Lecture5 - Petty Chapter 4 Reflection and Refraction(and...

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1 1 Petty Chapter 4 Reflection and Refraction (and Transmission) 2 Recall: – The electric vector describing EM radiation is given by: A(x,t) = A o exp[ i (kx - ω t) ] ν = frequency (cycles s -1 , Hz) = # crests that pass a given point each second ω = circular frequency (radians s -1 ) = 2 π ν (there are 2 π radians per cycle) k = 2 π / λ (radians cm -1 , sometimes called wavenumber, but don’t mix up with 1/ λ ). A o = maximum amplitude = constant Goal: Write the electric vector in terms of index of refraction: Complex index of refraction: N = n r + i n i c / c c is phase speed in any medium, n i is the imaginary part of the refractive index n r is the real part of the refractive index c = ν λ (cm s -1 , e.g., cycles per sec x cm per cycle). Since ν = ω /2 π and λ = 2 π /k, c = ω / k [or k = ω / c ] (In a vacuum, c = c = 3.0x10 10 cm s -1 is the phase speed of light.) For non-absorbing medium: N = n r so n r = c / c , so k = ω n r / c . Therefore, the electric vector in a non-absorbing, non-vacuum can be written as: A(x,t) = A o exp[ i ω (n r x/c - t) ]
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2 3 A(x,t) = A o exp[ i (kx - ω t) ] = A o exp[ i ω (n r x/c - t) ] (from last slide) Recall also that the “intensity” of the radiation is proportional to A* A : A*(x,t) A(x,t) = A o 2 exp[ i ω (n r x/c - t) ] exp[- i ω (n r x/c - t) ] = A o 2 This cannot represent a physically realistic wave because it cannot be attenuated if A o is a constant. Therefore, allow exponential attenuation (absorption) by re-writing A o as: A o = A o exp( – n i ω x / c) so that the electric vector can be re-written as: A = A o exp(– n i ω x / c) exp{ i ω ( (n r x / c) – t ) } A = A o exp { i ω (N x / c – t ) } Note that if we re-define k as complex so that k = ω N / c = k r + ik i (where k r = ω n r /c and k i = ω n i /c), then: A = A o exp {i (kx – ω t)} as before (top of slide). 4 (From last slide) The intensity is proportional to A*A: A*A = A o 2 exp(-2 ω n i x/c) = A o 2 exp(-4 π n i x/ λ ) Here we have used ω =2 π c/ λ , where λ is the vacuum wavelength. The attenuation is therefore given by exp(- 4 π n i x / λ ) . Thus for radiation that passes through an absorbing medium with incident intensity I o and outgoing intensity I, we find: I (x) / I o = exp (- 4 π n i x / λ ) In Chapter 2, Petty defines the absorption coefficient as: β a = 4 π n i / λ so I (x) / I o = exp ( – β a x ) = T(x) T = Transmission Notation & conventions: Here we have defined the complex index of refraction as N n r + i n i . This corresponds to a convention of A = A o exp [ i (kx – ω t) ]. N n r -i ±n i corresponds to a convention of A = A o exp [ i (kx + ω t) ]. A = A o exp { i ω (N x / c – t ) } A = A o exp { i ω [ (n r x / c) + (i n i x / c) – t ] } Beer’s Law
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3 5 Penetration Depth, D: Value of x where T = 1/e T = 1/e = exp(– 4 π n i x / λ ) –1 ±=±–4 ± π n i x / λ x = D = λ / (4 π n i ) (Petty 4.5) Next derive exponential attenuation of radiation as it passes through an absorbing medium following Bohren and Clothiaux, “Fundamentals of Atmospheric Radiation” (B&C).
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Lecture5 - Petty Chapter 4 Reflection and Refraction(and...

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