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Propagation of EM Waves in material media S.M.Lea 2017 1 Wave propagation As usual, we start with Maxwell’s equations with no free charges: · =0 ∇ · =0 ∇ × = − ∇ × = ∇ + = 0 (·−) (or equivalently, If we now assume that each field has the plane wave form we Fourier transform everything), the equations simplify: · =0 (1) · =0 (2) × = (3) = − + × (4) = = Then, for an LIH, conducting medium, we can eliminate and = to get: =0 · ³ ´ × = − − = − 1 + (5) For now let’s take = 0 (non-conducting medium). Then we get the wave equation: ³ ´ ³ ´ × × = − × − 2 = · 2 = 2 where we used equation (2). So the wave phase speed is 1 = = √ 1 (6) and the refractive index is: = = and = Then equation (3) becomes: r 0 0 (7) (8) = ̂ × (9) For most LIH materials in which = is a useful relation, ' 0 so the refractive index is primarily determined by the dielectric constant 0 2 Reflection and transmission of waves at a boundary Now let’s consider a wave incident on a flat boundary between two media with refractive indices 1 and 2 We choose coordinates so that the boundary is the − −plane. The field in the incident wave is ³ ´ = exp · − In general there will be a transmitted wave with ³ ´ = exp · − and a reflected wave with ³ ´ = exp · − The plane of incidence is the plane containing the normal to the boundary and the incident ray, that is ̂ and . We choose the axes so that the plane of incidence is the − −plane. The angle of incidence is the angle between and ̂ = ̂ Then · = ( sin + cos ) The boundary conditions we have to satisfy are (Notes 1 equations 5, 7, 8 and 10 with zero). and = is continuous ̂ · (10) = is continuous ̂ · (11) tan is continuous (12) tan is continuous (13) and For most ordinary materials 0 ' 1 so we shall ignore the difference between 0 In order to satisfy the boundary conditions for all times and at all and we must have · − = + − the same for each wave at the boundary = 0. That is, each wave has the same frequency and = sin is the same for each wave. From (8) with fixed we also have = and = 21 (eqn 8). So the angle of incidence equals the angle of reflection, and 2 sin = sin = or 2 sin 1 1 sin = 2 sin (14) which is Snell’s law. Notice that the physical argument that gives the laws of reflection and refraction (boundary conditions must hold for all time and everywhere on the boundary) is independent of the kind of wave and the specific form of the boundary conditions, and so these laws hold for waves of all kinds (sound waves, seismic waves, surface water waves, etc.). and However, two of the We now have four equations to solve for the unknowns equations (12 and 13) are vector equations with two components, so we actually have six is perpendicular to equations. Since we know the directions of the wave vectors and there are two components of and in the plane perpendicular to their respective so we have four unknowns. Our equations are not all independent. We can simplify by decomposing the incident light into two specific linear polarizations. 2.1 Polarization perpendicular to the plane of incidence is perpendicular to the plane of incidence, (that is, with our chosen In this polarization, = ̂). The vectors and in the waves are as shown in the diagram. The coordinates, is chosen so that the Poynting vector, is in the correct direction for each direction of wave. In this case there are only two unknowns, and but the boundary condition (10) is trivially satisfied, since = 0 The remaining conditions cannot all be independent. Eqn. 3 (12) has only one non-zero component: + = (15) and similarly from (13) we have: + = and from equation (9) we find = −̂ so we may rewrite this equation in terms of the electric field components. 1 ( − ) cos = 2 cos = 2 ( + ) cos (16) where we used equation (15) in the last step. The final boundary condition is (11): + 1 ( + ) sin = = 2 sin where we used = ̂ With Snell’s law, this relation duplicates eqn (15), so we have two equations for two unknowns. Rearranging eqn (16), we get (1 cos − 2 cos ) = (1 cos + 2 cos ) Solving for and using Snell’s law to eliminate , we have 1 cos − 2 cos = 1 cos + 2 cos r ³ ´2 1 cos − 2 1 − 12 sin2 r = ³ ´ 2 = 1 cos + 2 1 − 12 sin2 q 1 cos − 22 − 21 sin2 q 1 cos + 22 − 21 sin2 (17) The reflected amplitude depends on the angle of incidence, and on the ratio of the two refractive indices. From equation (17) we can conclude that has the opposite sign from if 2 1 independent of the angle If 2 1 µ ¶2 1 sin2 1 − sin2 = cos2 1− 2 s µ ¶2 1 sin2 1 cos 2 1 − 2 Thus the wave has a phase change of if 2 1 as we learn in elementary optics. Finally from equation (15), we find the transmitted amplitude: 21 cos q = 1 cos + 22 − 21 sin2 The time-averaged power transmitted is given by (waveguide notes eqn 30) ¸ ∙ ´ ³ = 1 Re × ( ̂ × ∗ ) = ̂ ||2 × ∗ = Re 2 20 20 4 (18) Thus the sum of power transmitted normally across the boundary plus power reflected normally is · ̂ = 1 | |2 cos + 2 | |2 cos · ̂ + 20 20 ⎧ ⎡ q ⎤2 ⎪ 2 2 2 | |2 ⎨ ⎣ 1 cos − 2 − 1 sin ⎦ q 1 = cos 20 ⎪ ⎩ 1 cos + 22 − 21 sin2 ⎫ ⎡ ⎤2 q 2 ⎪ 2 2 2 − 1 sin ⎬ 21 cos ⎦ q +2 ⎣ ⎪ 2 ⎭ 1 cos + 22 − 21 sin2 q 2 2 2 cos + 2 cos 22 − 21 sin2 + 22 − 21 sin2 1 1 | | = 1 cos ¶2 µ q 20 1 cos + 22 − 21 sin2 2 = 2.2 | | · ̂ 1 cos = 20 as expected! Polarization parallel to the plane of incidence and fields in this polarization look like this: The and boundary condition (11) is trivially satisfied. Boundary condition (13) with 1 = 2 = 0 together with eqn (9) gives + = 1 ( + ) = 2 5 (19) Then (12) gives: 1 ( − ) cos = cos = ( + ) 2 s 1− The final boundary condition (10) is µ 1 2 ¶2 sin2 (20) 1 ( + ) sin = 2 sin 21 and, since 1 ∝ (eqn 7), this duplicates eqn (19). Eqn (20) gives : ⎛ ⎛ ⎞ ⎞ s s µ ¶2 µ ¶2 1 1 1 1 1− sin2 ⎠ = ⎝cos + 1− sin2 ⎠ ⎝cos − 2 2 2 2 So = cos − 1 2 r ³ ´2 1 − 12 sin2 r ³ ´ 2 1 − 12 sin2 q 22 cos − 1 22 − 21 sin2 q = 22 cos + 1 22 − 21 sin2 cos + and then from (19): = 1 2 222 cos 1 q 2 2 cos + 2 − 2 sin2 1 2 2 1 21 2 cos q = 22 cos + 1 22 − 21 sin2 2.3 (21) (22) Polarization by reflection Since the reflected amplitudes (21) and (17) are not the same in the two different polarizations, the reflected light is always partially polarized when the incident light is unpolarized. Equation (21) shows that the reflected amplitude in the polarization parallel to the plane of incidence is zero if: q 22 cos − 1 22 − 21 sin2 = 0 or, squaring and writing 1 = cos2 + sin2 we have ¢ ¤ £ ¡ 42 cos2 = 21 22 cos2 + sin2 − 21 sin2 ¡ ¡ ¢ ¢ 22 cos2 22 − 21 = 21 sin2 22 − 21 Thus either 1 = 2 (no boundary), or tan = 6 2 1 (23) This is Brewster’s angle. Can this also happen for the other polarization? We would need: q 1 cos − 22 − 21 sin2 = 0 21 cos2 = 22 − 21 sin2 or 2 = 1 So the reflected field amplitude for this polarization is not zero unless there is no boundary. Thus when unpolarized light is incident at Brewster’s angle, the reflected light is 100% polarized perpendicular to the plane of incidence. At other angles of incidence, the reflected light is partially polarized. Why does this happen? At Brewster’s angle, 1 sin 1 sin = sin = = cos 2 tan Thus = − 2 The angle between the reflected and tramsmitted waves is = − + − = − + = 2 2 2 2 Thus electrons accelerated by the elelctric field in medium 2 would need to radiate along the direction of the acceleration in order to create the reflected wave. This is impossible. (wavemks notes, see eg eqn 38.) 3 Waves in a Conducting medium 3.1 Propagation If the medium is a conductor, (non-zero in (5)), the dispersion relation (6) becomes: ³ ´ 2 = 2 1 + (24) and thus the solution for is complex. where = || = || (cos + sin ) = + 2 2 = 1 + || tan 2 = Then r ³ ´2 · = ̂· − ̂· The imaginary part of indicates spatial attenuation of the wave, ∙ ³ ´2 ¸14 √ Im = = 1 + sin 7 (25) (26) while the real part of gives the wave phase speed ph = If is small ( ¿ 1), we may expand the functions (25) and (26) to first order: µ ¶ √ 1 ³ ´2 √ || = 1 + ' 4 = 2 and we get back the expected results as → 0 The imaginary part of is small because is small. r √ Im () ' = 2 2 and the wave phase speed is almost unchanged: 1 ph ' √ But if is large ( À 1), → 4, and r √ √ || ' = r 1 √ = Re () Im () ' √ = 2 2 The wave is damped within a short distance r 2 (27) ∼ 1 = The distance is called the skin depth. In a conducting medium the relations between the fields are (eqn 3): ³ ´ = || ̂ × so is also the phase shift between the fields. If is small ( ¿ low conductivity) then is almost the same as the amplitude of times √ and the phase shift the amplitude of ¯ ¯ ¯¯ is also small. But if is large ( ' high conductivity À 1) then ¯ ¯ is much larger ¯ ¯ √ ¯¯ than ¯ ¯ and the phase shift is almost 4 . Also the phase speed is given by r √ √ 0 0 0 0 20 = = p = ¿1 Re () 2 and are To summarize, in a good conductor the wave fields are primarily magnetic, out of phase by 4 and the wave phase speed is very slow. 3.2 Reflection and refraction at a boundary with a conducting medium How does this affect the reflection and refraction of a wave? Consider a wave with wave 8 number incident on a conducting medium. Inside the conductor the fields behave like s à ! 2 2 exp ( sin + cos − ) = exp sin + 1 − 2 sin − where is the incident wave number. For a good conductor, with r 1 '= = 2 we have r r µ ¶−12 2 21 √ 1 √ = 0 1 ' = 2 2 2 So is small when 2 is large. Then the coefficient of is q q q cos = 2 − 2 sin2 = 2 − 2 − 2 sin2 + 2 = −2 sin2 + 2 or cos = − sin r 1− 2 2 2 sin2 ¶2 #14 2 −2 = − sin 1 + 2 2 sin2 √ √ 2 −2 2 −2 ' − =− ( ¿ 1) where and thus ' 2 and −2 are proportional to " µ (28) 2 tan = 2 2 2 √ sin ' (1 − ) 2 Thus the field components in the conductor h i exp − (1 − ) This shows that the fields in the conducting medium propagate only a distance of order in the −direction (normal to the boundary). For polarization perpendicular to the plane of incidence (only non-zero), equation (15) still holds, but the second boundary condition (equation 13) becomes (using 28): √ 2 −2 ( − ) cos = cos = (1 + ) ' Combining with equation (15), we have: µ ¶ µ ¶ +1 1 +1 ' 2 = 1 + cos cos or cos = cos (1 − ) = 2 (1 + ) which is very small (| | ¿ ). Thus almost all the wave energy is reflected. A similar result holds for the other polarization. (See also Jackson problems 7.4 and 7.5.) 9