Plane waves in dielectrics, dielectric-dielectric boundaries

Problem:

In a purely classical model we consider a dielectric medium as a collection of uncoupled classical harmonic oscillators.  Assume that each oscillator consists of an electron connected to a fixed ion by a harmonic spring with frequency ω0.
(a)  Write down and solve the equation of motion for the electron when a monochromatic electric field with frequency ω is applied.
(b)  For an electron density n, calculate the electric polarization and the dielectric constant ε(ω).
(c)  For a free electron gas, at what frequency is ε = 0?  What is the physical significance of this frequency?

Solution:

• Concepts:
The driven harmonic oscillator, polarization, the plasma frequency
• Reasoning:
We model a medium interacting with an electromagnetic as a collection of driven harmonic oscillators.
• Details of the calculation:
(a)  Let E = E0 exp(-iωt).
The electric field and the spring exert a force on the electron.  The total force is
F = -qe E - mω02r =  md2r/dt2.
To solve this differential equation try a solution of the form
r = r0 exp(-iωt).  This yields
r0 = -(qeE0/m)/(ω02 - ω2).
(b)  The induced dipole moment is p = -qer = (qe2E0/m)/(ω02 - ω2) exp(-iωt).
Polarization = dipole moment per unit volume
P = np = n(qe2E/m)/(ω02 - ω2) = ε0χeE.
ε(ω) = ε0(1 + χe) = ε0 + n(qe2/m)/(ω02 - ω2).
(c)  For a free electron ω0 = 0.
We then have ε(ω) = ε0 - nqe2/(m ω2) = ε0(1 - nqe2/(ε0m ω2)) = ε0(1 - ωp22).
When ω = ωp the ε(ω) = 0.  ωp = (nqe2/(ε0m))½ is called the plasma frequency.  The plasma is opaque to EM waves with frequencies less than ωp and transparent to EM waves with frequencies greater than ωp.

Dielectric-dielectric boundaries

Problem:

A plane electromagnetic wave is incident normally from vacuum onto a plane (uniform, isotropic, non permeable, loss-less) dielectric interface.
(a)  Formulate the problem in terms of Maxwell's equations with the appropriate boundary conditions.
(b)  Determine the amplitude of reflected and transmitted waves.
(c)  Determine (i) the ratio of the reflected to the incident intensity, and (ii) the ratio of the transmitted to the incident intensity.

Solution:

• Concepts:
Maxwell's equations, boundary conditions for the electric and magnetic fields
• Reasoning:
The intensities of the reflected and transmitted waves are proportional to the squares of the corresponding electric field amplitudes.  We find the ratio of these amplitudes to the incident amplitude by applying boundary conditions.
• Details of the calculation:
(a)  Maxwell's equations in an ideal linear, isotropic, homogeneous dielectric medium are
∇∙E = ρf/ε,  ×E = -∂B/∂t,  ∇∙B = 0,  ×B = μjf + εμ∂E/∂t.
You may replace E by D/ε and B by μH.
Assume ρf and jf are zero in the medium.  Then
×(×E) = (∇∙E) - 2E = -(∂/∂ t)(×B) = -με∂2E/∂t2.
2E = με∂2E/∂t2.
×(×B) = (∇∙B) - 2B = με(∂/∂ t)(×E) = -με∂2B/∂t2.
2B = με∂2B/∂t2.
Plane wave solutions E = E((k/k)∙r - vt).  B = B((k/k)∙r - vt) exist.  (v = (με)).
We have  EB.  Ek.  Bk.  B = (με)½(k/k)×E.
k2 = μεω2   (SI units).
Sinusoidal plane waves are of the form  E = E0ei(k∙r - ωt)B = B0ei(k∙r - ωt).
Maxwell's equation yield the boundary conditions
 (SI units) (D2 - D1)∙n2 = σf (B2 - B1)∙n2 = 0 (E2 - E1)∙t = 0 (H2 - H1)∙t = kf∙n

At a dielectric-dielectric interface σf and kf are zero

If we set up the problem as shown in the figure above the boundary conditions yield
(E2 - E1)∙t = 0.  (E2 - E1)(x/x) = 0.  Ei - Er = Et.
(H2 - H1)∙t = kf∙n, (H2 - H1)(y/y) = 0, there are no free surface currents.
H2 = B22H1 = B11.
B = (1/v)(k/k)×EBi = (y/y)Ei(n1/c), Br = (y/y)Ei(n1/c), Bt = (y/y)Et(n2/c).
(n11)(Ei + Er)  = (n22)Et.
(b) (n11)(Ei + Er)  = (n22)(Ei - Er).
Er/Ei = (n2μ1 - n1μ2)/(n2μ1 + n1μ2) = r12  is the reflection coefficient.
Let β = (μ1n22n1) = (μ12)(ε2μ2)½/(ε1μ1)½ = (ε2μ1)½/(ε1μ2)½.
Then we can write
(β - 1)/(β + 1) = r12.

(n11)(2Ei - Et)  = (n22)Et.
Et/Ei = 2n1μ2/(n1μ2 + n2μ1) = t12  is the transmission coefficient.
We can write
2/(1 + β) = t12.
(c) The reflectance is  R = <Sr>/<Si> = |r12|2,
and the transmittance is  T = <St>∙n/<Si>∙n where n is a unit vector normal to the interface.
T = <St>/<Si> = (μ1n22n1)|t12|2.  R + T = 1.
[<S> = (|E02|/2)(ε/μ)½] =  (|E02|/2)(n/(cμ).]

Problem:

A plane electromagnetic wave is incident upon a a plane dielectric-dielectric interface as shown below.

The polarization is normal to the plane defined by the incident, reflected, and transmitted wave vector.
(a)  Show that Snell's law and the law of reflection result from application of the appropriate boundary conditions.
(b)  Show that the reflection coefficient is given by Er/Ei = sin(θt - θi)/sin(θt + θi).

Solution:

• Concepts:
Maxwell's equations, boundary conditions for the electric and magnetic fields
• Reasoning:
The intensities of the reflected and transmitted waves are proportional to the squares of the corresponding electric field amplitudes.  We find the ratio of these amplitudes to the incident amplitude by applying boundary conditions.
• Details of the calculation:
(a)  The y-direction is the direction of the polarization, we have s-polarization.
E
i(r,t) = j Ei exp(i(ki∙r - ωt)).
E
r(r,t) = j Er exp(i(kr∙r - ωt)).
E
t(r,t) = j Et exp(i(kt∙r - ωt)).
The tangential component of E is continuous across the boundary.  The y-direction is tangential to the interface.  Therefore Ei exp(i(ki∙r)) + Er exp(i(kr∙r)) = Et exp(i(kt∙r)) for all r in the z = 0 plane.
This is only possible if ki∙r = kr∙r = kt∙r for all r in the z = 0 plane.
Let r = i, then kix = krx = ktx.  or   kisin(θi) = krsin(θr) = ktsin(θt).
ki = kr = (με)½ω = n1ω/c.  kt = (με')½ω = n2ω/c.  Therefore
n1sin(θi) = n1sin(θr) = n2sin(θt).
This implies θi = θr (law of reflection) and n1sin(θr) = n2sin(θt) (Snell's law).
(b)  The tangential component of H = B/μ is continuous across the boundary.  The x-direction is tangential to the interface.  Since μ is the same on both sides of the interface, the tangential component of B is continuous across the boundary.
Bi = ki×Ei/ω.  Bi∙i = (1/ω)(ki×Ei)∙i = (1/ω)(Ei×i)∙ki = -(1/ω)Eikiz = -(1/ω)Eikicos(θi).
Similarly Br∙i = (1/ω)Erkrcos(θr).  Bt∙i = -(1/ω)Etktcos(θt).
We then have Eikicos(θi) - Erkrcos(θr) = Etktcos(θt),
or  n1Eicos(θi) - n1Ercos(θr) = n2Etcos(θt).
Combining this equation with Ei + Er = Et and θi = θr yields
Er/Ei = (n1cosθi - n2cosθt)/(n1cosθi + n2cosθt)
= (cosθi - (sinθi/sinθt)cosθt)/(cosθi + (sinθi/sinθt)cosθt)
= (1 - (tanθi/tanθt))/(1 + (tanθi/tanθt))
= (tanθt - tanθi)/(tanθt + tanθi)
= sin(θt - θi)/sin(θt + θi).
[sin(a - b)/sin(a + b) = (sina cosb - cosa sinb)/(sina cosb + cosa sinb)
= sina cosb/(sina cosb + cosa sinb) - cosa sinb/(sina cosb - cosa sinb)
= 1/(1 + tanb/tana) - 1/(1 + tana/tanb)
= (tana - tanb)/(tana + tanb)].

Problem:

Fresnel's reflectance formulas are given by

R = |sin(θi - θr)/sin(θi + θr)|2   or   R = |tan(θi - θr)/tan(θi + θr)|2

depending on the polarization of the incident wave with respect to the plane of incidence, and with θi and θr the angles of incidence and refraction, respectively.
(a)  Specify the boundary conditions across an interface.
(b)  For simplicity assume μ1 = μ2 = μ0, ε1 = ε0, ε2 = ε, and derive the reflectance formulas.
(c)  For polarization parallel to the plane of incidence, find the Brewster angle for an index of refraction of n2 = 1.50, and comment on the polarization of the reflected radiation for a wave of mixed polarization incident on a plane interface at the Brewster angle.

Solution:

• Concepts:
Maxwell's equations, boundary conditions for the electric and magnetic fields
• Reasoning:
The intensities of the reflected and transmitted waves are proportional to the squares of the corresponding electric field amplitudes.  We find the ratio of these amplitudes to the incident amplitude by applying boundary conditions.
• Details of the calculation:
(a)  At a dielectric-dielectric interface Maxwell's equation yield the boundary conditions
 (SI units) (D2 - D1)∙n2 = σf (B2 - B1)∙n2 = 0 (E2 - E1)∙t = 0 (H2 - H1)∙t = kf∙n

(b)  In the figure below, the y-direction is the direction of the polarization for s-polarization.

Ei(r,t) = j Ei exp(i(ki∙r - ωt)).
E
r(r,t) = j Er exp(i(kr∙r - ωt)).
E
t(r,t) = j Et exp(i(kt∙r - ωt)).
The tangential component of E is continuous across the boundary.
The y-direction is tangential to the interface.
Therefore Ei exp(i(ki∙r)) + Er exp(i(kr∙r)) = Et exp(i(kt∙r)) for all r in the z = 0 plane.
This is only possible if ki∙r = kr∙r = kt∙r for all r in the z = 0 plane.
Let r = i, then kix = krx = ktx,   or   kisin(θi) = krsin(θr) = ktsin(θt).
ki = kr = (μ0ε0)½ω = ω/c.  kt = (μ0ε')½ω = n2ω/c.  Therefore
sin(θi) = sin(θr) = n2sin(θt).
This implies θi = θr (law of reflection) and sin(θr) = n2sin(θt) (Snell's law) for s-polarization.

The tangential component of B = μ0H is continuous across the boundary.  The x-direction is tangential to the interface.
B = k×E/ω,  Bi∙i = (1/ω)(ki×Ei)∙i = (1/ω)(Ei × i)∙ki = -(1/ω)Eikiz = -(1/ω)Eikicos(θi).
Similarly Br∙i = (1/ω)Erkrcos(θr),  Bt∙i = -(1/ω)Etktcos(θt).
We then have Eikicos(θi) - Erkrcos(θr) = Etktcos(θt),
or  Eicos(θi) - Ercos(θr) = n2Etcos(θt).
Combining this equation with Ei + Er = Et and θi = θr from above yields
Er/Ei = (cosθi - n2cosθt)/(cosθi + n2cosθt)
= (cosθi - (sinθi/sinθt)cosθt)/(cosθi + (sinθi/sinθt)cosθt)
= (1 - (tanθi/tanθt))/(1 + (tanθi/tanθt))
= (tanθt - tanθi)/(tanθt + tanθi)
= sin(θt - θi)/sin(θt + θi).
R =|Er/Ei|2 = |sin(θt - θi)/sin(θt + θi)|2.
[sin(a - b)/sin(a + b) = (sina cosb - cosa sinb)/(sina cosb + cosa sinb)
= sina cosb/(sina cosb + cosa sinb) - cosa sinb/(sina cosb - cosa sinb)
= 1/(1 + tanb/tana) - 1/(1 + tana/tanb)
= (tana - tanb)/(tana + tanb)].

In the figure below, the direction of B is tangential to the interface for p-polarization.

Bi(r,t) = -j Bi exp(i(ki∙r - ωt)),
B
r(r,t) = -j Br exp(i(kr∙r - ωt)),
B
t(r,t) = -j Bt exp(i(kt∙r - ωt)).

The tangential component of E is continuous across the boundary.
The x-direction is tangential to the interface.
Therefore the x-component of E is continuous across the boundary.
Ei exp(i(ki∙r))cos(θi)  - Er exp(i(kr∙r))cos(θr) = Et exp(i(kt∙r))cos(θt) for all r in the z = 0 plane.
This is only possible if ki∙r = kr∙r = kt∙r for all r in the z = 0 plane.
Let r = i, then kix = krx = ktx,   or   kisin(θi) = krsin(θr) = ktsin(θt).
ki = kr = (μ0ε0)½ω = ω/c.  kt = (μ0ε')½ω = n2ω/c.  Therefore
sin(θi) = sin(θr) = n2sin(θt).
This implies θi = θr (law of reflection) and sin(θr) = n2sin(θt) (Snell's law) for p-polarization.

The tangential component of B = μH is continuous across the boundary.
The y-direction is tangential to the interface.
Bi + Br = Bt.
B = k×E/ω,  Bi = (1/ω)Eiki, = Ei/c. Br = (1/ω)Erkr = Er/c, Bt = (1/ω)Etkt = n2Et/c.
We then have Ei + Er = n2Et.
Combining this equation with Ei cos(θi)  - Er cos(θr) = Et cos(θt) and θi = θr  ffom above yields
Er/Ei = (n2cosθi - cosθt)/(n2cosθi + cosθt)
= ((sinθi/sinθt)cosθi - cosθt)/((sinθi/sinθt)cosθi + cosθt)
= (sinθicosθi - cosθtsinθt)/(sinθicosθi + cosθtsinθt)
= tan(θi - θt)/tan(θi + θt).
R =|Er/Ei|2 = |tan(θi - θt)/tan(θt + θi)|2.
[(sina cosa - cosb sinb)/(sina cosa + cosb sinb)
= (sina cosa(sin2b + cos2b) - cosb sinb(sin2a + cos2a))/(sina cosa(sin2b + cos2b) + cosb sinb(sin2a + cos2a))
= (sin(a - b)cos(a + b)/[sin(a + b)cos(a - b)] = tan(a - b)/tan(a  +b)].

(c)  When θt + θi = п/2, the reflectance is zero for p-polarization.
This happens when θi = θB, n2sinθB  = n2sin((п/2) - θB), tanθB = n2/n1.
θB is called the Brewster angle.
When θi = θB, then only light with s-polarization is reflected.  For a wave incident with mixed polarization incident with θi = θB, the reflected light is polarized.
With n1 = 1, n2 = 1.5 we have θB = 56.31o.

Problem:

An electromagnetic wave passes through a boundary between two media with n1 = 1 and n2 = 3 at near-normal incidence of θi = 0.5 degree.
(a)  Find the angle θt of the transmitted wave
(b)  Find the reflectance and the transmittance for the wave.

Solution:

• Concepts:
Snell's law, the Fresnel reflection coefficients
• Reasoning:
Snell's law lets us calculate  and the Fresnel reflection coefficients let us calculate the reflected intensity.
• Details of the calculation:
(a)  Snell's law:  n1sinθ1 = n2sinθ2.
Small angle approximation: n1θ1 = n2θ2.
θ1 = θi = 0.5o =  8.727 mrad.  θ2 = θt = (1/3)*8.727 mrad =  0.1667o.

(b)  Fresnel reflection coefficient for p-polarization:  r12p = tan(θ1 - θ2)/tan(θ1 + θ2).
Fresnel reflection coefficient for s-polarization:  r12s = sin(θ1 - θ2)/sin(θ1 + θ2).

Small angle approximation: r12 = (θ1 - θ2)/(θ1 + θ2), independent of polarization.
The reflectance R = |r12|2 , and the transmittance T = 1 - R.
r12 = 0.5,  R = 0.25 = 25%,  T = 0.75 = 75%.

Problem:

A beam of light with wavelength 450 nm in vacuum is incident on a prism as shown in the figure, and totally reflected through 90o.  The index of refraction of the prism is 1.6.

(a)  Compute the distance beyond the long side of the prism at which the electric field strength is reduced to 1/e of its value just at the surface.  Assume the light is polarized so that E is perpendicular to the plane of incidence.
(b)  Is your answer changed if E lies in the plane of incidence?

Solution:

• Concepts:
Snell's law, total internal reflection, the evanescent wave
• Reasoning:
If n2 < n1 , then for  θi > θc = sin-1(n2/n1)  we have total internal reflection.  The wave vector kt then has a component parallel to the interface which is real and a component perpendicular to the interface which is imaginary.  We use Snell's law to calculate the components of the wave vector of the transmitted wave.
• Details of the calculation:
(a)  Choose a coordinate system.

The interface lies in the x-y plane.  ki makes an angle of 45o with the z-axis.  The critical angle for total internal reflection is θc = sin-1(n2/n1) = sin-1(1/1.6) = 38.7o, so we have total internal reflection.
n sinθi = sinθt = 1.1313
ktx = ktsinθt.  ktz = ktcosθt.
cosθt = (1 - sin2θt)½ = i*0.5295.
ktcosθt = i*0.5295*(2π/450 nm) = iβ.
Et = E0texp(i(kt∙r - ωt)) = E0te-βzexp(i(ktxx - ωt))
The electric field strength is reduced to 1/e of its value just at the surface when z = 1/β = 135 nm.
(b)  Polarization does not enter into the above calculation, so the result is independent of the direction of polarization.