Transmission and Reflection of Electromagnetic Waves, Assignments of Physics

Download Transmission and Reflection of Electromagnetic Waves and more Assignments Physics in PDF only on Docsity! Physics 505 Fall 2005 Homework Assignment #11 — Solutions Textbook problems: Ch. 7: 7.3, 7.5, 7.8, 7.16 7.3 Two plane semi-infinite slabs of the same uniform, isotropic, nonpermeable, lossless dielectric with index of refraction n are parallel and separated by an air gap (n = 1) of width d. A plane electromagnetic wave of frequency ω is indicent on the gap from one of the slabs with angle of indicence i. For linear polarization both parallel to and perpendicular to the plane of incidence, a) calculate the ratio of power transmitted into the second slab to the incident power and the ratio of reflected to incident power; We introduce (complex) electric field vectors of the form ~Eiei ~k·~x and ~Ere−i ~k·~x on the incident side, ~E+ei ~k0·~x and ~E−e−i ~k0·~x in the air gap, and ~Etei ~k·(~x−~d) on the transmitted side. (We have removed an unimportant phase from the transmitted side by shifting ~x by the vector ~d pointing from the incident to the transmitted side of the air gap. If i is the incident angle, then the angle r from the normal in the air gap is given by Snell’s law, n sin i = sin r, and the transmitted angle is also i (because it is the same dielectric). We see that cos r = √ 1− sin2 r = √ 1− n2 sin2 i and that cos r is purely imaginary in the event that i is greater than the critical angle for total internal reflection. To obtain Et and Er in terms of Ei, we may match the parallel components of ~E as well as the parallel components of ~H. We consider two cases. For ~E perpendicular to the plane of incidence, the matching becomes first interface second interface E‖ : Ei + Er = E+ + E−, E+eiφ + E−e−iφ = Et H‖ : n(Ei − Er) cos i = (E+ − E−) cos r, (E+eiφ − E−e−iφ) cos r = nEt cos i where we have introduced the phase φ = ~k0 · ~d = k0d cos r = ωd cos r c The matching conditions at the first interface may be written as E+ = 12Ei(1 + α) + 1 2Er(1− α) E− = 12Ei(1− α) + 1 2Er(1 + α) (1) where we have defined α = n cos i cos r = n cos i√ 1− n2 sin2 i Similarly, the matching conditions at the second interface yield E+ = 12e −iφEt(1 + α) E− = 12e iφEt(1− α) (2) Equating (1) and (2) allows us to solve for the ratios Et Ei = 4α (1 + α)2e−iφ − (1− α)2eiφ = 2α 2α cos φ− i(1 + α2) sinφ Er Ei = (1− α2)(eiφ − e−iφ) (1 + α)2e−iφ − (1− α)2eiφ = i(1− α2) sinφ 2α cos φ− i(1 + α2) sinφ (3) where α = n cos i√ 1− n2 sin2 i , φ = ωd cos r c = ωd √ 1− n2 sin2 i c So long as i is below the critical angle, both α and φ are real. In this case, the transmission and reflection coefficients are T = ∣∣∣∣EtEi ∣∣∣∣2 = 4α24α2 cos2 φ + (1 + α2)2 sin2 φ = 4α24α2 + (1− α2)2 sin2 φ R = ∣∣∣∣ErEi ∣∣∣∣2 = (1− α2)2 sin2 φ4α2 cos2 φ + (1 + α2)2 sin2 φ = (1− α2)2 sin2 φ4α2 + (1− α2)2 sin2 φ (4) Note that T + R = 1, as expected. However, this exhibits a classic interference behavior, where T oscillates between (2α/(1 + α2))2 and 1 as the number of wavelengths in the gap vary. For ~E parallel to the plane of incidence, we find instead the matching conditions first interface second interface E‖ : (Ei − Er) cos i = (E+ − E−) cos r, (E+eiφ − E−e−iφ) cos r = Et cos i H‖ : n(Ei + Er) = (E+ + E−), E+eiφ − E−e−iφ = nEt These equations have the same structure as the perpendicular case, but with the index of refraction entering somewhat differently. We find the matching conditions n−1E+ = 12Ei(1 + β) + 1 2Er(1− β) n−1E− = 12Ei(1− β) + 1 2Er(1 + β) which is essentially equivalent to (3), up to redefining α → 1/n. (In fact, this problem can easily be generalized for incidence at an arbitrary angle i by taking 1/n→ cos i/n cos r.) We now take the limit where this is an excellent conductor, σ/ω0  1. In this case, the index of refraction (5) and phase change (6) may be approximated by n ≈ √ i σ ω0 = (1 + i) √ σ 20ω = 2 γ φ = ωD c n ≈ (1 + i)ωD c √ σ 20ω = (1 + i)D √ µ0σω 2 = iλ For |γ|  1 (equivalent to |n|  1) we drop terms of O(1/n2) compared to 1 to arrive at Et Ei = 2γe−λ (1− e−2λ) + γ(1 + e−2λ) Er Ei = −(1− e−2λ) (1− e−2λ) + γ(1 + e−2λ) (7) where we have kept the O(γ) term in the denominator which becomes important in the limit λ→ 0. b) Verify that for zero thickness and infinite thickness you obtain the proper limiting results. The zero thickness limit corresponds to λ→ 0. In this case, it is easy to see from (7) that λ→ 0 : Et Ei → 1, Er Ei → 0 In the infinite thickness limit, we find instead λ→∞ : Et Ei → 0, Er Ei → −1 1 + γ Note that the reflection coefficient does not go to unity, as some of the power is dissipated in the conductor. A perfect conductor (σ = ∞) has γ = 0, so all the power is reflected in the perfect conductor limit. c) Show that, except for sheets of very small thickness, the transmission coefficient is T = 8(<γ)2e−2D/δ 1− 2e−2D/δ cos(2D/δ) + e−4D/δ Sketch log T as a function of (D/δ), assuming <γ = 10−2. Define “very small thickness.” To compute the transmission coefficient from (7), we keep in mind that both γ and λ are complex. As long as we are not in the “very small thickness” limit, the O(γ) term in the denominator can be ignored. In this case Et Ei ≈ 2γe −λ (1− e−2λ) so that T = ∣∣∣∣EtEi ∣∣∣∣2 = 4|γ|2e−2<λ1− 2<(e−2λ) + e−4<λ Taking |γ|2 = 2(<γ)2 as well as e−2λ = e2iD/δe−2D/δ then gives T = 8(<γ)2e−2D/δ 1− 2e−2D/δ cos(2D/δ) + e−4D/δ The very small thickness limit corresponds to when the O(γ) term becomes im- portant. This occurs when |1− e−2λ| ' |γ(1 + e−2λ)| Expanding for small λ yields |2λ| ' |2γ| ⇒ D δ ' ωδ c Hence small thicknesses correspond to D < ωδ2 c 7.8 A monochromatic plane wave of frequency ω is indicdent normally on a stack of layers of various thicknesses tj and lossless indices of refraction nj . Inside the stack, the wave has both forward and backward moving components. The change in the wave through any interface and also from one side of a layer to the other can be described by means of 2× 2 transfer matrices. If the electric field is written as E = E+eikx + E−e−ikx in each layer, the transfer matrix equation E′ = TE is explicitly( E′+ E′− ) = ( t11 t12 t21 t22 ) ( E+ E− ) a) Show that the transfer matrix for propagation inside, but across, a layer of index of refraction nj and thickness tj is Tlayer(nj , tj) = ( eikjtj 0 0 e−ikjtj ) = I cos(kjtj) + iσ3 sin(kjtj) where kj = njω/c, I is the unit matrix, and σk are the Pauli spin matrices of quantum mechanics. Show that the inverse matrix is T ∗. Again, normal incidence makes this problem straightforward. For a right moving plane wave of the form eikjz passing through a layer of thickness tj , one picks up a phase eikjtj , while for a left moving wave, one picks up a phase e−ikjtj . More precisely E′+ = E+(z = tj) = E+(z = 0)e ikjtj = E+eikjtj E′− = E−(z =j) = E−(z = 0)e −ikjtj = E−e−ikjtj This directly leads to the transfer matrix Tlayer(nj , tj) = ( eikjtj 0 0 e−ikjtj ) where the inverse is obviously the complex conjugate. b) Show that the transfer matrix to cross an interface from n1 (x < x0) to n2 (x > x0) is Tinterface(2, 1) = 1 2 ( n + 1 −(n− 1) −(n− 1) n + 1 ) = I (n + 1) 2 − σ1 (n− 1) 2 where n = n1/n2. For the matching across layers, we again take the ~E perpendicular to plane of incidence conventions. This gives simply E‖ : E+ + E− = E′+ + E ′ − H‖ : n1(E+ − E−) = n2(E′+ − E′−) which may be solved to give E′+ = 1 2E+(1 + n) + 1 2E−(1− n) E′− = 1 2E+(1− n) + 1 2E−(1 + n) where n = n1/n2. This yields the transfer matrix Tinterface(2, 1) = 1 2 ( n + 1 −(n− 1) −(n− 1) n + 1 ) c) Show that for a complete stack, the incident, reflected, and transmitted waves are related by Etrans = det(T ) t22 Einc, Erefl = − t21 t22 Einc where tij are the elements of T , the product of the forward-going transfer ma- trices, including from the material filling space on the incident side into the first