sign in sign up
<<
Home , Circular polarization, Elliptical polarization, Geometrical optics, Linear polarization, Phase (waves), Plane of incidence

Section 3: Electromagnetic 1 EM waves in vacuum In regions of space where there are no charges and currents, Maxwell equations read

∇ ⋅E = 0 (3.1) ∇ ⋅B = 0 (3.2) ∂B ∇×E =− (3.3) ∂t ∂E ∇×B = µ ε (3.4) 0 0 ∂t They are a set of coupled, first order, partial differential equations for E and B. They can be decoupled by applying curl to eqs. (3.3) and (3.4): ∂B ∂ ∂ 2 E ∇×∇×()()EEE =∇∇⋅ −∇2 =−∇× =−() ∇× B =− µ ε (3.5) ∂t ∂ t0 0 ∂ t 2 ∂E  ∂ ∂ 2 B ∇×∇×()()BBB =∇∇⋅ −∇2 =∇×µε  = µε() ∇× E =− µε (3.6) 00∂t  00 ∂ t 00 ∂ t 2 Since ∇ ⋅E = 0 and ∇ ⋅B = 0 we have ∂2E ∇2E = µ ε (3.7) 0 0 ∂t 2 ∂2B ∇2B = µ ε (3.8) 0 0 ∂t 2 We now have separate equations for E and B, but they are of second order; that's the price you pay for decoupling them. In vacuum, then, each Cartesian component of E and B satisfies the three-dimensional equation ∂2 f ∇2 f = µ ε (3.9) 0 0 ∂t 2 The solution of this equation is a wave. So Maxwell’s equations imply that empty space supports the propagation of electromagnetic waves, traveling at a speed 1 c= =3.00 ⋅ 108 m / s (3.10) µ ε 0 0 which happens to be precisely the velocity of , c. The implication is astounding: light is an electromagnetic wave. Of course, this conclusion does not surprise anyone today, but imagine what a ε µ revelation it was in Maxwell's time! Remember how 0 and 0 came into the theory in the first place: they were constants in Coulomb's law and the Biot-Savart law, respectively. You measure them in experiments involving charged pith balls, batteries, and wires—experiments having nothing whatever to do with light. And yet, according to Maxwell's theory you can calculate c from these two numbers. Notice the crucial role played by Maxwell's contribution to Ampere's law; without it, the wave equation would not emerge, and there would be no electromagnetic theory of light.

1 Monochromatic waves Since different in the visible range correspond to different colors, such waves are called monochromatic . Suppose that the waves are traveling in the z direction and have no x or y dependence ; these are called plane waves, because the fields are uniform over every plane to the direction of propagation. We are interested, then, in fields of the form = ikz(−ω t ) E(z , t ) E 0 e (3.11) = i( kz−ω t ) B(z , t ) B 0 e , (3.12) where E0 and B0 are the (complex) (the physical fields, of course, are the real parts of E and B). Substituting eqs.(3.11) and (3.12) to eqs. (3.7) and (3.8) respectively we find that ω c = . (3.13) k Here, k is the wave number, which is related to the of the wave by the equation 2π λ = , (3.14) k and ω is the angular of EM wave.

Fig. 1 This is the paradigm for a monochromatic plane wave. The wave is polarized in the x direction (by convention, we use the direction of E to specify the of an electromagnetic wave).

Now, the wave equations for E and B were derived from Maxwell's equations. However, whereas every solution to Maxwell's equations (in empty space) must obey the wave equation, the converse is not true; Maxwell's equations impose extra constraints on E0 and B0. In particular, since ∇ ⋅E = 0 and ∇ ⋅B = 0 , it follows that = = E0z B 0 z 0 (3.15) That is, electromagnetic waves are transverse: the electric and magnetic fields are perpendicular to the ∂B direction of propagation. Moreover, Faraday's law, ∇×E =− , implies a relation between the electric ∂t and magnetic amplitudes: xˆ y ˆ zˆ ∇×=∇×() ikzt()−ω −×∇ ikzt () − ω =−×∇ ikzt () − ω =− ikzt ()− ω = EE0e E 0 e E 0 e EEe 00x y 0 (3.16) 0 0 ik −+i() kz−ω t =+ω ω i () kz− ω t ()()xyˆˆikE00yx ikEexy ˆˆ iB 00 xy iBe which results in

2 − =ω = ω kE0y B 0 x, kE 0 x B 0 y , (3.17) or, more compactly: k 1 B=() zEˆ × =() zEˆ × (3.18) 0ω 0c 0 Evidently, E and B are in phase and mutually perpendicular; their (real) amplitudes are related by k 1 B= E = E (3.19) 0ω 0c 0 ∂E The fourth of Maxwell's equations, ∇×B = µ ε , does not yield an independent condition; it simply 0 0 ∂t reproduces Eq. (3.16). There is nothing special about the z direction, of course—we can easily generalize to monochromatic plane waves traveling in an arbitrary direction. The notation is facilitated by the introduction of the , k , pointing in the direction of propagation, whose magnitude is the wave number k. The scalar product k⋅ r is the appropriate generalization of kz , so = i(k⋅ r − ω t ) E( r ,t ) E0 e e , (3.20)

1⋅ − ω 1 Br( ,t ) = E() ne × e i(k r t ) =×() nE . (3.21) c0 c k Where vector n = is the unit vector in the direction of propagation of the EM wave and e is the k polarization vector. Because E is transverse, n⋅ e = 0 . (3.22) Linear and circular polarizations The plane wave (3.20) and (3.21) is a wave with its vector always in the direction e. Such a = wave is said to be linearly polarized with polarization vector e1 e . Evidently there exists another wave ≠ which is linearly polarized with polarization vector e2 e 1 and is linearly independent of the first. Thus the two waves are E( r ,t ) = E e e i(k⋅ r − ω t ) 1 1 1 (3.23) = i(k⋅ r − ω t ) E2( r ,t ) E 2 e 2 e 1 with B=() n × E . 1,2c 1,2 They can be combined to give the most general homogeneous plane wave propagating in the direction k= k n : = + =+( ) i(k⋅ r − ω t ) Er(,)t Er1 (,) t Er 2 (,) tEEe 1122 e e . (3.24)

The amplitudes E1 and E2 are complex numbers, to allow the possibility of a phase difference between waves of different .

If E1 and E2 have the same phase wave (3.24) represents a linearly polarized wave, with its polarization θ = =2 + 2 vector making an tanE2 / E 1 with respect to e1 and a magnitude E E2 E 1 as shown in Fig.2a.

3

Fig. 2 (a) Electric field of a linearly polarized wave. (b) Electric field of a circularly polarized wave.

If E1 and E2 have different phases, the wave (3.24) is elliptically polarized. To understand what this means let us consider the simplest case, . Then E1 and E2 have the same magnitude, but differ in phase by 90°. The wave (3.24) becomes: =( ± ) i(k⋅ r − ω t ) Er( ,tE ) 0 ee 1 ie 2 (3.25) with E0 the common real . We imagine axes chosen so that the wave is propagating in the positive z direction, while e1 and e2 are in the x and y directions, respectively. Then the components of the actual electric field, obtained by taking the real part of (3.25), are = − ω Ex (,)r t E0 cos( kz t ) = − ω (3.26) Ey (,)r t∓ E0 sin( kz t ) At a fixed point in space, the fields (3.26) are such that the electric vector is constant in magnitude, but ω ( + ) sweeps around in a circle at a frequency , as shown in Fig.2b. For the upper sign e1i e 2 , the rotation is counterclockwise when the observer is facing into the oncoming wave. This wave is called left circularly polarized in . In the terminology of modern , however, such a wave is said to have positive helicity. The latter description seems more appropriate because such a wave has a positive ( − ) projection of angular momentum on the z axis. For the lower sign e1i e 2 , the rotation of E is clock- wise when looking into the wave; the wave is right circularly polarized (optics): it has negative helicity.

Figure 3. Electric field and magnetic induction for an elliptically polarized wave.

The two circularly polarized waves (3.26) form an equally acceptable set of basic fields for description of a general state of polarization. If we introduce the complex orthogonal unit vectors: 1 e± =() e ± i e , (3.27) 2 1 2 then a general representation, equivalent to (3.24), is

4 i(k⋅ r − ω t ) Er( ,t ) =( E++ e + E −− e ) e (3.28) where E+ and E− are complex amplitudes. If E+ and E− have different magnitudes, but the same phase,

(3.28) represents an elliptically polarized wave with principal axes of the ellipse in the directions of e1 + − = and e2 . The ratio of semimajor to semiminor axis is (1r ) /(1 r ) , where r E−/ E + . If the amplitudes iα have a phase difference between them, E−/ E + = re , then the ellipse traced out by the E vector has its axes rotated by an angle α/2. Fig.3 shows the general case of and the ellipse traced out by E at a given point in space. For r = ± 1 we get back a linearly polarized wave. Energy and momentum of electromagnetic waves Energy per unit volume stored in electromagnetic fields is

1 1  u=ε E2 + B 2  . (3.29) 0 µ 2 0  Note that here the E and B are real quantities (real pars of complex quantities used in the previous paragraph). In case of monochromatic plane wave 1 B2= E 2 = µ ε E 2 , (3.30) 0c2 0 000 so the magnetic and electric contribution are equal ==εε2 2 2 −+ ωδ u0 E 0 E 0 cos ( kz t ) . (3.31) As the wave travels, it carriers this energy alone with it. The energy flux density (energy per unit area, per unit time) transported by the fields is given by the Poynting vector: 1 S=() E × B . (3.32) µ 0 For monochromatic plane waves propagating in the z direction, 1 E 2 Szz=ˆˆ =cε E2 = c ε E 2cos 2 ( kz −+= ωδ t ) zzˆˆ cu . (3.33) µ 0 0 0 0 c

Notice that S is the energy density (u) times the velocity of the waves (c) – as it A ∆ ∆ should be. For in a time t, a length c t passes through area A (Fig.4), carrying c with it an energy uAc ∆t. The energy per unit time, per unit area, transported by the wave is therefore uc. Fig.4 c∆t Electromagnetic fields not only carry energy, they also carry momentum. The momentum density stored in the fields is 1 p= S . (3.34) em c2 For monochromatic plane waves, then, 1 1 p=εE2cos 2 ( kz −+= ω t δ ) zˆ u zˆ . (3.35) em c0 0 c In the case of light, the wavelength is so short (~ 5 x 10 -7 m), and the period so brief (~ 10 -15 s), that any

5 macroscopic measurement will encompass many cycles. Typically, therefore, we're not interested in the fluctuating cosine-squared term in the energy and momentum densities; all we want is the average value. Now, the average of cosine-squared over a complete cycle is ½ so 1 u= ε E 2 . (3.36) 2 0 0 1 S= cε E 2 zˆ . (3.37) 2 0 0 1 p= ε E 2 zˆ . (3.38) em 2c 0 0 We use brackets, <>, to denote the time average over a complete cycle. The average power per unit area transported by an electromagnetic wave is called the intensity : 1 I≡ S = cEε 2 . (3.39) 2 0 0 Electromagnetic waves in matter In regions of matter where there are no free charges and free currents, Maxwell equations are ∇ ⋅D = 0 (3.40) ∇ ⋅B = 0 (3.41) ∂B ∇×E =− (3.42) ∂t ∂D ∇×H = (3.43) ∂t If the matter is linear 1 D=ε EH, = B (3.44) µ and homogeneous ( ε and µ are constants), Maxwell equations reduce to ∇ ⋅E = 0 (3.45) ∇ ⋅B = 0 (3.46) ∂B ∇×E =− (3.47) ∂t ∂E ∇×B = εµ , (3.48) ∂t ε µ εµ which are different from Maxwell equations in vacuum only by the replacement of 0 0 to . Evidently electromagnetic waves propagate trough a linear homogeneous medium at a speed 1 c v = = , (3.49) εµ n where εµ n ≡ , (3.50) ε µ 0 0 µ µ is the index of . For most (non-ferromagnetic) materials is very close to 0 so

6 ≅ ε n r , (3.51) ε≡ ε ε ε where r / 0 is the dielectric constant. Since r is always greater than 1, light travels more slowly µ= µ through matter. In our consideration of EM waves below we assume for simplicity that 0 . In that ε ε case all the results we obtained above to EM waves in vacuum are valid with replacement 0 by . and transmission of EM waves at incidence If a wave passes from one transparent medium into another, there is a reflected wave and a transmitted wave. The details depend on the exact nature of the electrodynamic boundary conditions which are ε⊥= ε ⊥ 11E 22 E (3.52) ⊥= ⊥ B1 B 2 (3.53) = E1 E 2 (3.54) = B1 B 2 (3.55) Here we assumed that the two media which are characterized by indices 1 and 2 have different electric ε≠ ε µ= µ = µ permeabilities 1 2 but the same magnetic permeabilities 1 2 0 . These equations relate the electric and magnetic fields just to the left and just to the right of the interface between two linear media. Now use them to deduce the laws governing reflection and refraction of electromagnetic waves at normal incidence

Fig.5

Suppose the xy plane forms the boundary between two linear media. A plane wave of frequency ω, traveling in the z direction and polarized in the x direction, approaches the interface from the left as is shown in Fig.5: −ω = ikz(1 t ) Ei(zt , ) Ee0 i xˆ (3.56)

1 −ω = ikz(1 t ) Bi(zt , ) Ee0 i yˆ . (3.57) v1 It gives rise to the reflected wave − − ω = i( kz1 t ) Er(zt , ) Ee0 r xˆ (3.58)

1 − − ω = − i( kz1 t ) Br(zt , ) Ee0 r yˆ , (3.59) v1 which travels back to the left in medium (1), and a transmitted wave

−ω = ikz(2 t ) Et(zt , ) Ee0 t xˆ (3.60)

7 1 −ω = ikz(2 t ) Bt(zt , ) Ee0 i yˆ , (3.61) v2 which continues on the right in medium (2). Note that the minus sign in Br is required by Eq.(3.21).

At z = 0, the combined fields on the left, Ei + Er and Bi + Br, must join the fields on the right, Et and Bt, in accordance with the boundary conditions. In this case there are no components perpendicular to the surface, so (3.52) and (3.53) are trivial. However, (3.54) and (3.55) require that + = E0i E 0 r E 0 t (3.62)

1()− = 1 E0i E 0 r E 0 t (3.63) v1 v 2 Eq. (3.63) can be rewritten as follows − = β E0i E 0 r E 0 t , (3.64) where v n β =1 = 2 . (3.65) v2n 1 Equations (3.62) and (3.64) are easily solved for the outgoing amplitudes, in terms of the incident amplitude: 1− β 2 E= EE, = E (3.66) 0r1+β 00 it 1 + β 0 i In terms of the indices of refraction this result is as follows n− n2 n E=1 2 EE, = 1 E . (3.67) 0r+ 00 it + 0 i nn12 nn 12 In order to calculate the fraction of energy which is transmitted and reflected we need to find the intensity ε ε (average power per unit area) which is given by eq.(3.39) with 0 replaced by : 1 I= ε v E 2 . (3.68) 2 0 The ratio of the reflected intensity to the incident intensity is therefore

2 2 IE n− n R ≡r =0r = 1 2 , (3.69) + Ii E0 i nn 1 2 Whereas the ration of the transmitted intensity to the incident intensity is

2 Iε v E 4 n n T ≡t =220 r = 12 , (3.70) Iε v E ()+ 2 i1 1 0 i n1 n 2 R is called the reflection coefficient and T the transmission coefficient ; they measure the fraction of the incident energy that is reflected and transmitted, respectively. Notice that R+ T = 1, (3.71) as conservation of energy, of course, requires. For instance, when light passes from air ( n1 = 1) into (n2 = 1.5), R = 0.04 and T = 0.96. Not surprisingly, most of the light is transmitted.

8 Reflection and Transmission at Oblique Incidence Now we consider the more general case of oblique incidence, in which the incoming wave meets the θ boundary at an arbitrary angle i (Fig.6). The normal incidence considered in the previous section is θ = really just a special case of oblique incidence, with i 0 .

Fig.6

Suppose, then, that a monochromatic plane wave

⋅ − ω 1 =i(ki r t ) =ˆ × ErEi(,)t0 i e ; Br i (,) t () kE ii (3.72) v1 approaches from the left, giving rise to a reflected wave,

⋅ − ω 1 =i(kr r t ) =ˆ × ErEr(,)t0 r e ; Br r (,) t () kE rr (3.73) v1 and a transmitted wave

⋅ − ω 1 =i(kt r t ) =ˆ × ErEt(,)t0 t e ; Br t (,) t () kE tt . (3.74) v2 All three waves have the same frequency ω that is determined once and for all at the source. The three wave numbers are related so that v n ===ω ==2 = 1 kirtv1 k v 1 k v 2 orkk irtt k k . (3.75) v1n 2

The combined fields in medium (1), Ei + Er and B i + Br, must now be joined to the fields Et and Bt in medium (2), using the boundary conditions (3.52) - (3.55). These all share the generic structure

⋅ −ω ⋅ − ω ⋅ − ω ( )eit()ki r +( ) ei(kr r t ) = ( ) e it ()kt r (3.76) The important thing to notice is that the x, y, and t dependence is confined to the exponents. Because the boundary conditions must hold at all points on the plane, and for all times, these exponential factors must be equal at z = 0. The time factors are already equal. As for the spatial terms, evidently when z = 0 ⋅= ⋅= ⋅ kri k r r kr t (3.77) or, more explicitly, + = + = + ()kxkykxkykxkyix () iy () rx () ry () tx () ty (3.78) for all x and all y. Eq.(3.78) can only hold if the components are separately equal, for if x = 0, we get

9 = = ()kiy () k ry () k ty (3.79) while y = 0 gives = = ()kix () k rx () k tx (3.80) = = = We can always choose axes in such a way that ki lies in the xz plane ()kiy () k ry () k ty 0 . Eq. (3.79) leads to First Law : The incident, reflected, and transmitted wave vectors form a plane (called the plane of incidence ), which also includes the normal to the surface (here, the z axis). Meanwhile, Eq.(3.80) implies that θ= θ = θ kisin ir k sin rt k sin t (3.81) θ θ θ where i is the angle of incidence , r is the angle of reflection , and t is the angle of transmission, more commonly known as the angle of refraction , all of them measured with respect to the normal (Fig. 6). In view of Eq. (3.81), then, Second Law : The angle of incidence is equal to the angle of reflection, θ= θ i r (3.82) This is the law of reflection . As for the transmitted angle, there is Third Law : sin θ n t = 1 (3.83) θ sin i n2 This is the law of refraction or Snell's law . These are the three fundamental laws of geometrical optics. Now that we have taken care of the exponential factors – they cancel, given Eq. (3.77) – the boundary conditions (3.52) – (3.55) become: ε(E+ E) = ε ( E ) 10i 0 rz 20 t z (3.84) (B+ B) = ( B ) 0i 0 rz 0 t z (3.85) (E+ E) = ( E ) 0i 0 rxy, 0 t xy , (3.86) (B+ B) = ( B ) 0i 0 rxy, 0 t xy , (3.87) 1 where Br( ,t ) =() kEˆ × in each case. The last two represent pairs of equations, one for the x- 0v 0 component and one for the y-component. Suppose that the polarization of the incident wave is parallel to the plane of incidence (the xz plane in Fig. 7); then the reflected and transmitted waves are also polarized in this plane. Then Eq. (3.84) reads ε(− θ + θε) =−( θ ) 10Eiirrsin E 0 sin 20 E tt sin (3.88) Eq. (3.85) adds nothing (0 = 0), since the magnetic fields have no z components; Eq. (3.86) becomes θ+ θ = θ E0icos ir E 0 cos r E 0 t cos t ; (3.89) and Eq. (3.87) gives 1()− = 1 E0i E 0 r E 0 t ; (3.90) v1 v 2

10 Fig. 7

Given the laws of reflection and refraction [(3.82) and (3.83)], Eqs. (3.88) and (3.90) reduce to − = β E0i E 0 r E 0 t , (3.91) where as before v n β =1 = 2 . (3.92) v2n 1 Eq. (3.89) says + = α E0i E 0 r E 0 t , (3.93) where cos θ α = t . (3.94) θ cos i Solving Eqs. (3.91) and (3.93) for the reflected and transmitted amplitudes, we obtain α− β 2 E= EE; = E . (3.95) 0rαβ+ 00 it αβ + 0 i

These are known as Fresnel's equations , for the case of polarization in the plane of incidence. There are two other , giving the reflected and transmitted amplitudes when the polarization is perpendicular to the plane of incidence. These equations you are asked to derive at home. Notice that the transmitted wave is always in phase with the incident one; the reflected wave is either in phase (“right side up"), if α> β , or 180° out of phase (“upside down”), if α< β . The amplitudes of the transmitted and reflected waves depend on the angle of incidence, because α is a θ function of i : 2 1− sin 2 θ 1−[] (n / n )sin θ α =t = 1 2 i ; (3.96) θ θ cosi cos i θ α θ In the case of normal incidence ( i = 0), = 1, and we recover Eq. (3.67). At grazing incidence ( i = α θ 90°), diverges, and the wave is totally reflected. Interestingly, there is an intermediate angle, B (called Brewster's angle ), at which the reflected wave is completely extinguished. According to Eq. (3.95), this occurs when α = β or β 2 sin 2 θ = ; (3.97) B 1+ β 2 or equivalently

11 n θ= β = 2 tan B ; (3.98) n1 θ Figure 8 shows a plot of the transmitted and reflected amplitudes as functions of i , for light incident on glass (n2 = 1.5) from air (n1 = 1). On the graph, a negative number indicates that the wave is 180° out of phase with the incident beam – the amplitude itself is the absolute value.

Fig.8 If the wave is polarized perpendicular to the plane of incidence there is no Brewster's angle for any n. Therefore, if a plane wave of mixed polarization is incident on a plane interface at the Brewster angle, the reflected radiation is completely plane-polarized with polarization vector perpendicular to the plane of incidence. This behavior can be utilized to produce beams of plane-polarized light but is not as efficient as other means employing anisotropic properties of some dielectric media. Even if the unpolarized wave is reflected at other than the Brewster angle, there is a tendency for the reflected wave to be pre- dominantly polarized perpendicular to the plane of incidence. The power per unit area striking the interface is S⋅ zˆ . Thus the incident intensity is 1 I= εv E 2 cos θ , (3.99) i2 1 1 0 i i while the reflected and transmitted intensities are 1 I= εv E 2 cos θ , (3.100) r2 1 1 0 r r 1 I= εv E 2 cos θ . (3.101) t2 2 2 0 t t The cosines are there because the intensities represent the average power per unit area of interface, and the interface is at an angle to the wave front. The reflection and transmission coefficients for waves polarized parallel to the plane of incidence are

2 2 I E α− β  R ≡r =0r = (3.102) α+ β  Ii E 0 i  

2 2 Iε v E cos θ 2  T ≡t =2 2 0 r t = αβ (3.103) ε θ αβ+  Ii1v 1 E 0 i cos i  

12 Fig.9 They are plotted as functions of the angle of incidence in Fig. 9 (for the air/glass interface). R is the fraction of the incident energy that is reflected – naturally, it goes to zero at Brewster's angle; T is the θ fraction transmitted – it goes to 1 at B . Note that R + T = 1, as required by conservation of energy: the energy per unit time reaching a particular patch of area on the surface is equal to the energy per unit time leaving the patch. There is another important phenomenon which is called total internal reflection. The word "internal" implies that the incident and reflected waves are in a medium of larger index of refraction than the θ> θ refracted wave (n1 > n 2). Snell's law (7.36) shows that, if n1 > n 2, then t i . Consequently, there is a θ= θ 0 θ= π critical angle when i i at which t / 2 , n θ 0 = 2 sin i , (3.104) n1 i.e., the refracted wave is propagated parallel to the surface. There can be no energy flow across the θ> θ 0 surface. Hence at that angle of incidence there must be total reflection. What happens if i i ? To θ> θ 0 answer this we first note that, for i i , n sin θ sinθ=1 sin θ =i > 1 . (3.105) t i θ 0 n2 sin i θ This means that t is a complex angle with a purely imaginary cosine. 2 sin θ  cosθ =i i  − 1 (3.106) t θ 0 sin i  The meaning of these complex quantities becomes clear when we consider the propagation factor for the refracted wave: 2 θ sin θ  sin i −k z i  − 1 ik2 x 2  0  ⋅ ikx()sinθ+ z cos θ θ 0 sin θ eeikt r =2 t t = eesin i i  (3.107) θ> θ 0 This shows that, for i i , the refracted wave is propagated only parallel to the surface and is attenuated exponentially beyond the interface. The attenuation occurs within a very few of θ≅ θ 0 the boundary, except for i i .

13