PDF file: PART 1 Optics Notes

Lecture Notes ... Department of Physics, University of Bergen, 5007 Bergen. ..... no sources, so that J = 0 and ρ = 0, we can through differentiation obtain second- order ..... Now we put z = 0 i (5.7), take an inverse Fourier transform, and use (?

PHYS261-OPTICS PART, PART 1

PHYS 263 Physical Optics Lecture Notes

Jakob J. Stamnes

—————————————————————– Department of Physics, University of Bergen, 5007 Bergen. Tel: 55 58 28 18. Fax: 55 58 94 40. E-mail: [email protected]ﬁ.uib.no

Autumn 2004 Spring 2003

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Contents I

Elementary electromagnetic waves

1 Maxwell’s equations, the material 1.1 Maxwell’s equations . . . . . . . 1.2 The continuity equation . . . . . 1.3 The material equations . . . . . . 1.4 Boundary conditions . . . . . . .

3

equations, . . . . . . . . . . . . . . . . . . . . . . . . . . . .

and . . . . . . . . . . . .

boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 4 4 5 6

2 Poynting’s vector and the energy law

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3 The wave equation and the speed of light

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4 Scalar waves 4.1 Plane waves . . . . . . . . . . . . . . . . 4.2 Spherical waves . . . . . . . . . . . . . . 4.3 Harmonic (monochromatic) waves . . . 4.4 Complex representation . . . . . . . . . 4.5 Linearity and the superposition principle 4.6 Phase velocity and group velocity . . . . 4.7 Repetition . . . . . . . . . . . . . . . . .

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11 11 12 13 14 15 15 16

5 Pulse propagation in a dispersive medium

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6 General electromagnetic plane wave

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7 Harmonic electromagnetic waves of arbitrary form - Time averages

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8 Harmonic electromagnetic plane wave – Polarisation

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9 Reﬂection and refraction of a plane wave 9.1 Reﬂection law and refraction law (Snell’s law) . . . . . . . . . . 9.2 Generalisation of the reﬂection law and Snell’s law . . . . . . . 9.3 Reﬂection and refraction of plane electromagnetic waves . . . . 9.3.1 Reﬂectance and transmittance . . . . . . . . . . . . . . 9.3.2 Brewster’s law . . . . . . . . . . . . . . . . . . . . . . . 9.3.3 Unpolarised light (natural light) . . . . . . . . . . . . . 9.3.4 Rotation of the plane of polarisation upon reﬂection and 9.3.5 Total reﬂection . . . . . . . . . . . . . . . . . . . . . . .

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29 29 31 32 35 38 38 39 40

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List of Figures 1.1

ˆ separates two diﬀerent dielectric media. . . . . A plane interface with unit normal n

7

4.1

A plane wave that propagates in direction ˆs, has no variation in any plane that is normal to ˆs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12

A plane wave propagates in the positive z direction in a dispersive medium that ﬁlls the half space z ≥ 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

18

The vectors E, H, and ˆs for an electromagnetic plane wave represent a right-handed Cartesian co-ordinate system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

22

5.1 6.1 8.1 8.2 9.1

9.2 9.3 9.4 9.5 9.6 9.7

Instantaneous picture of the electric vector of a plane wave that propagates in the z direction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The end point of the electric vector describes an ellipse that is inscribed in a rectangle with sides 2a1 and 2a2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reﬂection and refraction of a plane wave at a plane interface between two diﬀerent media. Illustration of propagation directions and angles of incidence, reﬂection, and transmission. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reﬂection and refraction of a plane wave. Illustration of the co-ordinate system (ˆ n, ˆ ˆt). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . b, Generalisation of Snell’s law and the reﬂection law to include non-planar waves that are incident upon a curved interface. . . . . . . . . . . . . . . . . . . . . . . . . . . . Reﬂection and refraction of a plane electromagnetic wave at a plane interface between two diﬀerent media. Illustration of T E and T M components of the electric ﬁeld. . . Illustration of the angle αq between the electric vector Eq and the plane of incidence ˆT M q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . spanned by kq and e Illustration of Brewster’s law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Illustration of the refraction of a plane wave into an optically thinner medium, so that θi < θt . When θi → θic , then θt → π/2, and we get total reﬂection. . . . . . . . . . .

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Part I

Elementary electromagnetic waves

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Chapter 1

Maxwell’s equations, the material equations, and boundary conditions In this course we consider light to be electromagnetic waves of frequencies ν in the visible range, so that ν  (4 − 7.5) × 1014 Hz. Since λ = νc , where c is the speed of light in vacuum (c  3 × 108 m/s), we ﬁnd that the corresponding wavelength interval is λ  (0.4 − 0.75) µm. Thus, to study the propagation of light we must consider the propagation of the electromagnetic ﬁeld, which is represented by the two vectors E and B, where E is the electric ﬁeld strength and B is the magnetic induction or the magnetic ﬂux density. To enable us to describe the interaction of the electromagnetic ﬁeld with material objects we need three additional vector quantities, namely the current density J, the displacement D, and the magnetic ﬁeld strength H.

1.1

Maxwell’s equations

The ﬁve vectors mentioned above are linked together by Maxwell’s equations, which in Gaussian units are ∇×H=

1 ˙ 4π D+ J, c c

1˙ ∇ × E = − B. c In addition we have the two scalar equations

(1.1.1) (1.1.2)

∇ · D = 4πρ,

(1.1.3)

∇ · B = 0,

(1.1.4)

where ρ is the charge density. Equation (1.1.3) can be said to deﬁne the charge density ρ. Similarly, we can say that (1.1.4) implies that free magnetic charges do not exist.

1.2

The continuity equation

The charge density ρ and the current density J are not independent quantities. By taking the divergence of (1.1.1) and using that ∇ · (∇ × A) = 0 for an arbitrary vector A, we ﬁnd that ∇·J+

1 ˙ = 0, ∇·D 4π

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which on using (1.1.3) gives ∇ · J + ρ˙ = 0.

(1.2.1)

This equation is called the continuity equation, and it expresses conservation of charge. By integrating (1.2.1) over a closed volume V with surface S, we ﬁnd 

 ∇ · Jdv = −

V

∂ρ dv, ∂t

(1.2.2)

V

which by use of the divergence theorem gives   d d ˆ da = −  J·n ρdv = − Q. dt dt S

(1.2.3)

V

ˆ is the unit surface normal in the direction out of the volume V , so that (1.2.3) shows that Here n the integrated current ﬂux out of the closed volume V is equal to the loss of charge in the same volume.

Digression 1: Notation • Bold face is used to denote vector quantities, e.g. ˆ x + Ey e ˆ y + Ez e ˆz , E = Ex e ˆx , e ˆy , and e ˆz are unit vectors along the axes in a Cartesian co-ordinate system. where e • A dot above a symbol is used to denote the time derivative, e.g. ˙ = ∂ B. B ∂t • E, B, D, H, ρ, and J are functions of the position r and the time t, e.g. D = D(r, t). • The connection between Gaussian and other systems of units, e.g. MKS units, follows from J.D. Jackson, ”Classical Electrodynamics”, Wiley (1962), pp. 611-621. For conversion between Gaussian units and MKS units, we refer to the table on p. 621 in this book.

1.3

The material equations

Maxwell’s equations (1.1.1)-(1.1.4), which connect the fundamental quantities E, H, B, D, and J, are not suﬃcient to uniquely determine the ﬁeld vectors (E, B) from a given distribution of currents and charges. In addition we need the so-called material equations, which describe how the ﬁeld is inﬂuenced by matter. In general the material equations can be relatively complicated. But if the ﬁeld is time harmonic and the matter is isotropic and at rest, the material equations have the following simple form Jc = σE,

(1.3.1)

D = εE,

(1.3.2)

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B = µH,

(1.3.3)

where σ is the conductivity, ε is the permittivity or dielectric constant, and µ is the permeability. Equation (1.3.1) is Ohm’s law, and Jc is the conduction current density, which arises because the material has a non-vaninishing conductivity (σ = 0). The total current density J in (1.1.1) can in addition consist of an externally applied current density J0 , so that J = J0 + Jc = J0 + σE.

(1.3.4)

Digression 2: General material considerations • A material that has a non-negligible conductivity σ is called a conductor, while a material that has a negligible conductivity is called an insulator or a dielectric. • Metals are good conductors. • Glass is a dielectric; ε  2.25; σ = 0; µ = 1. • In anisotropic media (e.g. crystals) the relation in (1.3.2) is to be replaced by D = εE, where ε is a tensor, dyadic or matrix. • In a plasma (1.3.1) is to be replaced by J = σE, where the conductivity is a tensor. • There are also magnetically anisotropic media, in which (1.3.3) is to be replaced by B = µH. Thus, in this case the permeability is a tensor. Such materials are not important in optics. • In dispersive media ε is frequency dependent, i.e. ε = ε(ω). Maxwell’s equations and the material equations are still valid for each frequency component or time harmonic component of the ﬁeld. For a pulse consisting of many frequency components, one must apply Fourier analysis to solve Maxwell’s equations and the material equations separately for each time harmonic component, and then perform an inverse Fourier transformation. • In non-linear media there is no linear relation between D and E (equation (1.3.2) is not valid). Most media become non-linear when the electric ﬁeld strength becomes suﬃciently high.

1.4

Boundary conditions

Hitherto we have assumed that ε and µ are continuous functions of the position. But in optics we often have systems consisting of several diﬀerent types of glass. At the transition between air and glass or between two diﬀerent types of glass the material parameters are discontinuous. Let us therefore consider what happens to the electromagnetic ﬁeld at the boundary between two media. Consider two media that are separated by an interface, as illustrated in Fig. 1.1. From Maxwell’s equations, combined with Stokes’ and Gauss’ theorems, one can derive the following boundary conditions ˆ · (B(2) − B(1) ) = 0, n (1.4.1) ˆ · (D(2) − D(1) ) = 4πρs , n

(1.4.2)

ˆ × (E(2) − E(1) ) = 0, n

(1.4.3)

4π (1.4.4) Js , c ˆ is a unit vector along the surface normal. According to (1.4.1) the normal component of where n B is continuous across the boundary, while (1.4.2) says that if there exists a surface charge density ˆ × (H(2) − H(1) ) = n

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^ n

ε 1, µ 1 ε 2, µ 2 ˆ separates two diﬀerent dielectric media. Figure 1.1: A plane interface with unit normal n

ρs at the boundary, then the normal component of D is changed by 4πρs across the boundary between the two media. According to (1.4.3) the tangential component of E is continuous across the boundary, while (1.4.4) implies that if there exists a surface current density Js at the boundary, ˆ × H, is changed by 4π then the tangential component of H, i.e. of n c Js .

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Chapter 2

Poynting’s vector and the energy law The electric energy density we and the magnetic energy density wm are deﬁned by we =

1 E · D, 8π

1 H · B, 8π and the total energy density is the sum of these, i.e. wm =

w = we + wm .

(2.1) (2.2)

(2.3)

The energy ﬂux of the ﬁeld is represented by Poynting’s vector S, given by c E × H. (2.4) 4π Here S represents the amount of energy that per unit time crosses a unit area that is parallel with both E and H. In a non-conducting medium (σ = 0) we have the conservation law S=

∂w + ∇ · S = 0, ∂t

(2.5)

which expresses that the change of the energy density in a small volume is equal to the energy ﬂux out of the same volume [cf. (1.2.2) and (1.2.3)]. In optics the Poynting vector is very important, because its absolute value is proportional to the light intensity, i.e. |S| ∝ light intensity.

(2.6)

The direction of the Poynting vector, deﬁned by the unit vector ˆs = points in the direction of light propagation.

S , |S|

(2.7)

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Chapter 3

The wave equation and the speed of light The electric and magnetic ﬁelds E and H are connected through Maxwell’s equations (1.1.1)-(1.1.4), which are simultaneous, ﬁrst-order partial diﬀerential equations. But in those parts of space where there are no sources, so that J = 0 and ρ = 0, we can through diﬀerentiation obtain second-order partial diﬀerential equations that E and H satisfy individually. We assume that the medium is non-dispersive, so that D = εE, where ε˙ = 0, and B = µH, where µ˙ = 0. Then we have from (1.1.1) and (1.1.2) ∇×H=

1 ˙ 1 ˙ D = εE, c c

(3.1)

1˙ 1 ˙ ∇×E=− B = − µH. (3.2) c c Next, we assume that the medium is homogeneous, so that ε and µ do not vary with position. By taking the curl of (3.2) and combining the result with the time derivative of (3.1), we ﬁnd that 1 ¨ ˙ = − 1 µ 1 εE ¨ = − εµ E. ∇ × (∇ × E) = − µ∇ × H c c c c2 Now we use the vector relation ∇ × (∇ × A) = ∇(∇ · A) − ∇2 A,

(3.3)

(3.4)

which applies to an arbitrary vector A, to obtain ∇(∇ · E) − ∇2 E = −

εµ ¨ E, c2

(3.5)

which since ∇ · E = 0, gives ∇2 E −

εµ ¨ E = 0. c2

(3.6)

In a similar manner we ﬁnd εµ ¨ H = 0. c2 By comparing these results with the scalar wave equation ∇2 H −

∇2 V −

1 ¨ V = 0, v2

(3.7)

(3.8)

we see that in a source-free region of space each Cartesian component of E and H satisﬁes the scalar wave equation with phase velocity

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(3.9)

Note that this derivation is valid only in a non-dispersive medium in which both the permittivity and the permeability do not depend on the frequency.

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Chapter 4

Scalar waves Scalar waves are solutions of the scalar wave equation (3.8), which is given by ∇2 V (r, t) −

4.1

1 ∂2 V (r, t) = 0. v 2 ∂t2

(4.0.1)

Plane waves

Any solution of (4.0.1) of the form V (r, t) = V (r · ˆs, t),

(4.1.1)

is called a plane wave, since V at any time t is constant over any plane r · ˆs = constant,

(4.1.2)

which is normal to the unit vector ˆs (see Fig. 4.1). To show that (4.1.1) is a solution of (4.0.1), we introduce a new variable ζ = r · ˆs = xsx + ysy + zsz ,

(4.1.3)

so that ∂ζ = sx ; ∂x

∂ζ = sy ; ∂y

∂ζ = sz . ∂z

(4.1.4)

Further we ﬁnd that ∂V ∂V ∂V ∂ζ = · = sx . ∂x ∂ζ ∂x ∂ζ       2 ∂2V ∂V ∂ ∂V ∂ ∂V ∂ζ ∂ 2∂ V s = s = s = s = . x x x x ∂x2 ∂x ∂ζ ∂x ∂ζ ∂ζ ∂ζ ∂x ∂ζ 2

(4.1.5) (4.1.6)

In a similar way we ﬁnd ∂2V ∂2V = s2y 2 2 ∂y ∂y

;

∂2V ∂2V = s2z 2 . 2 ∂z ∂z

(4.1.7)

When we substitute (4.1.6) and (4.1.7) in (4.0.1) and take into account that s2x + s2y + s2z = 1, since ˆs is a unit vector, the wave equation becomes ∂2V 1 ∂2V − = 0. ∂ζ 2 v 2 ∂t2

(4.1.8)

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^ n

^e r

e^ r

z ^e φ

θ e^ φ

^n

θ y

^ n

^e θ

θ

e^ θ

φ

x

Figure 4.1: A plane wave that propagates in direction ˆs, has no variation in any plane that is normal to ˆs.

By introducing two new variables p and q, deﬁned by p = ζ − vt ; q = ζ + vt,

(4.1.9)

we ﬁnd (Exercise 2) that the wave equation in (p, q) variables can be written ∂2V = 0. ∂p∂q

(4.1.10)

This equation has the following general solution V = V1 (p) + V2 (q),

(4.1.11)

where V1 and V2 are arbitrary functions. By substitution from (4.1.3) and (4.1.9) in (4.1.11), we ﬁnd the following general plane-wave solution V (r, t) = V1 (r · ˆs − vt) + V2 (r · ˆs + vt).

(4.1.12)

ζ − vt = ζ + vτ − v(t + τ ),

(4.1.13)

V1 (ζ, t) = V1 (ζ + vτ, t + τ ).

(4.1.14)

Note that

so that

Equation (4.1.14) shows that during the time τ , V1 is displaced a length s = vτ in the positive ζ direction, i.e. V1 propagates with velocity v in the positive ζ direction. The conclusion is that V (ζ ± vt) represents a plane wave that propagates at velocity v in the positive ζ direction (lower sign) or in the negative ζ direction (upper sign).

4.2

Spherical waves

Consider now solutions of the scalar wave equation (4.0.1) of the form V = V (r, t),

(4.2.1)

where r = |r| =



x2 + y 2 + z 2 ,

(4.2.2)

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is the distance from the origin (0, 0, 0). Since we have no angular dependence in this case, the Laplacian operator has the following form in spherical coordinates (Exercise 3) 1 ∂2 (rV ), r ∂r2 which upon substitution in the wave equation (4.0.1) gives ∇2 V =

∂2 1 ∂2 (rV ) − (rV ) = 0. ∂r2 v 2 ∂t2 Since (4.2.4) is of the same form as (4.1.8), the solution becomes (cf. (4.1.12)) rV = V1 (r − vt) + V2 (r + vt).

(4.2.3)

(4.2.4)

(4.2.5)

Thus, we have obtained the following result: V (r±vt) represents a spherical wave that converges r towards the origin (upper sign) or diverges away from the origin (lower sign). Thus, V (r+vt) propr agates towards the origin with velocity v, whereas V (r−vt) propagates away from the origin with r velocity v.

4.3

Harmonic (monochromatic) waves

At a given point r in space the solution of the wave equation is a function only of time, i.e. V (r, t) = F (t),

(4.3.1)

where F (t) can be an arbitrary function. If F (t) has the simple form F (t) = a cos(ωt − δ),

(4.3.2)

then we have a harmonic wave in time. The quantities in (4.3.2) have the following meaning: a is the amplitude (positive), ω is the angular frequency, and ωt − δ is the phase. A harmonic wave is also called a monochromatic wave because it consists of only one frequency or wavelength component. The frequency ν and the period T follow from ω 1 = . 2π T The harmonic wave in (4.3.2) has period T because ν=

F (t + T ) = a cos(ω(t + T ) − δ) = a cos(ωt − δ + 2π) = F (t).

(4.3.3)

(4.3.4)

From (4.1.12) we see that the general expression for a wave that propagates in the ˆs direction can be written      r · ˆs r · ˆs  V = V1 (r · ˆs − vt) = V1 −v t − = V1 t − , (4.3.5) v v s where both V1 and V1 are arbitrary functions. By replacing t with t− r·ˆ v in (4.3.2) we get a harmonic plane wave     r · ˆs V (r, t) = a cos ω t − + δ = a cos[kr · ˆs − ωt + δ], (4.3.6) v

where

ω (4.3.7) v is the wave number. We see that that V (r, t) remains unchanged if we replace r · ˆs with r · ˆs + nλ, where n = 1, 2, . . ., and λ is given by k=

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2π 2π v =v = vT = . (4.3.8) k ω ν The quantity λ is called the wavelength. Note that for t = constant, V (r, t) in (4.3.6) is periodic with wavelength λ, i.e. λ=

V (r · ˆs, t) = V (r · ˆs + nλ, t) ; n = 1, 2, 3, . . . .

(4.3.9)

Now we introduce the wave vector or propagation vector k, deﬁned by k = kˆs.

(4.3.10)

so that the expression (4.3.6) for a plane, harmonic wave can be written V (r, t) = a cos(k · r − ωt + δ).

(4.3.11)

In a similar way the expression for a converging or a diverging harmonic spherical wave becomes cos(∓kr − ωt + δ) , (4.3.12) r where the upper sign corresponds to a converging spherical wave and the lower sign to a diverging spherical wave. Consider now a plane, harmonic wave that propagates in the positive z direction, so that [cf. (4.3.11)] V (r, t) = a

V (z, t) = a cos(kz − ωt + δ).

(4.3.13)

A wave front is deﬁned by the requirement that the phase shall be constant over it, i.e. φ = kz − ωt + δ = constant.

(4.3.14)

Hence it follows that on a wave front we have z = vt + constant ; v =

ω . k

(4.3.15)

Thus, the wave front propagates at the velocity v=

ω , k

(4.3.16)

which is called the phase velocity.

4.4

Complex representation

Alternatively we can express (4.3.11) and (4.3.12) in the following way V (r, t) = Re{U (r)e−iωt },

(4.4.1)

where Re{. . .} stands for the real part of {. . .}, and where the complex amplitude U (r) is given by U (r) = aei(k·r+δ) ,

(4.4.2)

for a plane wave, and by a i(±kr+δ) , e r for a diverging (upper sign) or converging (lower sign) spherical wave. U (r) =

(4.4.3)

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Note that when we perform linear operations, such as diﬀerentiation, integration or summation, we can drop the ’Re’ symbol during the operations, as long as we remember to take the real part of the result in the end. By substituting V (r, t) = U (r)e−iωt ,

(4.4.4)

(∇2 + k 2 )U (r) = 0,

(4.4.5)

in the wave equation (4.0.1), we get

which shows that the complex amplitude U (r) is a solution of the Helmholtz equation.

4.5

Linearity and the superposition principle

For any linear equation the sum of two or several solutions is also a solution. This is called the superposition principle. Since Maxwell’s equations are linear, the superposition principle is valid for electromagnetic waves as long as the material equations are linear. The superposition principle implies that we can construct general solutions of the wave equation or Maxwell’s equations by adding elementary solutions in the form of harmonic plane or spherical waves. We will discuss this in detail in Part II.

4.6

Phase velocity and group velocity

Consider a harmonic wave of the form [cf. (4.3.11)]   V (r, t) = Re U (r)e−iωt ,

(4.6.1)

where the complex amplitude U (r) is a solution of the Helmholtz equation (4.4.5), i.e. (∇2 + k 2 )U (r) = 0.

(4.6.2)

The wave number k can be written ω ω c

= = k0 n, v c v where k0 is the wave number in vacuum, i.e. k=

k0 =

ω , c

(4.6.3)

(4.6.4)

and n is the refractive index given by c √ = εµ. (4.6.5) v A general wave V (r, t) can always be expressed as a sum of harmonic components. We will return to this later. If ε depends on ω, i.e. ε = ε(ω), then the phase velocity also will depend on ω, since v = nc = v(ω). This means that diﬀerent harmonic components will propagate at diﬀerent phase velocities. A polychromatic wave or a pulse, which is comprised of many harmonic components, therefore will change its shape during propagation, and the energy will not propagate at the phase velocity, but at the group velocity, which is deﬁned as n=

dω . dk If n(ω) = constant, we have a non-dispersive medium. Since vg =

ω = vk,

(4.6.6)

(4.6.7)

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where the phase velocity v =

c n

now is constant, we have in this case

d (vk) = v. dk Thus, the phase velocity and the group velocity are equal in a non-dispersive medium where n = constant. In dispersive media we have vg =

d dv dv dv (vk) = v + k =v−λ =v+ν , dk dk dλ dν ω where the last two results follow from the relation k = 2π λ = v. vg =

4.7

(4.6.8)

Repetition

From Maxwell’s equations in source-free space (J = 0 ; ρ = 0) we ﬁnd εµ ¨ εµ ¨ E = 0 ; ∇2 H − 2 H = 0. c2 c Comparison of (4.7.1) with the scalar wave equation ∇2 E −

(4.7.1)

1 ¨ V = 0, (4.7.2) v2 shows that any Cartesian component of E and H satisﬁes the scalar wave equation with phase velocity v given by ∇2 V −

c c v=√ = . εµ n

(4.7.3)

The scalar wave equation (4.7.2) has simple solutions in the form of plane waves or spherical waves.

Plane waves For a plane wave V is given by V (r, t) = V1 (r · sˆ − vt) + V2 (r · sˆ − vt),

(4.7.4)

where V (ζ ∓ vt) represents a plane wave that propagates in the positive ζ direction (upper sign) or in the negative ζ direction (lower sign).

Spherical waves For a spherical wave V is given by V (r, t) =

V1 (r − vt) V2 (r + vt) + , r r

(4.7.5)

represents a spherical wave that propagates away from the origin (upper sign) or where V (r∓vt) r towards the origin (lower sign).

Harmonic (monochromatic) waves A plane harmonic wave that propagates in the direction k = kˆs is given by V (r, t) = a cos(k · r − ωt + δ),

(4.7.6)

and the corresponding spherical wave is a cos(±kr − ωt + δ), (4.7.7) r where the upper sign represents a diverging spherical wave and the lower sign represents a converging spherical wave. V (r, t) =

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Complex representation of harmonic waves In complex notation we have

V (r, t) = Re[U (r)e−iωt ].

(4.7.8)

For a plane wave the complex amplitude U (r) is given by U (r) = aei(k·r+δ) , and for a diverging or converging spherical wave it is given by U (r) =

a i(±kr+δ) e . r

By substituting (4.7.8) into the wave equation (4.7.2), we ﬁnd that U (r) satisﬁes the Helmholtz equation, i.e. (∇2 + k 2 )U (r) = 0.

(4.7.9)

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Chapter 5

Pulse propagation in a dispersive medium

z n(ω ) z=0 Figure 5.1: A plane wave propagates in the positive z direction in a dispersive medium that ﬁlls the half space z ≥ 0.

Consider a polychromatic, plane wave that propagates in the positive z direction in a linear, homogeneous, isotropic, and dispersive medium that ﬁlls the half space z > 0 (Fig. 5.1). The polychromatic, plane wave u(z, t) is comprised of diﬀerent harmonic components, which implies that we can represent u(z, t) by the following inverse Fourier transform  ∞ 1 u(z, t) = u ˜(z, ω)e−iωt dω, (5.1) 2π −∞ where the frequency spectrum u ˜(z, ω) is given as the Fourier transform of u(z, t), i.e.  ∞ u ˜(z, ω) = u(z, t)eiωt dt.

(5.2)

−∞

Thus, u(z, t) and u ˜(z, ω) constitute a Fourier transform pair. Since u ˜(z, ω) can be any Cartesian component of the frequency spectrum of the electric or magnetic ﬁeld, it satisﬁes the Helmholtz equation, i.e. [∇2 + k 2 (ω)]˜ u(z, ω) = 0,

(5.3)

where k(ω) =

ω ω c ω = = n(ω). v(ω) c v(ω) c

(5.4)

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Suppose now that u(z, t) is known for all values of t in the plane z = 0, and that u(0, t) vanishes for t < 0. Since there is no variation in the x and y directions, the Helmholtz equation (5.3) can be written as  2  d 2 + k (ω) u ˜(z, ω) = 0, (5.5) dz 2 which has the following general solution u ˜(z, ω) = u+ (ω)eik(ω)z + u− (ω)e−ik(ω)z .

(5.6)

If we consider propagation in the positive z direction only, then u− (ω) = 0, so that (5.1) gives  ∞ 1 u(z, t) = u+ (ω)ei(k(ω)z−ωt) dω. (5.7) 2π −∞ Now we put z = 0 i (5.7), take an inverse Fourier transform, and use (??) to obtain  ∞ u+ (ω) = u(0, t)eiωt dt = u ˜(0, ω),

(5.8)

−∞

so that (5.7) gives u(z, t) =

1 2π



u ˜(0, ω)ei(k(ω)z−ωt) dω,

(5.9)

−∞

or u(z, t) =

1 2π



z

u ˜(0, ω)ei c f (ω) dω,

(5.10)

−∞

where ct . (5.11) z c Consider ﬁrst the special case in which n(ω) = v(ω) = constant, which implies that we have a ω non-dispersive medium. Since k = v , where v now is constant, we have from (5.1) and (5.9)  ∞ z 1 z

u(z, t) = u ˜(0, ω)e−iω(− v +t) dω = u 0, t − . (5.12) 2π −∞ v f (ω) = ω[n(ω) − θ] ; θ =

This result shows that the pulse propagates in the positive z direction at velocity v without changing its shape. Suppose now that the medium is dispersive and that the frequency spectrum g˜(ω) of the pulse in (5.10) does not contain singularities and that it is suﬃciently wide. Then the main contribution to the pulse in (5.10) comes from frequencies ωs for which the phase f (ω) in (5.11) is stationary, i.e. from ωs that satisfy the equation f  (ωs ) = n(ωs ) − θ + ωs n (ωs ) = 0.

(5.13)

A model that is commonly used to study propagation in dispersive media, is the so called Lorentzmedium. For such a medium with one single resonance frequency the refractive index n(ω) is given by the following expression  n(ω) = 1 −

b2 ω 2 − ω02 + 2δiω

1/2 ,

(5.14)

where b is a constant, ω0 is the resonance frequency, and δ represents the damping (attenuation) in the medium.

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Equation (5.9) shows that when the medium is dispersive, then u(z, t) (for any z > 0) is a sum of harmonic plane waves of the form u ˜(0, ω)exp[i(k(ω)z − ωt)] = u ˜(0, ω)exp[−ki (ω)z]exp[i(kr (ω)z − ωt)], where kr (ω) and ki (ω) are the real and the imaginary part, repectively, of k(ω). Thus, the amplitude u ˜(0, ω)exp[−ki (ω)z], is damped exponentially as z increases, and the phase velocity is given by v(ω) = krω(ω) , where k(ω) = (ω/c)n(ω) = (ω/c)[nr (ω) + ini (ω)] = kr (ω) + iki (ω). Since the phase velocity v depends on the frequency ω, plane waves of diﬀerent frequencies will arrive at a given position z at diﬀerent times and thus cause a distortion of the pulse, i.e. the shape of the pulse will get changed. Also, the damping factor ki (ω) depends on ω, so that diﬀerent frequency components will have diﬀerent amplitudes when they arrive at a given position z.

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Chapter 6

General electromagnetic plane wave A general electromagnetic plane wave can be written in the form E = E(k · r − ωt) ; H = H(k · r − ωt),

(6.1)

where k = kˆs, with ˆs pointing in the direction of propagation. We introduce a new variable u = k · r − ωt, so that ∂u = kx ; ∂x

∂u = ky ; ∂y

∂u = kz ; ∂z

∂u = −ω. ∂t

(6.2)

In source-free space Maxwell’s equations (1.1.1)-(1.1.2) are given by ∇×H=

1 ˙ ε˙ D = E, c c

(6.3)

1˙ µ ˙ ∇×E=− B = H. (6.4) c c By using the chain rule, we ﬁnd that the x component of ∇ × E can be expressed as follows (∇ × E)x

where E =

dE du .

∂Ez ∂Ey dEz ∂u dEy ∂u − = − ∂y ∂z du ∂y du ∂z ω     = Ez ky − Ey kz = (k × E )x = (ˆs × E )x , v

= ∇ y E z − ∇z E y =

(6.5)

By proceeding in a similar manner, we ﬁnd that ∇×E=

ω ˆs × E , v

(6.6)

∇×H=

ω ˆs × H , v

(6.7)

where E =

dE dH ; H = . du du

(6.8)

Further, we have ˙ = ∂E = dE ∂u = −ωE ; H ˙ = −ωH . E ∂t du ∂t

(6.9)

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By substitution of (6.6)-(6.9) into Maxwell’s equations (6.3)-(6.4) the result is ε  ε ε c ˆs × H = (−v)E = − √ E = − E, c c εµ µ µ µ    ˆs × E = − (−v)H = H, c ε where we have used (3.9). Thus, we have µ ε    ˆs × H ; H = ˆs × E . E =− ε µ

(6.10) (6.11)

(6.12)

By integrating over u in (6.12) and setting the integration constant equal to zero, we get µ ε ˆs × H ; H = ˆs × E. E=− ε µ

(6.13)

Scalar multiplication of the equations in (6.13) with ˆs gives ˆs · E = ˆs · H = 0,

(6.14)

which shows that both E and H are transverse waves, i.e. both E and H are normal to the propagation direction ˆs, as illustrated in Fig. 6.1. Thus, the vectors ˆs, E, and H represent a right-handed Cartesian co-ordinate system.

E

^ s H Figure 6.1: The vectors E, H, and ˆs for an electromagnetic plane wave represent a right-handed Cartesian co-ordinate system.

For the electric and the magnetic energy density we ﬁnd we =

ε 2 1 E·D= E ; E = |E|, 8π 8π

(6.15)

µ 2 1 B·H= H ; H = |H|. 8π 8π √ √ Since µH = εE (cf. (6.13)), we get we = wm , and the total energy density becomes wm =

w = we + wm = 2we =

1 1 εE 2 = 2wm = µH 2 , 4π 4π

(6.16)

(6.17)

and the Poynting vector (2.4) becomes c c c S= E×H= EHˆs = E 4π 4π 4π

ε Eˆs = µ



1 εE 2 4π



c √ εµ

 sˆ = wvˆs.

(6.18)

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Thus, we have S = wvˆs,

(6.19)

which shows that the Poynting vector represents the energy ﬂow, both with respect to absolute value and direction. A dimensional analysis of (6.19) shows that Energi m Energi W = 2 = 2. · (6.20) m3 s m ·s m Thus, S represents the amount of energy per unit time that passes through a unit area of the plane that is spanned by E and H, as asserted previously in chapter 2. [|S|] = [w][v] =

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24

Chapter 7

Harmonic electromagnetic waves of arbitrary form - Time averages The E and H ﬁelds for a harmonic wave of arbitrary form can be written

E = Re E0 (r)e−iωt

; H = Re H0 (r)e−iωt ,

(7.1)

where E0 (r) and H0 (r) are complex vectors. Thus, we have I E0 (r) = ER 0 (r) + iE0 ((r),

(7.2)

I H(r) = HR 0 (r) + iH0 (r),

(7.3)

I R I where ER 0 , E0 , H0 , and H0 are real vectors. Since optical frequencies are very high (ω  1015 s−1 ), we can only observe averages of we , wm , and S, taken over a time interval −T  ≤ t ≤ T  , where T  is much larger than the period T = 2π ω . For the time average of the electric energy density we have [cf. (2.1)]

1 we = 2T  For any complex number z, we have Rez = Therefore, we may write



T

−T 

1 2 (z

E = Re[E0 (r)e−iωt ] =

ε |E|2 dt. 8π

(7.4)

+ z ∗ ), where z ∗ is the complex conjugate of z. 1 [E0 e−iωt + E∗0 e+iωt ], 2

so that we get |E|2 = E · E =

1 1 2iωt ]. (7.5) [E0 e−iωt + E∗0 eiωt ] · [E0 e−iωt + E∗0 eiωt ] = [E20 e−2iωt + 2E0 · E∗0 + E∗2 0 e 4 4

Further, we have 1 2T 



T

−2iωt

e −T 

   T  1 1 e−2iωt 1 1 T sin(2ωT  ) = dt = = sin(2ωT  ).   2T −2iω −T  2T ω 4π T 

(7.6)

Since T  T , the integral that includes the factor e−2iωt can be neglected. Similarly, the integral that includes the factor e2iωt can be neglected, and we get we =

ε E0 · E∗0 . 16π

(7.7)

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By proceeding in a similar manner, we ﬁnd that the time average of the magnetic energy density becomes µ H0 · H∗0 . 16π The time average of the Poynting vector is given by [cf. (2.4)] wm =

S =

1 2T 



T

−T 

c (E × H)dt, 4π

(7.8)

(7.9)

where E × H can be written E×H = =

1 1 [E0 e−iωt + E∗0 eiωt ] × [H0 e−iωt + H∗0 eiωt ] 2 2 1 −2iωt + E0 × H∗0 + E∗0 × H0 + E∗0 × H∗0 e2iωt }. {E0 × H0 e 4

(7.10)

By substituting (7.10) into (7.9) and performing time averaging, we ﬁnd that the time average of the Poynting vector becomes S =

c c {E0 × H∗0 + E∗0 × H0 } = Re(E0 × H∗0 ). 16π 8π

(7.11)

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Chapter 8

Harmonic electromagnetic plane wave – Polarisation For an electromagnetic plane wave that is time harmonic, each Cartesian component of E and H is of the form a cos(τ + δ) = Re[aei(τ +δ) ] ; a > 0,

(8.1)

τ = k · r − ωt.

(8.2)

where

Let the z axis point in the ˆs direction. Then only the x and y components of E and H are non-zero, since the electromagnetic ﬁeld of a plane wave is transverse. Now we want to determine that curve which the end point of the electric vector describes during propagation. This curve consists of points that have co-ordinates (Ex , Ey ) given by Ex = a1 cos(τ + δ1 ) ; a1 > 0,

(8.3)

Ey = a2 cos(τ + δ2 ) ; a2 > 0,

(8.4)

Ez = 0.

(8.5)

In order to determine that curve which E(τ ) describes (Fig. 8.1), we eliminate τ from (8.3)-(8.4). We let β = τ + δ1 and get Ex = a1 cos β,

(8.6)

Ey = a2 cos(β + δ) = a2 [cos β cos δ − sinβ sin δ],

(8.7)

where δ = δ2 − δ1 . We substitute from (8.6) into (8.7) and get   2 Ey Ex Ex = cos δ − 1 − sin δ, a2 a1 a1

(8.8)

which upon squaring gives 

Ex a1

2

 +

Ey a2

2 −2

Ex Ey cos δ = sin2 δ. a1 a2

(8.9)

This is the equation of a conic section. The cross term implies that it is rotated relative to the co-ordinate axes (x, y). By letting δ = π2 , we get

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27

x E x (τ ) E(τ )

z

E y (τ )

y

Figure 8.1: Instantaneous picture of the electric vector of a plane wave that propagates in the z direction.



Ex a1

2

 +

Ey a2

2 = 1,

(8.10)

which shows that the equation describes an ellipse. In a co-ordinate system (ξ, η), which coincides with the axes of the ellipse, the equations for the ﬁeld components become Eξ = a cos(τ + δ0 ),

(8.11)

Eη = ±b sin(τ + δ0 ),

(8.12)

which upon squaring gives 

Eξ a



2 +

Eη b

2 = 1.

(8.13)

When τ + δ0 = 0, we have Eξ = a; Eη = 0, and when τ + δ0 = π2 , we have Eξ = 0; Eη = ±b. This shows that when the upper or lower sign in (8.12) applies, the electric vector rotates against or with the clock, respectively, if we view the xy plane from the positive z axis. Rotation against the clock is called left-handed polarisation, and rotation with the clock is called right-handed polarisation. The relation between the two co-ordinate systems (x, y) and (ξ, η) is shown in Fig. 8.2, where (cf. Exercise 7) a2 + b2 = a21 + b22 , tan 2ψ = tan(2α) cos δ ; tan α =

(8.14) π a2 (0 ≤ α ≤ ), a1 2

(8.15)

b sin 2ψ = sin(2α) sin δ ; tan ψ = ± . (8.16) a Since sin δ < 0 when the upper sign in (8.12) applies, we have left-handed polarisation when sin δ < 0. We consider now some special cases of (8.6)-(8.7). Linear polarisation.

If the phase diﬀerence δ is equal to an integer times π, i.e. if δ = mπ (m = 1, ±1, ±2, . . .),

(8.17)

Ex = a1 cos β,

(8.18)

then we get from (8.6)-(8.7)

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28

y η

ξ a b

ψ

2a 2

x

2a1 Figure 8.2: The end point of the electric vector describes an ellipse that is inscribed in a rectangle with sides 2a1 and 2a2 .

Ey = a2 cos(β + mπ) = a2 (−1)m

Ex , a1

(8.19)

which shows that the ellipse degenerates into a straight line, i.e. Ey a2 = (−1)m . Ex a1 Circular polarisation. of 2π, i.e. if

(8.20)

If the amplitudes are equal and the phase diﬀerence is ± π2 plus a multiple a1 = a2 ,

π + 2mπ (m = 0, ±1, ±2, . . .), 2 then the ellipse in (8.6)-(8.7) degenerates into a circle, i.e. δ=±

Ex = a cos β, π

Ey = a cos β + 2mπ ± = ∓a sin β. 2 By squaring these two equations, we get Ex2 + Ey2 = a2 .

(8.21) (8.22)

(8.23) (8.24)

(8.25)

We have right-handed circular polarisation when Ey = −a sin β and left-handed circular polarisation when Ey = +a sin β.

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Chapter 9

Reﬂection and refraction of a plane wave

k

r

k

θr

t θt

^ n=^ ez θi

ki

ε2 , µ

ε1, µ 1

2

Figure 9.1: Reﬂection and refraction of a plane wave at a plane interface between two diﬀerent media. Illustration of propagation directions and angles of incidence, reﬂection, and transmission. We let a plane wave be incident upon a plane interface between two diﬀerent media, as shown in Fig. 9.1. The incident wave gives rise to a reﬂected wave and a transmitted wave, which we assume are plane waves as well. Thus, each component of E or H can be written q

Aqj = Re{aqj ei(k

·r−ωt)

} (j = x, y, z) ,

(9.0.1)

where A stands for E or H and q = i, r, t, so that ki , kr , and kt are the wave vectors of the incident, reﬂected, and transmitted waves, respectively.

9.1

Reﬂection law and refraction law (Snell’s law)

The existence of continuity conditions that E and H must satisfy at the interface between the two media in Fig. 9.1, implies that when r represents a point at the interface, the argument in the exponential function in (9.0.1) must be the same for the reﬂected and transmitted waves as for the incident wave. Thus, we have

FYS 263

30 ^t =

θ

^n x ^b

^n

i

k

i

^ ^ ki x n b = ___________ i ^| | k x n

ˆ Figure 9.2: Reﬂection and refraction of a plane wave. Illustration of the co-ordinate system (ˆ n, b, ˆt).

ki · r − ωt = kr · r − ωt = kt · r − ωt,

(9.1.1)

ki · r = kr · r = kt · r.

(9.1.2)

or

ˆ and ˆt represent ˆ , b, Now we introduce a Cartesian co-ordinate system in which the unit vectors n ˆ point along the interface normal into the medium of the a right-handed system (Fig. 9.2). Let n ˆ and ˆt be deﬁned by refracted wave, and let b i ˆ ˆ = k ×n ˆ ˆ × b. b ; ˆt = n i ˆ| |k × n

(9.1.3)

ˆ = 0, ˆ ; kbi = ki · b ki = ktiˆt + kni n

(9.1.4)

ˆ ˆ + kbr b, kr = ktr ˆt + knr n

(9.1.5)

ˆ ˆ + kbt b, kt = kttˆt + knt n

(9.1.6)

ˆ r = rtˆt + rb b.

(9.1.7)

In this co-ordinate system we have

ˆ i.e. b ˆ is normal Note that the co-ordinate system is deﬁned such that k has no component along b, i ˆ. to the plane of incidence, which is spanned by the vectors k and n Since i

ˆ = k i rt , ˆ ) · (rtˆt + rb b) ki · r = (ktiˆt + kni n t

(9.1.8)

ˆ · (rtˆt + rb b) ˆ = k r r t + k r rb , ˆ + kbr b) kr · r = (ktr ˆt + knr n t b

(9.1.9)

ˆ · (rtˆt + rb b) ˆ = k t rt + k t r b , ˆ + kbt b) kt · r = (kttˆt + knt n t b

(9.1.10)

it follows from the continuity condition (9.1.2) that kti rt = ktr rt + kbr rb = ktt rt + kbt rb .

(9.1.11)

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31

But since (9.1.11) shall apply to any point at the interface, i.e. to all values of rt and rb , we must have kbr = kbt = 0.

(9.1.12)

ˆ . Therefore, we have Thus, both kr and kt must lie in the plane of incidence spanned by ki and n kti = ktr = ktt = kt , i

r

(9.1.13)

t

which implies that the components of k , k , and k parallel to the interface are equal. By using ˆ t ; q = i, r, t, ˆ × kq = n ˆ ) = −bk ˆ × (ktˆt + knq n n

(9.1.14)

ˆ × ki = n ˆ × kr , n

(9.1.15)

ˆ × kt = n ˆ × ki . n

(9.1.16)

we ﬁnd that

Further, we use the relation |a × b| = |a||b| sin θ, where θ is the angle between the vectors a and b. Thus, we ﬁnd from (9.1.15) and Fig. 9.1 that k i sin θi = k r sin θr . i

(9.1.17)

r

Also, we know that k = k = n1 k0 , where n1 is the refractive index in medium 1, and k0 is the wave number in vacuum. The reﬂection law therefore becomes θi = θr ,

(9.1.18)

which in (9.1.15) is given in vectorial form. From (9.1.16) and Fig. 9.1 we get the refraction law or Snell’s law k i sin θi = k t sin θt .

(9.1.19)

which by using k i = n1 k0 and k t = n2 k0 , becomes n1 sin θi = n2 sin θt .

(9.1.20)

Equation (9.1.16) represents Snell’s law in vector form. Note that (9.1.15) and (9.1.16) contain more information than (9.1.18) and (9.1.20). From the vector equations it is clear that kr and kt lie in the plane of incidence.

9.2

Generalisation of the reﬂection law and Snell’s law

The reﬂection law and Snell’s law (the refraction law) can be generalised to include non-planar waves that are incident upon a non-planar interface. This is illustrated in Fig. 9.3, where the ﬁeld from a point source propagates towards a curved interface. Suppose now that the distance from the point source to the interface is much larger than the wavelength. Then at each point on the interface we may consider the incident wave to be a plane wave locally, and we may replace the interface locally by the tangent plane through the point in question. Then we can use Snell’s law and the reﬂection law as derived for a plane wave that is incident upon a plane interface, as illustrated in Fig. 9.3.

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32 kr

θ

local tangent plane

r kt

θ

i

θt ^n

ki n1

n2

Point source

Figure 9.3: Generalisation of Snell’s law and the reﬂection law to include non-planar waves that are incident upon a curved interface.

9.3

Reﬂection and refraction of plane electromagnetic waves

Note that the reﬂection law and the refraction law apply to all types of plane waves, i.e. to acoustic, electromagnetic, and elastic waves. In the derivation we have only used that kq · r − ωt (q = i, r, t) shall be the same for q = i, q = r, and q = t. Now we take a closer look at the reﬂection and refraction of plane electromagnetic waves in order to determine how much of the energy in the incident wave that is reﬂected and transmitted. We know that a plane electromagnetic wave is transverse, i.e. that both E and B = µH are normal to the propagation direction k = kˆs. In Fig. 9.1 we have chosen the z axis in the direction of the interface normal. If E is normal to the plane of incidence, we have s polarisation (from German, “Senkrecht”) or T E polarisation (“transverse electric” relative to the plane of incidence or the z axis). And if E is parallel with the plane of incidence, we have p polarisation or T M polarisation, since in this case B is normal to the plane of incidence or the z axis; hence the use of the term T M or “transverse magnetic”. A general time-harmonic, plane electromagnetic wave consists of both a T E and a T M component. With the time dependence e−iωt suppressed, we have for the spatial part of the ﬁeld E = ET E + ET M ; B = BT E + BT M , ˆz ik·r kt × e e , kt

(9.3.2)

ˆ)z ik·r k × (kt × e e , kkt

(9.3.3)

ET E = E T E ET M = E T M

ˆz ) ik·r k × (kt × e 1 k × ET E = E T E e , k0 k0 kt

(9.3.4)

1 1 ˆz )]eik·r . k × ET M = E T M k × [k × (kt × e k0 k0 kkt

(9.3.5)

BT E = BT M =

(9.3.1)

ˆz )] = k[k · (kt × e ˆz )] − (kt × e ˆz )k · k = −k 2 kt × e ˆz , we get But since k × [k × (kt × e BT M =

ˆz ik·r −k T M kt × e E e . k0 kt

(9.3.6)

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33

Note that the vectors ˆT E = e

ˆz kt × e kt

ˆT M = ; e

ˆz ) k × (kt × e , kkt

(9.3.7)

are unit vectors in the directions of ET E and ET M , respectively. We represent each of the incident, reﬂected, and transmitted ﬁelds in the manner given above, so that (q = i, r, t) Eq = ET Eq + ET M q ; Bq = BT Eq + BT M q , ˆz ikq ·r kt × e e , kt

(9.3.9)

ˆz ) ikq ·r kq × (kt × e e , k q kt

(9.3.10)

ET Eq = E T Eq ET M q = E T M q BT Eq =

(9.3.8)

ˆz ) ikq ·r k q T Eq kq × (kt × e E e , q k0 k kt

(9.3.11)

ˆz ikq ·r −k q T M q kt × e E e , k0 kt

(9.3.12)

BT M q = where

ˆz ; kt = kx e ˆx + ky e ˆy , ki = kt + kz1 e

(9.3.13)

ˆz ; kt = kt + kz2 e ˆz , kr = kt − kz1 e

(9.3.14)

 kq =

k1 = n1 k0 k2 = n2 k0

for q = i, r for q = t.

(9.3.15)

The continuity conditions that must be satisﬁed at the interface z = 0 are that the tangential components of E and H = µ1 B be continuous, i.e.

ˆz × ET Ei + ET Er − ET Et + ET M i + ET M r − ET M t = 0, e  ˆz × e

  1  T Ei 1  T Mi 1 T Et 1 T Mt + BT Er − B + + BT M r − B B B µ1 µ2 µ1 µ2

(9.3.16)

 = 0.

(9.3.17)

Further, we have ˆz × [kq × (kt × e ˆz )] = (kq · e ˆz ) e ˆ z × kt , e

(9.3.18)

ˆz × (kt × e ˆz ) = kt . e

(9.3.19)

By substituting from (9.3.9)-(9.3.12) into the boundary conditions (9.3.16)-(9.3.17) and using (9.1.2) and (9.3.18)-(9.3.19), we get  

kz1 T M i kz1 T M r kz2 T M t ˆz × kt kt E T Ei + E T Er − E T Et + e E − E − E = 0, (9.3.20) k1 k1 k2     1 kz1 T Ei kz1 T Er 1 kz2 T Et ˆz × kt e − E − E E µ1 k0 k0 µ2 k0     1 −k1 T M i k1 T M r −k2 T M t +kt − = 0. (9.3.21) E − E E µ1 k0 k0 k0

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34

ˆz × kt are orthogonal vectors, the expression inside each of the {} parentheses in Since kt and e (9.3.20) and (9.3.21) must vanish, i.e. E T Ei + E T Er = E T Et ,

(9.3.22)

  kz1 µ2 E T Ei − E T Er = kz2 µ1 E T Et ,

(9.3.23)

  kz1 k2 E T M i − E T M r = kz2 k1 E T M t ,

(9.3.24)

  k1 µ2 E T M i − E T M r = k2 µ1 E T M t .

(9.3.25)

Now we deﬁne reﬂection and transmission coeﬃcients as RT E =

E T Er E T Ei

; TTE =

E T Et , E T Ei

(9.3.26)

RT M =

ET Mr ET Mi

; TTM =

ET Mt , ET Mi

(9.3.27)

kz2 µ1 T E T , kz1 µ2

(9.3.28)

so that (9.3.22)-(9.3.25) give 1 + RT E = T T E ; 1 − RT E = 1 − RT M =

kz2 k1 T M k2 µ1 T M T ; 1 + RT M = T . kz1 k2 k1 µ2

(9.3.29)

The two equations in (9.3.28) have the following solution RT E =

µ2 kz1 − µ1 kz2 µ2 kz1 + µ1 kz2

; TTE =

2µ2 kz1 , µ2 kz1 + µ1 kz2

(9.3.30)

; TTM =

2k1 k2 µ2 kz1 . k22 µ1 kz1 + k12 µ2 kz2

(9.3.31)

whereas the two equations in (9.3.29) give k22 µ1 kz1 − k12 µ2 kz2 k22 µ1 kz1 + k12 µ2 kz2

RT M =

The interpretation of the reﬂection and transmission coeﬃcients follow from (9.3.26)-(9.3.27). Thus, the reﬂection coeﬃcient represents the amplitude ratio between the reﬂected and the incident E ﬁeld, whereas the transmission coeﬃcient represents the amplitude ratio between the transmitted and the incident E ﬁeld. Note that (9.3.22)-(9.3.23) and (9.3.28) contain only T E quantities, whereas equations (9.3.24)(9.3.25) and (9.3.29) contain only T M quantities. This implies that these two wave types are independent or de-coupled upon reﬂection and refraction. Thus, an incident T E plane wave produces a reﬂected T E plane wave and a transmitted T E plane wave, whereas an incident T M plane wave produces a reﬂected T M plane wave and a transmitted T M plane wave. Upon reﬂection and refraction there is no coupling between T E and T M waves. From Fig. 9.1 it follows that ˆz = k1 cos θi ; kz2 = kt · e ˆz = k2 cos θt , kz1 = ki · e

(9.3.32)

so that if µ1 = µ2 = 1 the reﬂection and transmission coeﬃcients become TTM =

2n1 cos θi n2 cos θi + n1 cos θt

TTE =

; RT M =

n2 cos θi − n1 cos θt , n2 cos θi + n1 cos θt

2n1 cos θi n1 cos θi − n2 cos θt ; RT E = , i t n1 cos θ + n2 cos θ n1 cos θi + n2 cos θt

(9.3.33) (9.3.34)

FYS 263

35 ^e TMt e^ TMr

k

r

kt

k

i

r i θ =θ

k

t

θt

^ ez θi

^e TMi

n1

ki

n2

^e TEi = ^e TEr = ^e TEt

=

^e TE

z=0

Figure 9.4: Reﬂection and refraction of a plane electromagnetic wave at a plane interface between two diﬀerent media. Illustration of T E and T M components of the electric ﬁeld.

These expressions are called the Fresnel formulas. By using Snell’s law (9.1.20), we can rewrite them as (Exercise 9) TTM =

2 sin θt cos θi tan(θi − θt ) ; RT M = , t i t + θ ) cos(θ − θ ) tan(θi + θt )

sin(θi

TTE =

2 sin θt cos θi sin(θi + θt )

; RT E = −

sin(θi − θt ) . sin(θi + θt )

(9.3.35) (9.3.36)

At normal incidence where θi = θt = 0, we get from (9.3.33) and (9.3.34) TTE = TTM =

n2 2 n−1 ; RT M = −RT E = ; n= . n+1 n+1 n1

(9.3.37)

The fact that RT M = −RT E at normal incidence follows from the way in which ET E and ET E are deﬁned. From Fig. 9.4 we see that these two vectors point in opposite directions at normal incidence.

9.3.1

Reﬂectance and transmittance

ˆT M q (q = i, r, t) and e ˆT E for T M and T E polarisation. Fig. 9.4 shows the polarisation vectors e These unit vectors are parallel with the electric ﬁeld and follow from (9.3.9)-(9.3.12) ˆT Ei = e ˆT Er = e ˆT Et = e ˆT E = e ˆT M q = e

ˆz kt × e kt

; |ˆ eT E | = 1,

ˆz ) kq × (kt × e ; |ˆ eT M q | = 1. k q kt

(9.3.38) (9.3.39)

ˆT M q , be αq [see Fig. 9.5], Let the angle between Eq and the plane of incidence spanned by kq and e so that ˆT E E q sin αq + e ˆT M q E q cos αq . Eq = e

(9.3.40)

FYS 263

36 E

q

^e TMq

^e TEq E TEq q α E TMq

q k

Figure 9.5: Illustration of the angle αq between the electric vector Eq and the plane of incidence ˆT M q . spanned by kq and e

Further, we let J i , J r , and J t denote the energy ﬂows of respectively the incident, reﬂected, and transmitted ﬁelds per unit area of the interface. Then we have J pq = S pq cos θq ; p = T E, T M ; q = i, r, t,

(9.3.41)

where S pq is the absolute value of the Poynting vector, given by q ε c pq c pq pq c 2 pq pq S = (E pq ) . (9.3.42) |E × H | = E H = 4π 4π 4π µq √ √ Here we have used the relation εq E pq = µq H pq . The reﬂectance Rp (p = T E, T M is the ratio between the reﬂected and incident energy ﬂows. From (9.3.41)-(9.3.42) we have RT M = RT E =

JT Mr |E T M r |2 = = (RT M )2 . T M i J |E T M i |2

(9.3.43)

J T Er |E T Er |2 = = (RT E )2 , J T Ei |E T Ei |2

(9.3.44)

Thus, the reﬂectance Rp is equal to the square of reﬂection coeﬃcient Rp . The transmittance T p (p = T E, T M ) is the ratio between the transmitted and incident energy ﬂows, and (9.3.41)-(9.3.42) give T TM =

JT Mt n2 µ1 cos θt T M 2 = (T ) , JT Mi n1 µ2 cos θi

(9.3.45)

T TE =

J T Et n2 µ1 cos θt T E 2 = (T ) . J T Ei n1 µ2 cos θi

(9.3.46)

Thus, the transmittance T p is proportional to the square of the transmission coeﬃcient T p (p = T E, T M ). When µ2 = µ1 = 1, we ﬁnd on substitution from (9.3.35)-(9.3.36) into (9.3.43)-(9.3.46) the following expressions for the reﬂectance and the transmittance RT M =

tan2 (θi − θt ) , tan2 (θi + θt )

(9.3.47)

FYS 263

37 sin2 (θi − θt ) , sin2 (θi + θt )

(9.3.48)

sin 2θi sin 2θt , sin2 (θi + θt ) cos2 (θi − θt )

(9.3.49)

sin 2θi sin 2θt . sin2 (θi + θt )

(9.3.50)

RT E = T TM =

T TE = By use of these formulas one can show that

RT M + T T M = 1 ; RT E + T T E = 1,

(9.3.51)

so that for each of the two polarisations the sum of the reﬂected energy and the transmitted energy is equal to the incident energy. From (9.3.41) and (9.3.42) we have ε1 pi 2 c J pi = |E | cos θi , (9.3.52) 4π µ1 which by the use of (9.3.40) gives

JT Mi

c 4π

ε1 T Ei 2 c |E | = cos θi µ1 4π

ε1 2 2 i E sin α , µ1 c ε1 T M i 2 c ε1 2 = cos θi |E | = cos θi E cos2 αi . 4π µ1 4π µ1

J T Ei = cos θi

(9.3.53) (9.3.54)

But since the total incident energy ﬂow is given by J i = cos θi

c 4π

ε1 2 E , µ1

(9.3.55)

we ﬁnd J T Ei = J i sin2 αi ; J T M i = J i cos2 αi .

(9.3.56)

Thus, we have R=

Jr J T M r + J T Er JT Mr J T Er = = T M i cos2 αi + T Ei sin2 αi , i i J J J J

(9.3.57)

which gives R = RT M cos2 αi + RT E sin2 αi ,

(9.3.58)

T = T T M cos2 αi + T T E sin2 αi .

(9.3.59)

and similarly we ﬁnd

i

t

At normal incidence, θ = θ = 0, and the distinction between T E and T M polarisation disappears. From (9.3.43)-(9.3.46) combined with (9.3.33)-(9.3.34), we ﬁnd (when µ1 = µ2 = 1)  2 R = RT M = RT E = (RT E )2 = RT M =



 2  2 T = T TM = T TE = TTE = TTM =

n−1 n+1

2

4n (n + 1)2

; n= ; n=

n2 , n1

n2 . n1

(9.3.60) (9.3.61)

When n → 1, we see that R → 0 and T → 1, as expected. Similarly, we ﬁnd from (9.3.47)-(9.3.50) that R → 0, R⊥ → 0, T → 1, T⊥ → 1 when n → 1.

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38

kr

ki

θ iB

θ iB

1 2 θ tB

kt

Figure 9.6: Illustration of Brewster’s law.

9.3.2

Brewster’s law

From (9.3.47) it follows that RT M = 0 when θi + θt = π2 , since then tan(θi + θt ) = ∞. We call this particular angle of incidence θiB and the corresponding refraction or transmission angle θtB . By using Snell’s law (9.1.20), we ﬁnd

π n2 sin θtB = n2 sin − θiB = n2 cos θiB = n1 sin θiB , (9.3.62) 2 so that RT M = 0 when θi = θiB , where θiB is given by tan θiB =

n2 = n. n1

(9.3.63)

The angle θiB is called the polarisation angle or the Brewster angle. When the angle of incidence is equal to θiB , the E vector of the reﬂected light has no component in the plane of incidence (Fig. 9.6). This fact is exploited in sunglasses with polarisation ﬁlter. The ﬁlter is oriented such that only light that is polarised vertically (Fig. 9.6) is transmitted. Thus, one avoids to a certain degree annoying reﬂections from e.g. a water surface. Note that kr · kt = 0, i.e. kr and kt are normal to one another when θi = θiB , as shown in Fig. 9.6.

9.3.3

Unpolarised light (natural light)

For natural light, e.g. light from an incandescent lamp, the direction of the E vector varies very rapidly in an arbitrary or irregular manner, so that no particular direction is given preference. The average reﬂectance R is obtained by averaging over all directions α. Since the average value of both sin2 α and cos2 α is 12 , we ﬁnd from (9.3.56) that J

T Mi

= J i cos2 αi = J

T Ei

= J i sin2 αi =

1 i J . 2

(9.3.64)

For the reﬂected components we ﬁnd J

T Mr

=

J J

T Mr T Mi

·J

T Mi

=

J J

T Mr T Mi

1 1 · J i = RT M J i , 2 2

(9.3.65)

FYS 263

39

J

T Er

=

J

T Er T Ei

1 1 · J i = RT E J i , 2 2

J which shows that the degree of polarisation for the reﬂected light can be deﬁned as   TM R − RT E  |J T M r − J T Er | r  P =  TM = T Mr . R + RT E  J + J T Er

(9.3.66)

(9.3.67)

The average reﬂectance is given by R=

J J

r i

=

J

T Mr

+J J

T Er

J

=

i

T Mr

2J

T Mi

+

J

T Er

2J

T Ei

=

 1  TM + RT E , R 2

(9.3.68)

so that the degree of polarisation becomes 1 1 TM − RT E |, |R R2 where |RT M − RT E | is called the polarised part of the reﬂected light. Similarly, we ﬁnd for the transmitted light Pr =

T =

9.3.4

1 TM 1 1 TM + T TE) ; P t = − T T E |. (T |T 2 T 2

(9.3.69)

(9.3.70)

Rotation of the plane of polarisation upon reﬂection and refraction

Note that if the incident light is linearly polarised, then also the reﬂected and the transmitted light will be linearly polarised, since the phases only change by 0 or π. This follows from the fact that the reﬂection and transmission coeﬃcients are real quantities [cf. (9.3.33)-(9.3.36)]. But the planes of polarisation for the reﬂected and the transmitted light are rotated in opposite directions relative to the polarisation plane of the incident light. The angles αi , αr , and αt that the planes of polarisation of the incident, reﬂected, and transmitted light form with the plane of incidence, are given by [cf. Fig. 9.5] tan αi =

E T Ei , ET Mi

(9.3.71)

E T Er tan α = T M r = E

E T Er E T Ei ET M r ET M i

E T Ei RT E = tan αi , ET Mi RT M

(9.3.72)

E T Et tan α = T M t = E

E T Et E T Ei ET M t ET M i

E T Ei TTE = tan αi . ET Mi TTM

(9.3.73)

r

t

By use of the Fresnel formulas (9.3.35)-(9.3.36) we can write tan αr = −

cos(θi − θt ) tan αi , cos(θi + θt )

tan αt = cos(θi − θt ) tan αi . Since 0 ≤ θi ≤

π 2

and 0 ≤ θt ≤

π 2,

(9.3.74) (9.3.75)

we get | tan αr | ≥ | tan αi |,

(9.3.76)

| tan αt | ≤ | tan αi |. i

(9.3.77) t

In (9.3.76) the equality sign applies at normal incidence (θ = θ = 0) and at grazing incidence (θi = π 2 ), whereas in (9.3.77) the equality sign applies only at normal incidence. These two inequalities

FYS 263

40

imply that upon reﬂection the plane of polarisation is rotated away from the plane of incidence, whereas upon transmission it is rotated towards the plane of incidence. Note that when θi = θiB , so that θiB + θtB = π2 , then tan αr = ∞. Thus, we have αr = π2 in accordance with Brewster’s law.

9.3.5

Total reﬂection

Snell’s law (9.1.20) can be written in the form sin θt =

sin θi n

; n=

n2 = n1

ε2 µ2 . ε1 µ1

(9.3.78)

Hence, it follows that if n < 1, then we get sin θt = 1 when θi = θic , where sin θic = n.

(9.3.79)

This implies that when θi = θic , we get θt = π2 , so that the transmitted light propagates along the interface. If θi ≥ θic , we have total reﬂection, i.e. no light will pass into the other medium. All light is then reﬂected. There exists a ﬁeld in the other medium, but there is no energy transport through the interface. When θi > θic , then sin θt > 1, which means that θt is complex. We have from (9.3.78)    sin2 θi ±i sin2 θi − n2 2 t t cos θ = ± 1 − sin θ = ±i −1= . (9.3.80) n2 n The lower sign in (9.3.80) must be discarded. Otherwise the ﬁeld in medium 2 would grow exponentially with increasing distance from the interface. The electric ﬁeld in medium 2 is t

ˆpt ei(k Ept = T p E pi e

·r−ωt)

(p = T E, T M ),

(9.3.81)

where kt · r = kx x + ky y + kz2 z,

(9.3.82)

with (cf. Fig. 9.1 and (9.3.80) with upper sign) kz2 = k2 cos θt = ik2

1 2 i n2 sin θ − n2 ; n = , n n1

(9.3.83)

so that k2  2 i sin θ − n2 . (9.3.84) n We see that Ept represents a wave that propagates along the interface and is exponentially damped with the distance z into medium 2. From ∇ · Dt = ε2 ∇ · Et = 0 it follows that t

eik

·r

= ei(kx x+ky y) e−|kz2 |z ; |kz2 | =

kt · Et = 0,

(9.3.85)

which gives Ezt = −

(kx Ext + ky Eyt ) . kz2

(9.3.86)

If we let the plane of incidence coincide with the xz plane, we have (cf. Fig. 9.7) kx = −k1 sin θi ; ky = 0,

(9.3.87)

Eyt = E T Et ei(kx x−ωt) e−|kz2 |z = T T E E T Ei ei(kx x−ωt) e−|kz2 |z ,

(9.3.88)

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41

E TMr kt kr

θi

θt

k

E TMt

i

θt

z θi E TMi

E TEi = E TEr = E TEt

ki

x Figure 9.7: Illustration of the refraction of a plane wave into an optically thinner medium, so that θi < θt . When θi → θic , then θt → π/2, and we get total reﬂection.

Ext = −E T M t cos θt ei(kx x−ωt) e−|kz2 | = −T T M E T M i cos θt ei(kx x−ωt) e−|kz2 |z , Ezt = −

kx t kx T M T M i i(kx x−ωt) −|kz2 |z Ex = T E e e . kz2 k2

(9.3.89) (9.3.90)

From these expressions for the components of Et and corresponding expressions for the components of Ht one can show (Exercise 11) that the time average of the z component of the Poynting vector is zero, which implies that there is no energy transport through the interface, as asserted earlier. The reﬂection coeﬃcients in (9.3.35)-(9.3.36) can be written as follows RT M =

sin θi cos θi − sin θt cos θt , sin θi cos θi + sin θt cos θt

(9.3.91)

sin θi cos θt − sin θt cos θi . sin θi cos θt + sin θt cos θi

(9.3.92)

RT E = −

By combining Snell’s law (9.3.78) and (9.3.80) with the upper sign with (9.3.91)-(9.3.92), we get  n2 cos θi − i sin2 θi − n2 TM  R = , (9.3.93) n2 cos θi + i sin2 θi − n2  i − i sin2 θi − n2 cos θ  RT E = . (9.3.94) cos θi + i sin2 θi − n2 Since both reﬂection coeﬃcients are of the form z/z ∗ , where z is a complex number, it follows that |RT M | = |RT E | = 1,

(9.3.95)

FYS 263

42

which shows that for each polarisation the intensity of the totally reﬂected light is equal to the intensity of the incident light. But the phase is altered upon total reﬂection. Letting Rp =

p p E pr zp = eiδ = p = e2iα (p = T E, T M ), pi E z ∗

(9.3.96)

where [cf. (9.3.93)-(9.3.94)]  TM z T M = n2 cos θi − i sin2 θi − n2 = |z T M |eiα ,

(9.3.97)

 TE z T E = cos θi − i sin2 θi − n2 = |z T E |eiα ,

(9.3.98)

we ﬁnd  tan α

TM

= tan 

tan α

TE

= tan

1 TM δ 2 1 TE δ 2



 =−



 =−

sin2 θi − n2 , n2 cos θi

(9.3.99)

sin2 θi − n2 . cos θi

(9.3.100)

The relative phase diﬀerence δ = δT E − δT M ,

(9.3.101)

is determined by  tan

1 δ 2



    tan 12 δ T E − tan 12 δ T M    , = 1 + tan 12 δ T E tan 12 δ T M

which upon substitution from (9.3.99)-(9.3.100) gives    1 cos θi sin2 θi − n2 tan . δ = 2 sin2 θi

(9.3.102)

(9.3.103)

We see that δ = 0 for θi = π2 (grazing incidence) and θi = θic (critical angle of incidence). Between these two values there is an angle of incidence θi = θim which gives a maximum phase diﬀerence δ = δ m , where θim is determined by  dδ  = 0. (9.3.104) dθi θim From (9.3.104) we ﬁnd 2n2 , 1 + n2 which upon substitution in (9.3.103) gives (Exercise 10)   1 m 1 − n2 tan = δ . 2 2n sin2 θim =

(9.3.105)

(9.3.106)

If the phase diﬀerence δ is equal to ± π2 and in addition E T M i = E T Ei , the totally reﬂected light will be circularly polarised. By choosing the angle αi between the polarisation plane and the plane of incidence equal to 45◦ , we make E T M i equal to E T Ei . In order to obtain δ = π2 we must have    δm  δm π ≥ tan π4 = 1, which according to (9.3.106) implies that 2 ≥ 4 . This means that tan 2 n2 + 2n − 1 ≤ 0. By completing the square on the left-hand side of (9.3.107), we ﬁnd that

(9.3.107)

FYS 263

43

n≤

2−1 ;

√ 1 n1 ≥ 2 + 1 = 2.41. = n n2

Thus, nn12 must exceed 2.41 in order that we shall obtain a phase diﬀerence of reﬂection.

(9.3.108) π 2

in one single