ \sigma/i \omega[/tex] where sigma is the conductivity. If epsilon is real, the current is out of phase by 90 degree with the driving field and will do no work. If epsilon has an imaginary component, the current and the driving E field will be partly in phase so that due to ohms law energy is dispersed.



#3
Feb811, 03:26 PM

P: 3

How do you calculate the imaginary part of the dielectric function?



#4
Feb911, 02:01 AM

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P: 3,593

Imaginary Part of Dielectric function
Not much differently from the real part. E.g. in a simple classical model of a solid consisting of damped harmonic oscillators, the damping would automatically lead to an imaginary part of polarization.



#5
Feb1011, 01:04 PM

P: 15

The plane wave solution of the wave equation (from Maxwell's equations) in a homogeneous conducting media is given clearly in Panofsky & Phillips, p 200 [1]. For the plane wave
E(z,t) = Eo*exp(beta*z)*exp(i*[alpha*zomega*t]), the real and imaginary parts of the propagation constant K = i*(alpha+i*beta) are, alpha = k*sqrt([sqrt{1+(sigma/[epsilon*omega])^2}+1]/2) beta = k*sqrt([sqrt{1+(sigma/[epsilon*omega])^2}1]/2) where k = the angular wavenumber = 2*pi/wavelength epsilon = permittivity sigma = conductivity omega =2*pi*f = angular frequency. Here, alpha is the real part of the angular wavenumber. beta is the attenuation coefficient and attributes to "optical loss" in terms of the BeerLambert Law [2]. In perfect dielectric, sigma = 0, and alpha = k, and beta = 0, as expected. [1] Panofsky & Phillips, Classical Electricity and Magnetism 2nd Ed., AddisonWesley, 1962. [2] http://en.wikipedia.org/wiki/Attenuation_coefficient 


#6
Feb1011, 03:14 PM

P: 3

Thank you all for your responses.
How is the polarization of the EM radiation waves relevant to the imaginary part of the dielectric function? 


#7
Feb1011, 09:23 PM

P: 15

E``(z)+(mu*epsilon*omega^2+i*mu*omega*sigma)*E(z)=0. But what if the permittivity, epsilon, is complex, thus has real and imaginary parts, epsilonr and epsiloni, epsilon=epsilonr+i*epsiloni, and and the media is conductive? If so then, E``(z)+(mu*epsilonr*omega^2+i*mu*omega*(sigma+ epsiloni *omega))*E(z)=0, so that the "effective conductivity" becomes, sigmaeff = sigma+epsiloni*omega, and alpha = k*sqrt([sqrt{1+(sigma/[epsilonr*omega]+epsiloni/epsilonr)^2}+1]/2) beta = k*sqrt([sqrt{1+(sigma/[epsilonr*omega]+epsiloni/epsilonr)^2}1]/2). I never seen an analysis that includes both conductivity and complex permittivity, but my search has not been exhaustive. This treatment has been nagging me for quite some time. Any comments? 


#8
Feb1111, 01:54 AM

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#9
Feb1111, 02:09 AM

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P: 3,593

See, e.g. Landau Lifgarbagez, Electrodynamics of Continuous media, or, for the specialist, Agranovich and Ginzburg, Crystal Optics with Spatial Dispersion, and Excitons. 


#10
Feb1111, 08:27 AM

P: 15




#11
Feb1111, 09:30 AM

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P: 3,593




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