Propagation of EM Waves in material media S.M.Lea 017 1 Wave propagation As usual, we start with Maxwell s euations with no free charges: =0 =0 = = + If we now assume that each field has the plane wave form = 0 ( ) (or euivalently, we Fourier transform everything), the euations simplify: =0 (1) =0 () = (3) = + (4) Then, for an LIH, conducting medium, we can eliminate = = and = to get: =0 = = 1+ (5) For now let s take =0 (non-conducting medium). Then we get the wave euation: = = = (6) where we used euation (). So the wave phase speed is = = 1 1
and the refractive index is: = r = 0 0 (7) and Then euation (3) becomes: = (8) ˆ = (9) For most LIH materials in which = is a useful relation, ' 0 so the refractive index is primarily determined by the dielectric constant 0 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 We choose coordinates so that the boundary is the plane. The field in the incident wave is = exp In general there will be a transmitted wave with = exp and a reflected wave with = exp The plane of incidence is the plane containing the normal to the boundary and the incident ray, that is ˆ and. We choose the axes so that the plane of incidence is the plane. The angle of incidence istheanglebetween and ˆ =ˆ Then = ( sin + cos ) The boundary conditions we have to satisfy are (Notes 1 euations 5, 7, 8 and 10 with and zero). ˆ = is continuous (10) ˆ = is continuous (11) tan is continuous (1) tan is continuous (13) For most ordinary materials 0 ' 1 so we shall ignore the difference between and 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.Thatis, each wave has the same freuency and = sin is the same for each wave. From (8) with fixed we also have = and = 1 (en 8). So the angle of incidence euals theangleofreflection, and
sin = sin = sin 1 or 1 sin = 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.). We now have four euations to solve for the unknowns and However, two of the euations (1 and 13) are vector euations with two components, so we actually have six euations. Since we know the directions of the wave vectors and is perpendicular to there are two components of and in the plane perpendicular to their respective so we have four unknowns. Our euations are not all independent. We can simplify by decomposing the incident light into two specific linear polarizations..1 Polarization perpendicular to the plane of incidence In this polarization, is perpendicular to the plane of incidence, (that is, with our chosen coordinates, = ˆ). The vectors and in the waves are as shown in the diagram. The direction of is chosen so that the Poynting vector, is in the correct direction for each 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. En. 3
(1) has only one non-zero component: + = (15) and similarly from (13) we have: + = and from euation (9) we find = ˆ so we may rewrite this euation in terms of the electric field components. 1 ( )cos = cos = ( + )cos (16) where we used euation (15) in the last step. The final boundary condition is (11): + = 1 ( + )sin = sin whereweused = ˆ With Snell s law, this relation duplicates en (15), so we have two euations for two unknowns. Rearranging en (16), we get ( 1 cos cos )= ( 1 cos + cos ) Solving for and using Snell s law to eliminate,wehave = 1 cos cos 1 cos + cos = 1 cos r1 sin 1 cos + r1 sin = 1 cos 1 sin (17) 1 cos + 1 sin The reflected amplitude depends on the angle of incidence, and on the ratio of the two refractive indices. From euation (17) we can conclude that has the opposite sign from if 1 independent of the angle If 1 1 sin 1 sin =cos s1 sin 1 cos Thus the wave has a phase change of if 1 as we learn in elementary optics. Finally from euation (15), we find the transmitted amplitude: 1 cos = 1 cos + 1 sin The time-averaged power transmitted is given by (waveguide notes en 30) = 1 Re =Re ( ˆ ) = ˆ 0 0 (18) 4
Thus the sum of power transmitted normally across the boundary plus power reflected normally is ˆ+ ˆ = 1 cos + cos 0 0 = 0 1 cos 1 sin 1 cos 1 cos + 1 sin + 1 cos 1 sin 1 cos + 1 sin = 1 cos 1 cos + 1 cos 1 sin + 1 sin 0 1 cos + 1 sin = 0 1 cos = ˆ as expected!. Polarization parallel to the plane of incidence The and fields in this polarization look like this: and boundary condition (11) is trivially satisfied. 1 = = 0 together with en (9) gives Boundary condition (13) with + = 1 ( + ) = (19) 5
Then (1) gives: s ( )cos = cos = 1 ( + ) 1 The final boundary condition (10) is 1 ( + )sin = sin and, since 1 1 (en 7), this duplicates en (19). En (0) gives : s cos s 1 1 sin = cos + 1 1 So sin (0) sin and then from (19): cos r1 = sin cos + r1 1 sin cos 1 = 1 sin cos + 1 1 sin = 1 cos cos + 1 1 sin 1 cos = cos + 1 1 sin (1) ().3 Polarization by reflection Since the reflected amplitudes (1) and (17) are not the same in the two different polarizations, the reflected light is always partially polarized when the incident light is unpolarized. Euation (1) shows that the reflected amplitude in the polarization parallel to the plane of incidence is zero if: cos 1 1 sin =0 or, suaring and writing 1=cos +sin we have 4 cos = 1 cos +sin 1 sin cos 1 = 1 sin 1 Thus either 1 = (no boundary), or tan = (3) 1 6
This is Brewster s angle. Can this also happen for the other polarization? We would need: 1 cos 1 sin = 0 1 cos = 1 sin or = 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, sin = 1 sin = sin 1 =cos tan Thus = Theanglebetweenthereflected and tramsmitted waves is = + = + = Thus electrons accelerated by the elelctric field in medium would need to radiate along the direction of the acceleration in order to create the reflected wave. This is impossible. (wavemks notes, see eg en 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: = 1+ and thus the solution for is complex. = = (cos + sin ) = + where r = 1+ tan = Then = ˆ ˆ The imaginary part of indicates spatial attenuation of the wave, Im = = 14 1+ sin (4) (5) (6) 7
while the real part of gives the wave phase speed ph = If is small ( 1), we may expand the functions (5) and (6) to first order: = 1+ 1 ' 4 = and we get back the expected results as 0 The imaginary part of is small because is small. Im () ' = r and the wave phase speed is almost unchanged: ph ' 1 But if is large ( À 1), 4,and ' r = Im () ' 1 r = =Re() The wave is damped within a short distance r 1 = (7) The distance is called the skin depth. In a conducting medium the relations between the fields are (en 3): = ˆ so is also the phase shift between the fields. If is small ( low conductivity) then the amplitude of is almost the same as the amplitude of times and the phase shift 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 = 0 0 = r 0 0 0 p Re () = 1 To summarize, in a good conductor the wave fields are primarily magnetic, and are out of phase by 4 and the wave phase speed is very slow. 3. 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 à s! exp ( sin + cos ) =exp sin + 1 sin where is the incident wave number. For a good conductor, with r ' = = 1 we have = r r 0 1 ' = 1 1 So is small when is large. Then the coefficient of is cos = sin = sin + = sin + or r cos = sin 1 sin " # 14 = sin 1+ sin ' = ( 1) (8) where tan = sin and thus ' and ' (1 ) Thus the field components in the conductor are proportional to exp h i (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), euation (15) still holds, but the second boundary condition (euation 13) becomes (using 8): ( )cos = cos = ' (1 + ) Combining with euation (15), we have: = 1+ +1 cos ' 1 +1 cos or cos = (1 + ) = cos (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