Today in Physics 218: electromagnetic waves in linear media

Transcription

1 Today in Physics 218: electromagnetic waves in linear media Their energy and momentum Their reflectance and transmission, for normal incidence Their polarization Sunrise over Victoria Falls, Zambezi River (Photo by Galen Rowell) 2 February 2004 Physics 218, Spring

2 Energy and momentum in electromagnetic waves in linear media Unlike the wave equations, we can t get the correct results for energy- and momentum-related quantities just by replacing c with c µε Reverting to complex amplitudes and plane waves, we can obtain from the linear-media form of Faraday s law, y ˆ x ˆ ikz ( ωt) ( ) 0 ˆ ikz ωt = - x+ y = ik ye x+ ik 0xe yˆ z z 1 B iω ( ) ( ) 0 ˆ 0 ˆ ikz ωt = = B xx+ B yy e c t c Now, k = ω v = ω µε c, so µε E B µε E B B µε z E 0x = 0y, 0y = 0x = ˆ 2 February 2004 Physics 218, Spring

3 Energy and momentum in electromagnetic waves in linear media (continued) To get the correct form for energy density in such waves, we need to go through some of the derivation of Poynting s theorem again (see lecture notes for 23 January): c 1 D J = H dw mech dt 4π 4π t 1 D = E Jdτ = dτ ce ( H) E 4π t V 1 B D = dτ c ( E H) H E 4π t t V D ε ε 2 1 But E = ( E E) = E = ( E D) t 2 t 2 t 2 t V Use product rule #6 2 February 2004 Physics 218, Spring

4 Energy and momentum in electromagnetic waves in linear media (continued) Similarly, H B = t dwmech 1 1 = dτ c + dt 4π 2 t ( E H) ( E D B H) c 1 d = + 4π 8π dt = S da S V S 1 2 t ( B H) ( E H) da ( E D B H) 1 d 8π dt c c S = E H = E B, 4π 4πµ V, so V udτ, where dτ 2 February 2004 Physics 218, Spring

5 Energy and momentum in electromagnetic waves in linear media (continued) and u ( ) 8 8 E B = E D+ B H = ε + π π µ To get the momentum density properly in linear media, we would similarly have to retrace our steps through the derivation of momentum conservation that involved the Maxwell stress tensor That s not hard, but not very illuminating, so we ll take a short cut to the answer: εµ ε g = εµ S = E 4πc H = E 4πc B This is one of the few instances in which we could have saved ourselves writing by starting with the MKS expressions 2 February 2004 Physics 218, Spring

6 Reflection and transmission of electromagnetic waves Consider plane electromagnetic waves in a universe consisting of two non-conducting linear media with different permittivity and permeability The propagation speed is different, and thus reflection can occur at their interface For strings, we worked out the amplitudes of reflected and transmitted waves by applying the boundary conditions implied by string continuity and smoothness For fields, we have other boundary conditions (cf the lecture notes for 16 January); labelling the media 1 and 2, ε E ε E = 4πσ = 0 B B = 0 1,1 2,2 f,1, π E E = 0 B B = Kf nˆ = 0 c,1,2,1,2 µ 1 µ 2 2 February 2004 Physics 218, Spring

7 Reflection and transmission of electromagnetic waves (continued) Suppose an infinite plane (z = 0) separates the two media, and that a plane wave with E polarized in the x direction is incident along z (that is, incident normally) What are the reflected and transmitted amplitudes of and B? See next slide, and write ikz ( 1 ωt) ikz ( 1 ωt) ikz ( 1 ωt ) I = xˆe0ie B I = yˆb0ie = yˆ µε 1 1E0Ie i ˆ ( k1z ωt) i( k1z ωt ) R = xe0re B R = yˆ µε 1 1E0Re ikz ˆ ( 2 ωt) ikz ˆ ( 2 ωt = xe e B = y µε E e ) T 0T T 2 2 0T Note that B must point along x, because 0R B = µε kˆ E 2 February 2004 Physics 218, Spring

8 Reflection and transmission of electromagnetic waves (continued) E 0I x k 1 E 0T k 2 B 0I µ 1, ε1 B 0T z E 0R y µ 2, ε2 -k 1 B 0R 2 February 2004 Physics 218, Spring

9 Reflection and transmission of electromagnetic waves (continued) Since there are no components of either field perpendicular to the surface, only the parallel-component continuity condition is helpful At z = 0, continuity of gives us Continuity of gives us E ( 0, ) + ( 0, ) = ( 0, ) EIx t ERx t ETx t e e e i t i t i t 0I ω 0R ω + = 0T ω H H ( 0, t) + H ( 0, t) = H ( 0, t) Iy Ry Ty µ µ µ iωt iωt iωt µε 1 1E0Ie µε 1 1E0Re = µε 2 2E0Te February 2004 Physics 218, Spring

10 Reflection and transmission of electromagnetic waves (continued) Cancel common factors, divide the H equations through by ε 1 µ 1, and define β = ε 2 µ 1 ε 1 µ 2, to get 0I + 0R = 0T, = β 0I 0R 0T Hey, that s exactly like reflection in a vibrating two-part string (lecture notes, 28 January); the solution is 2 0T = 0I, 1 + β 1 β 0R = 0I 1 + β 2 February 2004 Physics 218, Spring

11 Reflection and transmission of electromagnetic waves (continued) Only one kind of magnetic material has relative permeability much different from unity: ferromagnets But ferromagnets are all conductors, so since we are considering nonconductors at the moment we can take µ = 1 Then β = ε 2 ε 1 = n 2 n 1 = v 1 v 2, so 2v2 2n1 0T = 0I = 0I, v2 + v1 n1 + n2 (µ = 1) v2 v1 n1 n2 0R = 0I = 0I v + v n + n Just as for strings, the reflected wave is in phase with incident if v2 > v1, and 180 out of phase (upside down) if not 2 February 2004 Physics 218, Spring

12 Reflection and transmission of electromagnetic waves (continued) That s for electric field amplitude; what about intensity? If µ = 1, then c c ε 2ˆ c ε 2 cn 2 S= E B = E k I = E = E, 4π 4π 8π 8π so 2 2 IT n2 E0T n2 2n1 4n1n2 T = = =, 2 = II n1 E n1 n1 + n2 ( n1 + n2) 0I 2 0R I IR E n n R = = = I I E n n Of course (energy conservation), 2 2 4nn 1 2 n1 n2 n1 + n2 T + R = + = = 1 ( n1 + n2) n1 + n2 n1 + n2 2 February 2004 Physics 218, Spring

13 Polarization of transmitted and reflected light An interesting point is raised by problem 914: In writing 976 and 977, I tacitly assumed that the reflected and transmitted waves must have the same polarization as the incident wave along the x direction Prove that this must be so [Hint: let the polarization vectors of the transmitted and reflected waves be nˆ = cosθ xˆ + sin θ yˆ, nˆ = cosθ xˆ + sin θ yˆ T T T R R R and prove from the boundary conditions that θt = θr = 0 ] He means i( k1z ωt) i( k1z ωt ˆ ) R = xe0re B R = yˆ µε 1 1E0Re ikz ˆ ( 2 ωt) ikz ˆ ( 2 ωt = xe e B = y µε E e ) T 0T T 2 2 0T 2 February 2004 Physics 218, Spring

14 Polarization of transmitted and reflected light (continued) for which we now have to write i( k1z ωt) i( k1z ωt ˆ ) R = nre0re B R = ( zˆ nˆr) µε 1 1E0Re ikz ˆ ( 2 ωt) ikz ˆ ˆ ( 2 ωt = n E e B = z n µε E e ) ( ) T T 0T T T 2 2 0T though ˆ ( 1 ) ( 1 ) 0 ω t and ˆ ω I = xe Ie B I = y µε E Ie t still The boundary conditions now give us 0Ixˆ + 0Rnˆ = R 0Tnˆ T, 0Iyˆ 0R( zˆ nˆ ) = β R 0T ( zˆ nˆ T ) In Cartesian components, the first of these is 0I + 0R cosθr = 0T cos θt, x sinθ = sin θ y 0R R 0T T 2 February 2004 Physics 218, Spring

15 Polarization of transmitted and reflected light (continued) The Cartesian components of the second one are 0RsinθR = β 0T sin θt, cosθ = β cos θ 0I 0R R 0T T Compare the first y component to the second x component: 0RsinθR = 0T sin θt, 0RsinθR = β 0T sin θt Just add them: 0T sinθt ( 1 + β) = 0 sinθt = 0 sinθ = 0 sinθ = 0 0R R R Thus θt = θr = 0; that is, the incident, reflected and transmitted light have the same polarization x y 2 February 2004 Physics 218, Spring