Chapter 7. Plane Electromagnetic Waves and Wave Propagation

Chapter 7. Plane Electromagnetic Waves and Wave Propagation 7.1 Plane Monochromatic Waves in Nonconducting Media One of the most important consequences of the Maxwell equations is the equations for electromagnetic wave propagation in a linear medium. In the absence of free charge and current densities the Maxwell equations are 𝛁⋅𝐃=0 𝛁⋅𝐁=0

𝜕𝐃 𝜕𝑡 𝜕𝐁 𝛁×𝐄= − 𝜕𝑡 𝛁×𝐇=

(7.1)

The wave equations for 𝐄 and 𝐁 are derived by taking the curl of 𝛁 × 𝐇 and 𝛁 × 𝐄 𝜕𝐁 𝜕𝑡 𝜕𝐃 𝛁×𝛁×𝐇=𝛁× 𝜕𝑡

𝛁 × 𝛁 × 𝐄 = −𝛁 ×

(7.2)

For uniform isotropic linear media we have 𝐃 = 𝜀𝐄 and 𝐁 = 𝜇𝐇, where 𝜀 and 𝜇 are in general complex functions of frequency 𝜔. Then we obtain 𝜕 2𝐄 𝜕𝑡 2 𝜕 2𝐁 𝛁 × 𝛁 × 𝐁 = −𝜀𝜇 2 𝜕𝑡 𝛁 × 𝛁 × 𝐄 = −𝜀𝜇

(7.3)

Since 𝛁 × 𝛁 × 𝐄 = 𝛁(𝛁 ⋅ 𝐄) − 𝛁 𝟐 𝐄 = −𝛁 𝟐 𝐄 and, similarly, 𝛁 × 𝛁 × 𝐁 = −𝛁 𝟐 𝐁, 𝜕 2𝐄 𝜕𝑡 2 𝜕 2𝐁 𝛁 2 𝐁 = 𝜀𝜇 2 𝜕𝑡 𝛁 2 𝐄 = 𝜀𝜇

(7.4)

Monochromatic waves may be described as waves that are characterized by a single frequency. Assuming the fields with harmonic time dependence 𝑒 −𝑖𝜔𝑡 , so that 𝐄(𝐱, 𝑡) = 𝐄(𝐱)𝑒 −𝑖𝜔𝑡 and 𝐁(𝐱, 𝑡) = 𝐁(𝐱)𝑒 −𝑖𝜔𝑡 we get the Helmholtz wave equations 𝛁 2 𝐄 + 𝜀𝜇𝜔2 𝐄 = 0 𝛁 2 𝐁 + 𝜀𝜇𝜔 2 𝐁 = 0

1

(7.5)

Plane waves in vacuum Suppose that the medium is vacuum, so that 𝜀 = 𝜀0 and 𝜇 = 𝜇0 . Further, suppose 𝐄(𝐱) varies in only one dimension, say the 𝑧-direction, and is independent of 𝑥 and 𝑦. Then Eq. 7.5 becomes 𝑑 2 𝐄(z) (7.6) + 𝑘 2 𝐄(𝑧) = 0 2 𝑑𝑧 where the wave number 𝑘 = 𝜔/𝑐. This equation is mathematically the same as the harmonic oscillator equation and has solutions (7.7)

𝐄𝑘 (𝑧) = 𝓔𝑒 ±𝑖𝑘𝑧 where 𝓔 is a constant vector. Therefore, the full solution is 𝑧

𝐄𝑘 (𝑧, 𝑡) = 𝓔𝑒 ±𝑖𝑘𝑧−𝑖𝜔𝑡 = 𝓔𝑒 −𝑖𝜔(𝑡∓𝑐 )

(7.8)

This represents a sinusoidal wave traveling to the right or left in the 𝑧-direciton with the speed of light 𝑐. Using the Fourier superposition theorem, we can construct a general solution of the form 𝐄(𝑧, 𝑡) = 𝐅(𝑧 − 𝑐𝑡) + 𝐆(𝑧 + 𝑐𝑡) (7.9) Plane waves in a nonconducting, nonmagnetic dielectric In a nonmagnetic dielectric, we have 𝜇 = 𝜇0 and the index of refraction 𝑛(𝜔) = √

𝜀(𝜔) 𝜀0

(7.10)

We see that the results are the same as in vacuum, except that the velocity of wave propagation or the phase velocity is now 𝑣 = 𝑐/𝑛 instead of 𝑐. Then the wave number is 𝑘(𝜔) = 𝑛(𝜔)

𝜔 𝑐

(7.11)

Electromagnetic plane wave of frequency 𝜔 and wave vector Suppose an electromagnetic plane wave with direction of propagation 𝐧 to be constructed, 𝐧 where is a unit vector. Then the variable 𝑧 in the exponent must be replaced by 𝐧 ⋅ 𝐱, the projection of 𝐱 in the 𝐧 direction. Thus an electromagnetic plane wave with direction of propagation 𝐧 is described by 𝐄(𝐱, 𝑡) = 𝓔𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 = 𝓔𝑒 𝑖𝑘𝐧⋅𝐱−𝑖𝜔𝑡 (7.12) 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 𝑖𝑘𝐧⋅𝐱−𝑖𝜔𝑡 𝐁(𝐱, 𝑡) = 𝓑𝑒 = 𝓑𝑒 where 𝓔 and 𝓑 are complex constant vector amplitudes of the plane wave. 𝐄 and 𝐁 satisfy the wave equations (Eq. 7.5), therefore the dispersion relation is given as 𝜔 2 𝜔 (7.13) 𝑘 2 = 𝜀𝜇𝜔2 = (𝑛 ) → 𝑘=𝑛 𝑐 𝑐 2

Let us substitute the plane wave solutions (Eq. 7.12) into the Maxwell equations. This substitution will impose conditions on the constants, 𝐤, 𝓔 and 𝓑, for the plane wave functions to be solutions of the Maxwell equations. For the plane waves, one sees that the operators 𝜕 = −𝑖𝜔, 𝜕𝑡

𝛁 = 𝑖𝐤

Thus the Maxwell equations become 𝛁⋅𝐃=0 𝛁⋅𝐁=0

𝜕𝐃 𝜕𝑡 𝜕𝐁 𝛁×𝐄=− 𝜕𝑡

(7.14)

𝛁×𝐇 =

→

𝐤 ⋅ 𝓔 = 0 𝐤 × 𝓑 = −𝜀𝜇𝜔𝓔 𝐤⋅𝓑=0 𝐤 × 𝓔 = 𝜔𝓑

where 𝐤 = 𝑘𝐧. The direction 𝐧 and frequency 𝜔 are completely arbitrary. The divergence equations demand that 𝐧 ⋅ 𝓔 = 0 and 𝐧 ⋅ 𝓑 = 0 (7.15) This means that 𝐄 and 𝐁 are both perpendicular to the direction of propagation 𝐧. The magnitude of 𝐤 is determined by the refractive index of the material 𝑘=𝑛

𝜔 𝑐

(7.16)

Then 𝓑 is completely determined in magnitude and direction 𝓑 = √𝜀𝜇 𝐧 × 𝓔 =

𝑛 𝐧×𝓔 𝑐

(7.17)

Note that in vacuum (𝑛 = 1), 𝐸 = 𝑐𝐵 in SI units. The phase velocity of the wave is 𝑣 = 𝑐/𝑛. Energy density and flux The time averaged energy density (see Eq. 6.94) is 𝑢=

1 1 1 (𝐄 ⋅ 𝐃∗ + 𝐁 ⋅ 𝐇 ∗ ) = (𝜀𝐄 ⋅ 𝐄∗ + 𝐁 ⋅ 𝐁 ∗ ) 4 4 𝜇

This gives 𝑢=

𝜀 2 1 2 2 |𝓔| = 𝑛 |𝓔| 2 2

(7.18)

The time averaged energy flux is given by the real part of the complex Poynting vector 𝐒=

1 (𝐄 × 𝐇 ∗ ) 2

Thus the energy flow is 𝐒=

1 𝜀 2 1 √ |𝓔| 𝐧 = 𝑛2 |𝓔|2 ⋅ 𝑣𝐧 = 𝑢𝐯 2 𝜇 2

3

(7.19)

7.2 Polarization and Stokes Parameters There is more to be said about the complex vector amplitudes 𝓔 and 𝓑. We introduce a righthanded set of orthogonal unit vectors (𝛜𝟏 , 𝛜𝟐 , 𝐧), as shown in Fig. 7.1, where we take 𝐧 to be the propagation direction of the plane wave. In general, the electric field amplitude 𝓔 can be written as 𝓔 = 𝛜𝟏 𝐸1 + 𝛜𝟐 𝐸2 (7.20) where the amplitudes 𝐸1 and 𝐸2 are arbitrary complex numbers. The two plane waves 𝐄𝟏 = 𝛜𝟏 𝐸1 𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 𝐄𝟐 = 𝛜𝟐 𝐸2 𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 with

𝑛 𝐁𝟏 = 𝛜𝟐 𝑐 𝐸1 𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 𝑛 𝐁𝟐 = −𝛜𝟏 𝑐 𝐸2 𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡

(7.21)

(7.22)

(if the index of refraction 𝑛 is real, 𝓔 and 𝓑 have the same phase) are said to be linearly polarized with polarization vectors 𝛜𝟏 and 𝛜𝟐 . Thus the most general homogeneous plane wave propagating in the direction 𝐤 = 𝑘𝐧 is expressed as the superposition of two independent plane waves of linear polarization: 𝐄(𝐱, 𝑡) = 𝓔𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 = (𝛜𝟏 𝐸1 + 𝛜𝟐 𝐸2 )𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡

(7.23)

Fig 7.1

It is convenient to express the complex components in polar form. Let 𝐸2 = 𝑎2 𝑒 𝑖𝛿2

(7.24)

𝐸1 𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 = 𝑎1 𝑒 𝑖(𝐤⋅𝐱−𝜔𝑡+𝛿1 ) ;

(7.25)

𝐸1 = 𝑎1 𝑒 𝑖𝛿1 , Then, for example,

that is, 𝛿1 is the phase of the 𝐸-field component in the 𝛜𝟏 -direction. It is no restriction to let 𝛿 = 𝛿2 − 𝛿1 , 4

𝛿1 = 0

(7.26)

since 𝛿1 = 0 merely dictates a certain choice of the origin of 𝑡. With this choice, 𝐄(𝐱, 𝑡) = 𝛜𝟏 𝑎1 𝑒 𝑖(𝐤⋅𝐱−𝜔𝑡) + 𝛜𝟐 𝑎2 𝑒 𝑖(𝐤⋅𝐱−𝜔𝑡+𝛿)

(7.27)

𝐄𝑟 (𝐱, 𝑡) = 𝛜𝟏 𝑎1 cos(𝐤 ⋅ 𝐱 − 𝜔𝑡) + 𝛜𝟐 𝑎2 cos(𝐤 ⋅ 𝐱 − 𝜔𝑡 + 𝛿)

(7.28)

or the real part is The 𝐸-field is resolved into components in two directions, with real amplitudes 𝑎1 and 𝑎2 , which may have any values. In addition the two components may be oscillating out of phase by 𝛿, that is, at any given point 𝐱, the maximum of 𝐸 in the 𝛜𝟏 -direction may be attained at a different time from the maximum of 𝐸 in the 𝛜𝟐 -direction. Polarization A detailed picture of the oscillating 𝐸-field at a certain point, e.g., 𝐱 = 0, is best seen by considering some special cases. 𝐄(0, 𝑡) = 𝛜𝟏 𝑎1 𝑒 −𝑖𝜔𝑡 + 𝛜𝟐 𝑎2 𝑒 −𝑖(𝜔𝑡−𝛿) (7.29) or 𝐄𝑟 (0, 𝑡) = 𝛜𝟏 𝑎1 cos 𝜔𝑡 + 𝛜𝟐 𝑎2 cos(𝜔𝑡 − 𝛿) Linearly polarized wave If 𝐸1 and 𝐸2 have the same phase, i.e., 𝛿 = 0, 𝐄(0, 𝑡) = (𝛜𝟏 𝑎1 + 𝛜𝟐 𝑎2 )𝑒 −𝑖𝜔𝑡 or 𝐄𝑟 (0, 𝑡) = (𝛜𝟏 𝑎1 + 𝛜𝟐 𝑎2 ) cos 𝜔𝑡

(7.30)

represents a linearly polarized wave, with its polarization vector 𝛜 = 𝛜𝟏 cos 𝜃 + 𝛜𝟐 sin 𝜃 with 𝜃 = tan−1 (𝑎2 /𝑎1 ) and a magnitude 𝐸 = √𝑎12 + 𝑎22 , as shown in Fig. 7.2.

Fig 7.2 𝑬-field of a linearly polarized wave

If 𝑎1 = 0 or 𝑎2 = 0, we also have linear polarization. For 𝛿 = 𝜋, 𝐄(0, 𝑡) = (𝛜𝟏 𝑎1 − 𝛜𝟐 𝑎2 )𝑒 −𝑖𝜔𝑡 or 𝐄𝑟 (0, 𝑡) = (𝛜𝟏 𝑎1 − 𝛜𝟐 𝑎2 ) cos 𝜔𝑡 is again linearly polarized.

5

(7.31)

Elliptically polarized wave If 𝐸1 and 𝐸2 have different phases, the wave of Eq. 7.27 is elliptically polarized. The simplest case is circular polarization. Then 𝑎1 = 𝑎2 and 𝛿 = ±𝜋/2: (7.32)

𝐄(0, 𝑡) = 𝑎1 (𝛜𝟏 ± 𝑖𝛜𝟐 )𝑒 −𝑖𝜔𝑡 or 𝐄𝑟 (0, 𝑡) = 𝛜𝟏 𝑎1 cos 𝜔𝑡 ± 𝛜𝟐 𝑎1 sin 𝜔𝑡

At a fixed point in space, the fields are such that the electric vector is constant in magnitude, but sweeps around in a circle at a frequency 𝜔, as shown in Fig. 7.3. For 𝛿 = +𝜋/2, 𝛜+ = 𝟏 √𝟐

(𝛜𝟏 + 𝑖𝛜𝟐 ), the tip of the 𝐸-vector traces the circular path counterclockwise. This wave is

called left circularly polarized (positive helicity) in optics. For 𝛿 = −𝜋/2, 𝛜− =

𝟏 √𝟐

(𝛜𝟏 − 𝑖𝛜𝟐 ),

same path but traced clockwise, then the wave is called right circularly polarized (negative helicity). For other values of 𝛿, we have elliptical polarization for the trace being an ellipse.

Fig 7.3 Trace of the tip of the 𝑬-vector (𝒂𝟏 = 𝒂𝟐 ) at a given point in space as a function of time. The propagation direction is point toward us. The traces for 𝜹 = 𝟎 and 𝝅 are linearly polarized. The traces for 𝜹 = 𝝅/𝟐 and −𝝅/𝟐 are left and right circularly polarized, respectively.

Stokes Parameters The two circularly polarized waves form a basis set for a general state of polarization. We introduce the complex orthogonal unit vectors: 𝟏 (7.33) (𝛜𝟏 ± 𝑖𝛜𝟐 ) 𝛜± = √𝟐 6

They satisfy the orthonormal conditions,

𝛜∗± ⋅ 𝛜∓ = 0 (7.34) { 𝛜∗± ⋅ 𝛜3 = 0 𝛜∗± ⋅ 𝛜± = 1 Then the most general homogeneous plane wave propagating in the direction 𝐤 = 𝑘𝐧 (Eq. 7.23) can be expressed as the superposition of two circularly polarized waves: 𝐄(𝐱, 𝑡) = (𝛜𝟏 𝐸1 + 𝛜𝟐 𝐸2 )𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 = (𝛜+ 𝐸+ + 𝛜− 𝐸− )𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡

(7.35)

where 𝐸+ and 𝐸− are complex amplitudes.

Fig 7.4 Electric field for an elliptically polarized wave.

When the ratio of the amplitudes is expressed as (7.36) 𝐸− = 𝑟𝑒 𝑖𝛼 𝐸+ the trace of the tip of the 𝐸-vector is an ellipse as shown in Fig. 7.4. For 𝛼 = 0, the ratio of semimajor to semiminor axis is |(1 + 𝑟)/(1 − 𝑟)|. Stokes parameters The polarization state of the general plane wave (Eq. 7.35) 𝐄(𝐱, 𝑡) = (𝛜𝟏 𝐸1 + 𝛜𝟐 𝐸2 )𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 = (𝛜+ 𝐸+ + 𝛜− 𝐸− )𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡

(7.37)

can be expressed by either (𝐸1 , 𝐸2 ) or (𝐸+ , 𝐸− ). We can determine these complex coefficients using Stokes parameters obtained by intensity measurements using polarizers and wave plates. We express the complex components in polar form: 𝐸1 = 𝑎1 𝑒 𝑖𝛿1 𝐸+ = 𝑎+ 𝑒 𝑖𝛿+ , (7.38) 𝐸2 = 𝑎2 𝑒 𝑖𝛿2 𝐸− = 𝑎− 𝑒 𝑖𝛿− The Stokes parameters of the linear polarization basis 𝑠0 = |𝐸1 |2 + |𝐸2 |2 = 𝑎12 + 𝑎22 𝑠1 = |𝐸1 |2 − |𝐸2 |2 = 𝑎12 − 𝑎22 𝑠2 = 2Re[𝐸1∗ 𝐸2 ] = 2𝑎1 𝑎2 cos(𝛿2 − 𝛿1 ) 𝑠3 = 2Im[𝐸1∗ 𝐸2 ] = 2𝑎1 𝑎2 sin(𝛿2 − 𝛿1 )

7

(7.39)

and of the circular polarization basis 𝑠0 = |𝐸+ |2 + |𝐸− |2 = 𝑎+2 + 𝑎−2 𝑠1 = 2Re[𝐸+∗ 𝐸− ] = 2𝑎+ 𝑎− cos(𝛿− − 𝛿+ ) 𝑠2 = 2Im[𝐸+∗ 𝐸− ] = 2𝑎+ 𝑎− sin(𝛿− − 𝛿+ ) 𝑠3 = |𝐸+ |2 − |𝐸− |2 = 𝑎+2 − 𝑎−2

(7.40)

The four parameters are not independent and satisfy the relation 𝑠02 = 𝑠12 + 𝑠22 + 𝑠32

(7.41)

7.3 Plane Monochromatic Waves In Conducting Media In a conducting medium there is an induced current density in response to the 𝐸-field of the wave. The current density J is linearly proportional to the electric field (Ohm’s law, Eq. 5.21): 𝐉 = 𝜎𝐄 The constant of proportionality 𝜎 is called the conductivity. For an electromagnetic plane wave with direction of propagation 𝐧 (Eq. 7.12) described by 𝐄(𝐱, 𝑡) = 𝓔𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 = 𝓔𝑒 𝑖𝑘𝐧⋅𝐱−𝑖𝜔𝑡 𝐁(𝐱, 𝑡) = 𝓑𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 = 𝓑𝑒 𝑖𝑘𝐧⋅𝐱−𝑖𝜔𝑡 the Maxwell equation 𝛁×𝐇 = becomes

𝜕𝐃 + 𝜎𝐄 𝜕𝑡

𝜎 𝜔 𝐤 × 𝓑 = −𝜔𝜇 (𝜀 + 𝑖 ) 𝓔 = − 2 𝜀𝑟 (𝜔)𝓔 𝜔 𝑐

(7.42)

(7.43)

where we define a complex dielectric constant 𝜀𝑟 =

1 𝜎 (𝜀 + 𝑖 ) 𝜀0 𝜔

(7.44)

Comparing Eq. 7.44 with Eq. 7.14, we can see that the transverse dispersion relation results in 𝑘 = √𝜀𝑟

𝜔 𝜔 =𝑛 𝑐 𝑐

(7.45)

where we define a complex refractive index 𝑛 = 𝑛𝑟 + 𝑖𝑛𝑖 = √𝜀𝑟

(7.46)

To interpret the wave propagation in the conducting medium, it is useful to express the complex propagation vector 𝐤 as (7.47) 𝐤 = 𝐤 + 𝑖𝐤 𝑟

8

𝑖

Then the plane wave is expressed as 𝐄(𝐱, 𝑡) = 𝓔𝑒 𝑖𝐤⋅𝐱−𝑖𝜔𝑡 = (𝓔𝑒 −𝐤𝑖 ⋅𝐱 )𝑒 𝑖(𝐤𝑟 ⋅𝐱−𝜔𝑡)

(7.48)

This is a plane wave propagating in the direction 𝐤 𝑟 with wavelength 𝜆 = 2𝜋/𝑘𝑟 ; but it decreases in amplitude, most rapidly in the direction 𝐤 𝑖 .

7.4 Reflection and Refraction of Electromagnetic Waves at a Plane Interface between Dielectrics Normal Incidence We begin with the simplest possible case: a plane wave normally incident on a plane dielectric interface. We will see that the boundary conditions are satisfied only if reflected and transmitted waves are present.

Fig 7.5 Reflection and transmission at normal incidence

Fig. 7.5 describes the incident wave (𝐄, 𝐁) travelling in the z-direction, the reflected wave (𝐄′′ , 𝐁 ′′ ) travelling in the minus z-direction, and the transmitted wave (𝐄′ , 𝐁 ′ ) travelling in the zdirection. The interface is taken as coincident with the 𝑥𝑦-plane at 𝑧 = 0, with two dielectric media with the indices of refraction, 𝑛 for 𝑧 < 0 and 𝑛′ for 𝑧 > 0. The electric fields, which are assumed to be linearly polarized in the 𝑥-direction, are described by (7.49)

𝐄 = 𝐞𝑥 𝐸𝑒 𝑖(𝑘𝑧−𝜔𝑡) ′

{ 𝐄′ = 𝐞𝑥 𝐸 ′ 𝑒 𝑖(𝑘 𝑧−𝜔𝑡) 𝐄′′ = −𝐞𝑥 𝐸 ′′ 𝑒 𝑖(−𝑘𝑧−𝜔𝑡) where 𝑘=𝑛

𝜔 , 𝑐

From Eq. 7.17, 𝓑=

𝑘 ′ = 𝑛′ 𝑛 𝐤×𝓔 𝑐𝑘

9

𝜔 𝑐

(7.50)

Therefore, the magnetic fields associated with the electric fields of Eq. 7.49 are given by 𝑐𝐁 = 𝐞𝑦 𝑛𝐸𝑒 𝑖(𝑘𝑧−𝜔𝑡) { 𝑐𝐁 ′ = 𝐞𝑦 𝑛′𝐸 ′ 𝑒 𝑖(𝑘

′ 𝑧−𝜔𝑡)

(7.51)

𝑐𝐁 ′′ = 𝐞𝑦 𝑛𝐸 ′′ 𝑒 𝑖(−𝑘𝑧−𝜔𝑡) Clearly the reflected and transmitted waves must have the same frequency 𝜔 as the incident wave if boundary conditions at 𝑧 = 0 are to be satisfied for all 𝑡. The 𝐸-field must be continuous at the boundary, (7.52) 𝐸 − 𝐸 ′′ = 𝐸 ′ The 𝐻-field must also be continuous, and for nonmagnetic media (𝜇 = 𝜇 ′ = 𝜇0 ), so must be the 𝐵-field: (7.53) 𝑛(𝐸 + 𝐸 ′′ ) = 𝑛′ 𝐸 ′ ′ ′′ Eqs. 7.52 and 7.53 can be solved simultaneously for the amplitudes 𝐸 and 𝐸 in terms of the incident amplitude 𝐸: 𝑛′ − 𝑛 2𝑛 (7.54) ′′ 𝐸 = ′ 𝐸, 𝐸′ = ′ 𝐸 𝑛 +𝑛 𝑛 +𝑛 The Fresnel coefficients for normal incidence reflection and transmission are defined as 𝑟=

𝐸 ′′ 𝑛′ − 𝑛 = ′ , 𝐸 𝑛 +𝑛

𝑡=

𝐸′ 2𝑛 = ′ 𝐸 𝑛 +𝑛

(7.55)

For 𝑛′ > 𝑛, there is a phase reversion for the reflected wave. What is usually measureable is the reflected and transmitted average energy fluxes per unit area (a.k.a., the intensity of EM wave) given by the magnitude of the Poynting vector (7.56) 1 1 ∗ 2 𝑆 = |𝐄 × 𝐇 | = 𝑛𝑐𝜀0 |𝐸| 2 2 We define the reflectance 𝑅 and the transmittance 𝑇 for normal incidence by the ratios of the intensities 2

𝑆 ′′ 𝑛′ − 𝑛 𝑅= = |𝑟|2 = ( ′ ) , 𝑆 𝑛 +𝑛

𝑇=

𝑆 ′ 𝑛′ 2 4𝑛𝑛′ = |𝑡| = ′ (𝑛 + 𝑛)2 𝑆 𝑛

(7.57)

With the Fresnel coefficients given by Eq. 7.55, 𝑅 and 𝑇 satisfy 𝑅+𝑇 =1

(7.58)

for any pair of nonconducting media. This is an expression of energy conservation at the interface.

10

Oblique incidence We consider reflection and refraction at the boundary of two dielectric media at oblique incidence. The discussion will lead to three well-known optical laws: Snell’s law, the law of reflection, and Brewster’s law governing polarization by reflection. Fig. 7.6 depicts the situation that the wave vectors, 𝐤, 𝐤 ′ , and 𝐤 ′′ , are coplanar and lie in the 𝑥𝑧-plane. The media for 𝑧 < 0 and 𝑧 > 0 have the indices of refraction, 𝑛 and 𝑛′ , respectively. The unit normal to the boundary is 𝐧. The plane defined by 𝐤 and 𝐧 is called the plane of incidence, and its normal is in the direction of 𝐤 × 𝐧.

Fig 7.6 Reflection and transmission at oblique incidence. Incident wave 𝐤 strikes plane interface between different media, giving rise to a reflected wave 𝐤 ′′ and refracted wave 𝐤 ′ .

The three plane waves are: Incident 𝐄 = 𝐄0 𝑒 𝑖(𝐤⋅𝐱−𝜔𝑡) 𝑛 𝐁= 𝐤×𝐄 𝑐𝑘

(7.59)

Refracted ′

𝐄′ = 𝐄0′ 𝑒 𝑖(𝐤 ⋅𝐱−𝜔𝑡) 𝑛′ 𝐁′ = ′ 𝐤 ′ × 𝐄′ 𝑐𝑘

(7.60)

Reflected ′′

𝐄 = 𝐄0′′ 𝑒 𝑖(𝐤 ⋅𝐱−𝜔𝑡) 𝑛 ′′ 𝐁= 𝐤 × 𝐄′′ 𝑐𝑘 where 𝑘=𝑛

𝜔 , 𝑐

𝑘 ′ = 𝑛′

(7.61)

𝜔 𝑐

(7.62)

Phase matching on the boundary Not only must the refracted and reflected waves have the same frequency as the incident wave, but also the phases must match everywhere on the boundary to satisfy boundary conditions at all points on the plane at all times: (7.63) (𝐤 ⋅ 𝐱)𝑧=0 = (𝐤 ′ ⋅ 𝐱)𝑧=0 = (𝐤 ′′ ⋅ 𝐱)𝑧=0 11

This condition has three interesting consequences. Using the vector identity 𝐧 × (𝐧 × 𝐱) = (𝐧 ⋅ 𝐱)𝐧 − 𝐱 and 𝐧 ⋅ 𝐱 = 0 on the boundary, we obtain 𝐱 = −𝐧 × (𝐧 × 𝐱) We substitute this into Eq. 7.63, 𝐤 ⋅ 𝐱 = −𝐤 ⋅ [𝐧 × (𝐧 × 𝐱)] = −(𝐤 × 𝐧) ⋅ (𝐧 × 𝐱)

(7.64) (7.65) (7.66)

and similarly for the other members of Eq. 7.63. Since 𝐱 is an arbitrary vector on the boundary, Eq. 7.63 can hold if and only if (7.67) 𝐤 × 𝐧 = 𝐤 ′ × 𝐧 = 𝐤 ′′ × 𝐧 This implies that (i) All three vectors, 𝐤, 𝐤 ′ and 𝐤 ′′ , lie in a plane, i.e., 𝐤 ′ and 𝐤 ′′ lie in the plane of incidence; (ii) Law of reflection: |𝐤 × 𝐧| = |𝐤 ′′ × 𝐧| → 𝑘 sin 𝜃𝑖 = 𝑘 sin 𝜃𝑟 , thus (7.68) 𝜃𝑖 = 𝜃𝑟 (iii) Snell’s Law: |𝐤 × 𝐧| = |𝐤 ′ × 𝐧| → 𝑘 sin 𝜃𝑖 = 𝑘 ′ sin 𝜃𝑡 , thus (7.69) 𝑛 sin 𝜃𝑖 = 𝑛′ sin 𝜃𝑡 Boundary conditions and Fresnel coefficients At all points on the boundary, normal components of 𝐃 and 𝐁 and tangential components of 𝐄 and 𝐇 are continuous. The boundary conditions at 𝑧 = 0 are (i) (ii)

[𝜀(𝐄0 + 𝐄0′′ ) − 𝜀 ′ 𝐄0′ ] ⋅ 𝐧 = 0 [𝐤 × 𝐄0 + 𝐤 ′′ × 𝐄0′′ − 𝐤 ′ × 𝐄0′ ] ⋅ 𝐧 = 0

(7.70)

[𝐄0 + 𝐄0′′ − 𝐄0′ ] × 𝐧 = 0 1 1 (iv) [ (𝐤 × 𝐄0 + 𝐤 ′′ × 𝐄0′′ ) − ′ 𝐤 ′ × 𝐄0′ ] × 𝐧 = 0 𝜇 𝜇

(iii)

In applying the boundary conditions it is convenient to consider two separate situations: the incident plane wave is linearly polarized with its polarization vector (a) perpendicular (spolarization) and (b) parallel (p-polarization) to the plane of incidence (see Fig. 7.7). For simplicity, we assume the dielectrics are nonmagnetic (𝜇 = 𝜇 ′ = 𝜇0 ). (a) s-polarization The 𝐸-fields are normal to 𝐧, therefore (i) in Eq. 7.70 is automatically satisfied. (iii) and (iv) give 𝐸0 + 𝐸0′′ − 𝐸0′ = 0

(7.71)

𝑛(𝐸0 − 𝐸0′′ ) cos 𝜃𝑖 − 𝑛′ 𝐸0′ cos 𝜃𝑡 = 0

(7.72)

and while (ii), using Snell’s law, duplicates (iii). With Eqs. 7.71 and 7.72, we obtain the s-pol Fresnel coefficients, 12

𝐸0′ 2𝑛 cos 𝜃𝑖 2𝑛 cos 𝜃𝑖 𝑡𝑠 = = = 𝐸0 𝑛 cos 𝜃𝑖 + 𝑛′ cos 𝜃𝑡 𝑛 cos 𝜃𝑖 + √𝑛′2 − 𝑛2 sin2 𝜃𝑖

(7.73)

and 𝐸0′′ 𝑛 cos 𝜃𝑖 − 𝑛′ cos 𝜃𝑡 𝑛 cos 𝜃𝑖 − √𝑛′2 − 𝑛2 sin2 𝜃𝑖 𝑟𝑠 = = = 𝐸0 𝑛 cos 𝜃𝑖 + 𝑛′ cos 𝜃𝑡 𝑛 cos 𝜃𝑖 + √𝑛′2 − 𝑛2 sin2 𝜃𝑖

(7.74)

where, using Snell’s law, we could write cos 𝜃𝑡 = √1 − (𝑛/𝑛′)2 sin2 𝜃𝑖

(7.75)

(b) p-polarization The 𝐵-fields are normal to 𝐧, therefore (ii) in Eq. 7.70 is automatically satisfied. (iii) and (iv) give cos 𝜃𝑖 (𝐸0 − 𝐸0′′ ) − cos 𝜃𝑡 𝐸0′ = 0

(7.76)

𝑛(𝐸0 + 𝐸0′′ ) − 𝑛′ 𝐸0′ = 0

(7.77)

and while (i), using Snell’s law, duplicates (iv). With Eqs. 7.76 and 7.76, we obtain the p-pol Fresnel coefficients, 𝐸0′ 2𝑛 cos 𝜃𝑖 2𝑛𝑛′ cos 𝜃𝑖 (7.78) 𝑡𝑝 = = ′ = 2 𝐸0 𝑛 cos 𝜃𝑖 + 𝑛 cos 𝜃𝑡 𝑛′ cos 𝜃𝑖 + 𝑛√𝑛′2 − 𝑛2 sin2 𝜃𝑖 and 2

𝐸0′′ 𝑛′ cos 𝜃𝑖 − 𝑛 cos 𝜃𝑡 𝑛′ cos 𝜃𝑖 − 𝑛√𝑛′2 − 𝑛2 sin2 𝜃𝑖 𝑟𝑝 = = = 𝐸0 𝑛′ cos 𝜃𝑖 + 𝑛 cos 𝜃𝑡 𝑛′ 2 cos 𝜃𝑖 + 𝑛√𝑛′2 − 𝑛2 sin2 𝜃𝑖

(7.79)

For normal incidence, 𝑟𝑝 = −𝑟𝑠 = −(𝑛 − 𝑛′ )/(𝑛 + 𝑛′ ), because we assign opposite directions for 𝐄 and 𝐄′′ for p-polarization.

Fig 7.7 Reflection and refraction with polarization (a) perpendicular (s-polarization) and (b) parallel (ppolarization) to the plane of incidence 13

For certain purposes, it is more convenient to express the Fresnel coefficients in terms of the incident and refraction angles, 𝜃𝑖 and 𝜃𝑡 only. Using the Snell’s law, 𝑛 sin 𝜃𝑖 = 𝑛′ sin 𝜃𝑡 , we can write 2𝑛 cos 𝜃𝑖 2 cos 𝜃𝑖 2 cos 𝜃𝑖 sin 𝜃𝑡 𝑡𝑠 = = = ′ ′ 𝑛 𝑛 cos 𝜃𝑖 + 𝑛 cos 𝜃𝑡 cos 𝜃 + cos 𝜃 sin 𝜃𝑡 cos 𝜃𝑖 + cos 𝜃𝑡 sin 𝜃𝑖 𝑖 𝑡 𝑛 then 2 sin 𝜃𝑡 cos 𝜃𝑖 (7.80) 𝑡𝑠 = sin(𝜃𝑡 + 𝜃𝑖 ) Similarly, sin(𝜃𝑡 − 𝜃𝑖 ) (7.81) 𝑟𝑠 = sin(𝜃𝑡 + 𝜃𝑖 ) 𝑡𝑝 =

2 sin 𝜃𝑡 cos 𝜃𝑖 sin(𝜃𝑡 + 𝜃𝑖 ) cos(𝜃𝑡 − 𝜃𝑖 )

(7.82)

𝑟𝑝 =

tan(𝜃𝑡 − 𝜃𝑖 ) tan(𝜃𝑡 + 𝜃𝑖 )

(7.83)

and

Brewster’s angle and total internal reflection We next consider the dependence of 𝑅 and 𝑇 on the angle of incidence, using the Fresnel coefficients. Brewster angle We see that 𝑟𝑝 in Eq. 7.88 vanishes when 𝜃𝑡 + 𝜃𝑖 = 𝜋/2. Using Snell’s law, we can determine Brewster’s angle 𝜃𝐵 = 𝜃𝑖 at which the p-polarized reflected wave is zero: 𝜋 𝑛 sin 𝜃𝐵 = 𝑛′ sin ( − 𝜃𝐵 ) = 𝑛′ cos 𝜃𝐵 2 or 𝑛′ tan 𝜃𝐵 = 𝑛

(7.84)

Polarization at the Brewster angle is a practical means of producing polarized radiation. If a plane wave of mixed polarization is incident on a plane interface at the Brewster angle, the reflected radiation is completely s-polarized. The generally lower reflectance for p-polarized lights accounts for the usefulness of polarized sunglasses. Since most outdoor reflecting surfaces are horizontal, the plane of incidence for most reflected glare reaching the eyes is vertical. The polarized lenses are oriented to eliminate the strongly reflected s-component. Fig. 7.8 shows 𝑅𝑠 and 𝑅𝑝 as a function of 𝜃𝑖 with 𝑛 = 1 and 𝑛′ = 1.5, as for an air-glass interface. The Brewster angle is 𝜃𝐵 = 56∘ for this case. 14

Fig 7.8 Reflectance for s- and p-polarzation at an air-glass interface. Brewster’s angle is 𝜽𝑩 = 𝟓𝟔∘ 𝒏=𝟏 𝒏′ = 𝟏. 𝟓

Total internal reflection There is another case in which 𝑅𝑠 = 𝑅𝑝 = 1. Eqs. 7.74 and 7.79 indicates that perfect reflection occurs for 𝜃𝑡 = 𝜋/2. The incident angle for which 𝜃𝑡 = 𝜋/2 is called the critical angle, 𝜃𝑖 = 𝜃𝑐 . From Snell’s law 𝑛′ (7.85) sin 𝜃𝑐 = 𝑛 𝜃𝑐 can exist only if 𝑛 > 𝑛′, i.e., the incident and reflected waves are in a medium of larger index of refraction than the refracted wave.

Fig 7.9 Reflectance for s- and p-polarzation at an air-glass interface. Brewster’s angle is 𝜽𝑩 = 𝟑𝟒∘ and the critical angle is 𝜽𝒄 = 𝟒𝟐∘ 𝒏 = 𝟏. 𝟓 𝒏′ = 𝟏

15

For waves incident at 𝜃𝑐 , the refracted wave is propagated parallel to the surface. There can be no energy flow across the surface. Hence at that angle of incidence there must be total reflection. For incident angles greater than the critical angle 𝜃𝑖 > 𝜃𝑐 , Snell’s law gives sin 𝜃𝑡 =

𝑛 𝑛 sin 𝜃 > sin 𝜃𝑐 = 1 𝑖 𝑛′ 𝑛′

This means that 𝜃𝑡 is a complex angle with a purely imaginary cosine. sin 𝜃𝑖 2 √ cos 𝜃𝑡 = 𝑖 ( ) −1 sin 𝜃𝑐

(7.86)

Then Eqs. 7.74 and 7.79 indicates that 𝑟𝑠 and 𝑟𝑝 both take the form 𝑟= where 𝑎 and 𝑏 are real, therefore, 𝑅=

|𝑟|2

𝑎 − 𝑖𝑏 𝑎 + 𝑖𝑏

𝑎 − 𝑖𝑏 2 =| | =1 𝑎 + 𝑖𝑏

The result is that 𝑅𝑠 = 𝑅𝑝 = 1 for all 𝜃𝑖 > 𝜃𝑐 . This perfect reflection is called total internal reflection. The meaning of this total internal reflection becomes clear when we consider the propagation factor for the refracted wave: 𝑒 𝑖𝐤

′ ⋅𝐱

= 𝑒 𝑖𝑘

′ (𝑥 sin 𝜃 +𝑧 cos 𝜃 ) 𝑡 𝑡

𝑧 𝑖𝑘 ′ (sin 𝜃𝑖 )𝑥 sin 𝜃𝑐

= 𝑒− 𝛿𝑒

(7.87)

where 1 2𝜋𝑛 √sin2 𝜃𝑖 − sin2 𝜃𝑐 = −𝑖𝑘 ′ cos 𝜃𝑡 = 𝑘√sin2 𝜃𝑖 − sin2 𝜃𝑐 = 𝛿 𝜆

(7.88)

with the wavelength of the radiation 𝜆 in vacuum. This shows that, for 𝜃𝑖 > 𝜃𝑐 , the refracted wave is propagating only parallel to the surface and is attenuated exponentially beyond the interface. The attenuation occurs within a few wavelengths of the boundary except for 𝜃𝑖 ≈ 𝜃𝑐 . Goos-Hänchen effect An important side effect of total internal reflection is the propagation of an evanescent wave across the boundary surface. Essentially, even though the entire incident wave is reflected back into the originating medium, there is some penetration into the second medium at the boundary. The evanescent wave appears to travel along the boundary between the two materials. The penetration of the wave into the “forbidden” region is the physical origin of the Goos-Hänchen effect: If a beam of radiation having a finite transverse extent undergoes total internal reflection, the reflected beam emerges displaced laterally with respect to the prediction of a geometrical ray refected at the boundary.

16

Fig 7.10 Geometrical interpretation of the Goos-Hänchen effect, the lateral displacement of a totally internallyreflected beam of radiation because of the penetration of the evanescent wave into the region of smaller index of refraction.

Fig. 7.10 shows a geometrical interpretation of the Goos-Hänchen effect. We can estimate the displacement 𝐷 ≈ 2𝛿 sin 𝜃𝑖 . Rigorous calculation shows that 𝐷 depends on the polarization of the incident radiation: −1

𝐷𝑠 = 2𝛿 sin 𝜃𝑖 ,

sin2 𝜃𝑖 𝐷𝑝 = 𝐷𝑠 ( 2 − cos 2 𝜃𝑖 ) sin 𝜃𝑐

(7.89)

7.5 Frequency Dispersion in Materials How an EM wave propagates in a linear material medium is determined entirely by the optical constants, 𝑛𝑅 and 𝑛𝐼 , where the complex index of refraction is 𝑛 = 𝑛𝑅 + 𝑖𝑛𝐼 depending only on 𝜀 and 𝜇. In general, 𝜀(𝜔) and 𝜇(𝜔) depend on the frequency of the wave, varying widely in the range from d-c to x-rays. Dispersion is the phenomenon in which the phase velocity of a wave depends on its frequency. Media having such a property are termed dispersive media. The dispersion relation of an EM wave in a dispersive medium is expressed as 𝑘 = 𝑛(𝜔)

𝜔 𝜔 = 𝑐 𝑣𝑝ℎ (𝜔)

(7.90)

Drude-Lorentz harmonic oscillator model All ordinary matters are composed of electrons and nuclei. The bound electrons can be treated as harmonic oscillators. For generality we make it a damped harmonic oscillator. When an EM wave is present, the oscillator is driven by the electric field of the wave. The response of the medium is obtained by adding up the motions of the electrons. The equation of motion for an electron of charge −𝑒 and acted on by an electric field 𝐄(𝐱, 𝑡) is 𝑑2𝐱 𝑑𝐱 𝑒 2 + 𝛾 + 𝜔 𝐱 = − 𝐄(𝐱, 𝑡) 0 𝑑𝑡 2 𝑑𝑡 𝑚

(7.91)

where the damping constant 𝛾 has the dimensions of frequency. The amplitude of oscillation is small compared to the spatial variation of the field (e.g., size of atom ∼ 1 Å ≪ 𝜆𝑜𝑝𝑡 ∼ 500 − 700 nm). Assuming that the field varies harmonically in time with frequency 𝜔 as 𝑒 −𝑖𝜔𝑡 , the dipole moment contributed by one electron is 𝑒2 𝐄 𝐩 = −𝑒𝐱 = 2 𝑚 𝜔0 − 𝜔 2 − 𝑖𝜔𝛾 17

(7.92)

If there are 𝑁 molecules per unit volume with 𝑍 electrons per molecule, and there are 𝑍 electrons per molecule with binding frequency 𝜔𝑗 and damping constant 𝛾𝑖 , then the dielectric constant is given by 𝑓𝑗 𝜀(𝜔) 𝑁𝑒 2 (7.93) 𝜀𝑟 (𝜔) = = 1 + 𝜒𝑒 (𝜔) = 1 + ∑ 2 2 𝜀0 𝜀0 𝑚 𝜔𝑗 − 𝜔 − 𝑖𝜔𝛾𝑗 𝑗

where the oscillator strengths 𝑓𝑗 satisfy the sum rule, (7.94)

∑ 𝑓𝑗 = 𝑍 𝑗

Resonant absorption and anomalous dispersion In a dispersive medium (nonmagnetic), plane waves are expressed as 𝐄(𝑧, 𝑡) = 𝐄0 𝑒 𝑖𝑘(𝜔)𝑧−𝑖𝜔𝑡 with the complex wave number 𝑘(𝜔) = √𝜀𝜇0 𝜔 = 𝑛(𝜔)

𝜔 𝑐

(7.95)

(7.96)

Writing 𝑘 in terms of its real and imaginary parts,

𝛼 2 with the attenuation constant or absorption coefficient 𝛼, Eq. 7.95 becomes 𝑘 =𝛽+𝑖 1

𝐄(𝑧, 𝑡) = 𝐄0 𝑒 −2𝛼𝑧 𝑒 𝑖(𝛽𝑧−𝜔𝑡)

(7.97)

(7.98)

Evidently the wave is exponentially attenuated because the damping absorbs energy. The intensity of the wave ( ∝ |𝐄(𝑧, 𝑡)|2 ) falls off as 𝑒 −𝛼𝑧 . The relation between (𝛼, 𝛽) and 𝜀𝑟 is 𝛼 2 𝜔2 𝛽 2 − 4 = 2 Re 𝜀𝑟 𝑐 𝜔2 𝛽𝛼 = 2 Im 𝜀𝑟 𝑐

(7.99)

Fig 7.11 Real and imaginary parts of the dielectric constant in the neighborhood of a resonance. The region of anomalous dispersion is the frequency interval where absorption occurs.

18

The general features of the real and imaginary parts of 𝜀𝑟 (𝜔) around a resonant frequency are shown in Fig. 7.11. Most of the time Re 𝜀(𝜔) (or the index of refraction with small 𝛼) rises gradually with increasing frequency (normal dispersion). However, in the immediate neighborhood of a resonance Re 𝜀(𝜔) drops sharply. Because this behavior is atypical, it is called anomalous dispersion. Notice that the region of anomalous dispersion coincides with the region of maximum absorption. Drude model: Electric conductivity at low frequencies If the density of free electrons (i.e., 𝜔0 = 0 in Eq. 7.93) is 𝑁𝑒 , 𝑁𝑒 𝑒 2 𝜀(𝜔) = 𝜀𝑏 (𝜔) + 𝑖 𝑚𝜔(𝛾0 − 𝑖𝜔)

(7.100)

where 𝜀𝑏 (𝜔) is the contribution of the bound electrons. With the Ohm’s law 𝐉 = 𝜎𝐄 and 𝐃 = 𝜀𝑏 𝐄 where the fields are harmonic in terms of 𝑒 −𝑖𝜔𝑡 , the Maxwell-Ampere equation 𝛁×𝐇 =𝐉+ becomes

𝜕𝐃 𝜕𝑡

𝜎 𝛁 × 𝐇 = −𝑖𝜔 (𝜀𝑏 + 𝑖 ) 𝐄 𝜔

(7.101)

Comparing Eq. 7.101 with Eq. 7.100, we obtain an expression for the Drude conductivity: 𝜎(𝜔) =

𝑁𝑒 𝑒 2 𝜎0 = 𝑚(𝛾0 − 𝑖𝜔) 1 − 𝑖𝜔𝜏0

(7.102)

where the scattering time 𝜏0 =

1 𝛾0

(7.103)

and the d-c conductivity 𝑁𝑒 𝑒 2 𝜏0 𝜎0 = 𝜎(𝜔 = 0) = 𝑚

(7.104)

The scattering times of the common metals are on the order of 𝜏0 ∼ 10−14 s, thus 𝜎(𝜔) ≈ 𝜎0 for 𝜔 < 1012 Hz. High-frequency limit: plasma frequency At frequencies far above the highest resonant frequency Eq. 7.93 becomes 𝜀𝑟 (𝜔) ≈ 1 −

𝜔𝑝2 𝜔2

(7.105)

where the plasma frequency is defined as 𝜔𝑝2 =

𝑁𝑒 𝑒 2 𝜀0 𝑚

19

(7.106)

Some typical electron densities and plasma frequencies are listed below. 𝜔𝑝 (s −1 ) 𝑁𝑒 (m−3 ) Metal 1028 1016 24 Semiconductor (doped) 10 1014 Semiconductor (pure) 1020 1012 Ionosphere 107 1011 The dispersion relation is 𝑐𝑘 = √𝜀𝑟 (𝜔)𝜔 = √𝜔 2 − 𝜔𝑝2

(7.107)

𝜔2 = 𝜔𝑝2 + 𝑐 2 𝑘 2

(7.108)

or For 𝜔 < 𝜔𝑝 , 𝑘 is pure imaginary, therefore the light exponentially decays and penetrates only a very short distance in the medium. The plasma frequencies of common metals are in the UV, and hence the visible light is almost entirely reflected from metal surfaces and the metals suddenly become transparent in UV.

7.6 Wave Propagation in a Dispersive Medium Wave packet A wave packet or a pulsed electromagnetic wave is spatially and temporally localized.

Fig 7.12

Fourier integral transform From the basic solutions of Eq. 7.12 a plane wave takes the form 𝑢(𝑥, 𝑡) = 𝐴𝑒 𝑖𝑘𝑥−𝑖𝜔(𝑘)𝑡

(7.109)

and the superposition principle leads to the a general solution 𝑢(𝑥, 𝑡) =

1 √2𝜋

∞

∫ 𝐴(𝑘)𝑒 𝑖𝑘𝑥−𝑖𝜔(𝑘)𝑡 𝑑𝑘

(7.110)

−∞

The amplitude 𝐴(𝑘) is given by 𝐴(𝑘) =

1 √2𝜋

∞

∫ 𝑢(𝑥, 0)𝑒 −𝑖𝑘𝑥 𝑑𝑥 −∞

20

(7.111)

Form of the wave packets (i)

Square wave packet

Fig 7.13

The amplitude 𝐴(𝑘) of the normalized square wave shown in Fig. 7.13 is 𝐴(𝑘) =

1 √2𝜋

∞

1

−∞

√2𝜋𝑎

∫ 𝑢(𝑥)𝑒 −𝑖𝑘𝑥 𝑑𝑥 =

𝑎 2

∫ 𝑒 −𝑖𝑘𝑥 𝑑𝑥 −

𝑎 2

𝑘𝑎 𝑘𝑎 (7.112) 2 sin ( 2 ) 𝑎 sin ( 2 ) = [ ]=√ =√ 𝑘𝑎 𝜋𝑎 𝑘 2𝜋 √2𝜋𝑎 2 As the pulse length 𝑎 becomes small, i.e., more tightly localized, then Δ𝑘 = 4𝜋/𝑎 , which is the bandwidth of 𝐴(𝑘), becomes larger. The pulse length and the bandwidth have the relation 1 1 (7.113) Δ𝑥 Δ𝑘 ~ 𝑎 ⋅ ≥ 𝑎 2 (ii) Gaussian wave packet 1

𝑒−

𝑖𝑘𝑎 2

−𝑒 −𝑖𝑘

𝑖𝑘𝑎 2

Fig 7.14

The normalized Gaussian wave packet shown in Fig 7.14 is expressed as 𝑢(𝑥) =

1 √𝑎𝜋1/2

𝑒

−

𝑥2 2𝑎2

(7.114)

The amplitude 𝐴(𝑘) of the normalized square wave shown in Fig. 7.14 is 2 2 ∞ ∞ 2 1 1 1 (𝑥+𝑖𝑎2 𝑘) −𝑎 𝑘 − 𝐴(𝑘) = ∫ 𝑢(𝑥)𝑒 −𝑖𝑘𝑥 𝑑𝑥 = ∫ 𝑒 2𝑎2 𝑒 2 𝑑𝑥 √2𝜋 −∞ √2𝜋 3/2 𝑎 −∞ =

1 √2𝜋 3/2 𝑎

𝑒

−

𝑎2 𝑘 2 2

∞

∫ 𝑒

2

−

𝑥′ 2𝑎2

𝑎

𝑑𝑥 ′ = √

−∞

21

𝜋

𝑒− 1/2

𝑎2 𝑘 2 2

(7.115)

The pulse length and the bandwidth have the inequality relation 1 1 Δ𝑥 Δ𝑘 ~ 𝑎 ⋅ ≥ 𝑎 2 (iii) Gaussian pulse in the time domain

(7.116)

Fig 7.15

The time-bandwidth product is Δ𝑡 Δ𝜔 ~

1 √Γ

⋅ √Γ ≥

1 2

(7.117)

Phase vs. Group velocity If the distribution 𝐴(𝑘) is sharply peaked around some value 𝑘0 , the frequency 𝜔(𝑘) can be expanded around 𝑘0 : 𝑑𝜔 (7.118) 𝜔(𝑘) = 𝜔0 + | (𝑘 − 𝑘0 ) + ⋯ 𝑑𝑘 𝑘0 Then the field amplitude takes the form 𝑢(𝑥, 𝑡) ≅

𝑒 𝑖[𝑘0 (𝑑𝜔/𝑑𝑘)|0 −𝜔0 ]𝑡 √2𝜋

≅ 𝑢 (𝑥 − 𝑡 (

∞

∫ 𝐴(𝑘)𝑒 𝑖𝑘[𝑥−(𝑑𝜔/𝑑𝑘)|0 𝑡] 𝑑𝑘 −∞

𝑑𝜔 )| , 0) 𝑒 𝑖[𝑘0 (𝑑𝜔/𝑑𝑘)|0 −𝜔0 ]𝑡 𝑑𝑘 0

(7.119)

The pulse travels with a velocity, called the group velocity: 𝑣𝑔 =

𝑑𝜔 | 𝑑𝑘 0

22

(7.120)

The phase velocity is the speed of the individual wave crests, whereas the group velocity is the speed of the wave packet as a whole, i.e., the speed of the envelope propagation. For light waves the dispersion relation between 𝜔 and 𝑘 is given by 𝑐𝑘 (7.121) 𝜔(𝑘) = 𝑛(𝑘) The phase velocity is 𝜔(𝑘) 𝑐 (7.122) 𝑣𝑝 = = 𝑘 𝑛(𝑘) The group velocity is 𝑣𝑝 (𝜔0 ) 𝑑𝜔 1 1 1 𝑣𝑔 = | = = = = 𝑑 𝜔 𝜔 𝑑𝑛(𝜔) 𝑛(𝜔0 ) 𝜔 𝑑𝑛(𝜔) 𝑑𝑘 𝜔0 𝑑𝑘 | ( 𝑛(𝜔))| 1 + | + | 𝑑𝜔 𝜔0 𝑑𝜔 𝑐 𝑐 𝑐 𝑑𝜔 𝜔 𝑛(𝜔0 ) 𝑑𝜔 𝜔0 𝜔0 0

(7.123) Gaussian pulse propagation through a uniform, lossless, and dispersive medium We assume the dispersion relation 𝜔(𝑘) = 𝛽𝑘 2 The group and the phase velocities at 𝑘0 are 𝑣𝑔 = 2𝛽𝑘0 and 𝑣𝑝 = 𝛽𝑘0 A Gaussian pulse 𝑢(𝑥, 0) =

1 √𝑎𝜋1/2

𝑥2 𝑖𝑘0 𝑥 −2𝑎2 𝑒 𝑒

located at 𝑥 = 0 at 𝑡 = 0 is propagating in the 𝑥 direction. The corresponding Fourier amplitude is 𝑎2 𝑎 (𝑘−𝑘0 )2 𝐴(𝑘) = √ 1/2 𝑒 − 2 𝜋 The Gaussian pulse at a later time 𝑡 is 𝑢(𝑥, 𝑡) =

1

∞

∫ 𝐴(𝑘) 𝑒 𝑖(𝑘𝑥−𝜔𝑡) 𝑑𝑘

√2𝜋 −∞ ∞ 𝑎2 𝑎 − (𝑘−𝑘0 )2 𝑖(𝑘𝑥−𝜔𝑡) √ = ∫ 𝑒 2 𝑒 𝑑𝑘 2𝜋 3/2 −∞ ∞ 𝑎2 𝑎 −( +𝑖𝛽𝑡)(𝑘−𝑘0 )2 𝑖(𝑥−𝑣 𝑡)(𝑘−𝑘 ) 𝑖(𝑘 𝑥−𝜔 𝑡) 𝑔 0 𝑑𝑘 0 ∫ = √ 3/2 𝑒 0 𝑒 2 𝑒 2𝜋 −∞ 2

=𝑒

𝑖(𝑘0 𝑥−𝜔0 𝑡)

1/2 (𝑥 − 𝑣𝑔 𝑡) 𝑎 1 ( 1/2 ) exp [− ] 1/2 𝑎2 2𝜋 𝑎2 4 ( 2 + 𝑖𝛽𝑡) ( 2 + 𝑖𝛽𝑡)

23

The pulse envelop is 2

|𝑢(𝑥, 𝑡)| =

1 𝜋1/2

2

(𝑥 − 𝑣𝑔 𝑡) exp [− ] 1/2 𝛽2𝑡 2 𝛽2𝑡 2 2 𝑎 (1 + 4 4 ) 𝑎 (1 + 4 4 ) 𝑎 𝑎 1

Fig 7.16

  

The peak moves with group velocity. The packet width becomes larger with time. The pulse energy is preserved during the propagation.

Fig 7.17. Optical pulse broadening through propagating. 24

7.7 Causality and Kramers-Kronig Relations

25