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 ω_{0}.

(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**=**E**_{0}exp(-iωt).

The electric field and the spring exert a force on the electron. The total force is

**F**= -q_{e}**E**- mω_{0}^{2}**r**= md^{2}**r**/dt^{2}.

To solve this differential equation try a solution of the form

**r**=**r**_{0}exp(-iωt). This yields**r**_{0}= -(q_{e}**E**_{0}/m)/(ω_{0}^{2}- ω^{2}).

(b) The induced dipole moment is**p**= -q_{e}**r**= (q_{e}^{2}**E**_{0}/m)/(ω_{0}^{2}- ω^{2}) exp(-iωt).

Polarization = dipole moment per unit volume**P**= n**p**= n(q_{e}^{2}**E**/m)/(ω_{0}^{2}- ω^{2}) = ε_{0}χ_{e}**E**.

ε(ω) = ε_{0}(1 + χ_{e}) = ε_{0}+ n(q_{e}^{2}/m)/(ω_{0}^{2}- ω^{2}).

(c) For a free electron ω_{0}= 0.

We then have ε(ω) = ε_{0}- nq_{e}^{2}/(m ω^{2}) = ε_{0}(1 - nq_{e}^{2}/(ε_{0}m ω^{2})) = ε_{0}(1 - ω_{p}^{2}/ω^{2}).

When ω = ω_{p}the ε(ω) = 0. ω_{p}= (nq_{e}^{2}/(ε_{0}m))^{½}is called the plasma frequency. The plasma is opaque to EM waves with frequencies less than ω_{p}and transparent to EM waves with frequencies greater than ω_{p}.

__Dielectric-dielectric boundaries__

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**= ρ_{f}/ε,**∇**×**E**= -∂**B**/∂t,**∇∙B**= 0,**∇**×**B**= μ**j**_{f}+ εμ∂**E**/∂t.

You may replace**E**by**D**/ε and**B**by μ**H**.

Assume ρ_{f}and**j**_{f}are zero in the medium. Then

**∇**×(**∇**×**E**) =**∇**(**∇∙E**) -**∇**^{2}**E**= -(∂/∂ t)(**∇**×**B**) = -με∂^{2}**E**/∂t^{2}.

**∇**^{2}**E**= με∂^{2}**E**/∂t^{2}.

**∇**×(**∇**×**B**) =**∇**(**∇∙B**) -**∇**^{2}**B**= με(∂/∂ t)(**∇**×**E**) = -με∂^{2}**B**/∂t^{2}.

**∇**^{2}**B**= με∂^{2}**B**/∂t^{2}.

Plane wave solutions**E**=**E**((**k**/k)**∙r**- vt).**B**=**B**((**k**/k)**∙r**- vt) exist. (v = (με)^{-½}).

We have**E**⊥**B**.**E**⊥**k**.**B**⊥**k**.**B**= (με)^{½}(**k**/k)×**E**.

k^{2}= μεω^{2}(SI units).

Sinusoidal plane waves are of the form**E**=**E**_{0}e^{i(k∙r - ωt)}.**B**=**B**_{0}e^{i(k∙r - ωt)}.

Maxwell's equation yield the boundary conditions

(SI units) ( **D**_{2}-**D**_{1})**∙n**_{2}= σ_{f}( **B**_{2}-**B**_{1})**∙n**_{2}= 0( **E**_{2}-**E**_{1})**∙t**= 0( **H**_{2}-**H**_{1})**∙t**=**k**_{f}**∙n**At a dielectric-dielectric interface σ

_{f}and**k**_{f}are zero

If we set up the problem as shown in the figure above the boundary conditions yield

(**E**_{2}-**E**_{1})**∙t**= 0. (**E**_{2}-**E**_{1})**∙**(**x**/x) = 0. E_{i}- E_{r}= E_{t}.

(**H**_{2}-**H**_{1})**∙t**=**k**_{f}**∙n**, (**H**_{2}-**H**_{1})**∙**(**y**/y) = 0, there are no free surface currents.

**H**_{2}=**B**_{2}/μ_{2}.**H**_{1}=**B**_{1}/μ_{1}.

**B**= (1/v)(**k**/k)×**E**.**B**_{i}= (**y**/y)E_{i}(n_{1}/c),**B**_{r}= (**y**/y)E_{i}(n_{1}/c),**B**_{t}= (**y**/y)E_{t}(n_{2}/c).

(n_{1}/μ_{1})(E_{i}+ E_{r}) = (n_{2}/μ_{2})E_{t}.

(b) (n_{1}/μ_{1})(E_{i}+ E_{r}) = (n_{2}/μ_{2})(E_{i}- E_{r}).

E_{r}/E_{i}= (n_{2}- n_{1}μ_{2}/μ_{1})/(n_{2}+ n_{1}μ_{2}/μ_{1}) = r_{12}is the**reflection coefficient**.

Let β = (μ_{1}ε_{2}/μ_{2}ε_{1})^{½}= (μ_{1}ε_{2}μ_{2}/μ_{2}ε_{1}μ_{1})^{½}(μ_{1}/μ_{2})^{½}= (μ_{1}/μ_{2})(n_{2}/n_{1}) = (μ_{1}n_{2}/μ_{2}n_{1}).

Then we can write

(β - 1)/(β + 1) = r_{12}.

(n_{1}/μ_{1})(2E_{i}- E_{t}) = (n_{2}/μ_{2})E_{t}.

E_{t}/E_{i}= 2n_{1}/(n_{1}+ n_{2}μ_{1}/μ_{2}) = t_{12}is the**transmission coefficient**.

We can write

2/(1 + β) = t_{12}.

(c) The**reflectance**is R = <S_{r}>/<S_{i}> = |r_{12}|^{2},

and the**transmittance**is T = <**S**_{t}>**∙n**/<**S**_{i}>**∙n**where**n**is a unit vector normal to the interface.

T = <S_{t}>/<S_{i}> = (μ_{1}n_{2}/μ_{2}n_{1})|t_{12}|^{2}. R + T = 1.

[<S> = (|E_{0}^{2}|/2)(ε/μ)^{½}] = (|E_{0}^{2}|/2)(n/(cμ)]

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 E_{r}/E_{i}
= sin(θ_{t} - θ_{i})/sin(θ_{t}
+ θ_{i}).

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**_{i }exp(i(**k**_{i}**∙r**- ωt)).

E_{r}(**r**,t) =**j**_{r }exp(i(**k**_{r}**∙r**- ωt)).

E_{t}(**r**,t) =**j**_{t }exp(i(**k**_{t}**∙r**- ωt)).

The tangential component of**E**is continuous across the boundary. The y-direction is tangential to the interface. Therefore E_{i }exp(i(**k**_{i}**∙r**)) + E_{r }exp(i(**k**_{r}**∙r**)) = E_{t }exp(i(**k**_{t}**∙r**)) for all**r**in the z = 0 plane.

This is only possible if**k**_{i}**∙r**=**k**_{r}**∙r**=**k**_{t}**∙r**for all**r**in the z = 0 plane.

Let**r**= i, then k_{ix}= k_{rx}= k_{tx}. or k_{i}sin(θ_{i}) = k_{r}sin(θ_{r}) = k_{t}sin(θ_{t}).

k_{i}= k_{r}= (με)^{½}ω = n_{1}ω/c. k_{t}= (με')^{½}ω = n_{2}ω/c. Therefore

n_{1}sin(θ_{i}) = n_{1}sin(θ_{r}) = n_{2}sin(θ_{t}).

This implies θ_{i}= θ_{r}(law of reflection) and n_{1}sin(θ_{r}) = n_{2}sin(θ_{t}) (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.

**B**_{i}=**k**_{i}×**E**_{i}/ω.**B**_{i}**∙i**= (1/ω)(**k**_{i}×**E**_{i})**∙i**= (1/ω)(**E**_{i}×**i**)**∙k**_{i}= -(1/ω)E_{i}k_{iz}= -(1/ω)E_{i}k_{i}cos(θ_{i}).

Similarly**B**_{r}**∙i**= (1/ω)E_{r}k_{r}cos(θ_{r}).**B**_{t}**∙i**= -(1/ω)E_{t}k_{t}cos(θ_{t}).

We then have E_{i}k_{i}cos(θ_{i}) - E_{r}k_{r}cos(θ_{r}) = E_{t}k_{t}cos(θ_{t}),

or n_{1}E_{i}cos(θ_{i}) - n_{1}E_{r}cos(θ_{r}) = n_{2}E_{t}cos(θ_{t}).

Combining this equation with E_{i}+ E_{r}= E_{t}and θ_{i}= θ_{r}yields

E_{r}/E_{i}= (n_{1}cosθ_{i}- n_{2}cosθ_{t})/(n_{1}cosθ_{i}+ n_{2}cosθ_{t})

= (cosθ_{i}- (sinθ_{i}/sinθ_{t})cosθ_{t})/(cosθ_{i}+ (sinθ_{i}/sinθ_{t})cosθ_{t})

= (1 - (tanθ_{i}/tanθ_{t}))/(1 + (tanθ_{i}/tanθ_{t}))

= (tanθ_{t}- tanθ_{i})/(tanθ_{t}+ tanθ_{i})

= sin(θ_{t}- θ_{i})/sin(θ_{t}+ θ_{i}).

[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)].

Fresnel's reflectance formulas are given by

R = |sin(θ_{i }- θ_{r})/sin(θ_{i }+
θ_{r})|^{2}
or R = |tan(θ_{i }- θ_{r})/tan(θ_{i }+
θ_{r})|^{2}

depending on the polarization of the incident wave with respect to
the plane of incidence, and with θ_{i} and
θ_{r} the angles of
incidence and refraction, respectively.

(a) Specify the boundary conditions across an interface.

(b) For simplicity assume μ_{1 }= μ_{2 }= μ_{0},
ε_{1 }=
ε_{0}, ε_{2 }= 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 n_{2 }= 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) ( **D**_{2}-**D**_{1})**∙n**_{2}= σ_{f}( **B**_{2}-**B**_{1})**∙n**_{2}= 0( **E**_{2}-**E**_{1})**∙t**= 0( **H**_{2}-**H**_{1})**∙t**=**k**_{f}**∙n**(b) In the figure below, the y-direction is the direction of the polarization for

**s-polarization**.

**E**_{i}(**r**,t) =**j**_{i }exp(i(**k**_{i}**∙r**- ωt)).

E_{r}(**r**,t) =**j**_{r }exp(i(**k**_{r}**∙r**- ωt)).

E_{t}(**r**,t) =**j**_{t }exp(i(**k**_{t}**∙r**- ωt)).

The tangential component of**E**is continuous across the boundary.

The y-direction is tangential to the interface.

Therefore E_{i }exp(i(**k**_{i}**∙r**)) + E_{r }exp(i(**k**_{r}**∙r**)) = E_{t }exp(i(**k**_{t}**∙r**)) for all**r**in the z = 0 plane.

This is only possible if**k**_{i}**∙r**=**k**_{r}**∙r**=**k**_{t}**∙r**for all**r**in the z = 0 plane.

Let**r**= i, then k_{ix}= k_{rx}= k_{tx}, or k_{i}sin(θ_{i}) = k_{r}sin(θ_{r}) = k_{t}sin(θ_{t}).

k_{i}= k_{r}= (μ_{0}ε_{0})^{½}ω = ω/c. k_{t}= (μ_{0}ε')^{½}ω = n_{2}ω/c. Therefore

sin(θ_{i}) = sin(θ_{r}) = n_{2}sin(θ_{t}).

This implies θ_{i}= θ_{r}(law of reflection) and sin(θ_{r}) = n_{2}sin(θ_{t}) (Snell's law) for s-polarization.

The tangential component of**B**= μ_{0}**H**is continuous across the boundary. The x-direction is tangential to the interface.

**B**=**k**×**E**/ω,**B**_{i}**∙i**= (1/ω)(**k**_{i}×**E**_{i})**∙i**= (1/ω)(**E**_{i}×**i**)**∙k**_{i}= -(1/ω)E_{i}k_{iz}= -(1/ω)E_{i}k_{i}cos(θ_{i}).

Similarly**B**_{r}**∙i**= (1/ω)E_{r}k_{r}cos(θ_{r}),**B**_{t}**∙i**= -(1/ω)E_{t}k_{t}cos(θ_{t}).

We then have E_{i}k_{i}cos(θ_{i}) - E_{r}k_{r}cos(θ_{r}) = E_{t}k_{t}cos(θ_{t}),

or E_{i}cos(θ_{i}) - E_{r}cos(θ_{r}) = n_{2}E_{t}cos(θ_{t}).

Combining this equation with E_{i}+ E_{r}= E_{t}and θ_{i}= θ_{r}from above yields

E_{r}/E_{i}= (cosθ_{i}- n_{2}cosθ_{t})/(cosθ_{i}+ n_{2}cosθ_{t})

= (cosθ_{i}- (sinθ_{i}/sinθ_{t})cosθ_{t})/(cosθ_{i}+ (sinθ_{i}/sinθ_{t})cosθ_{t})

= (1 - (tanθ_{i}/tanθ_{t}))/(1 + (tanθ_{i}/tanθ_{t}))

= (tanθ_{t}- tanθ_{i})/(tanθ_{t}+ tanθ_{i})

= sin(θ_{t}- θ_{i})/sin(θ_{t}+ θ_{i}).

R =|E_{r}/E_{i}|^{2}= |sin(θ_{t}- θ_{i})/sin(θ_{t}+ θ_{i})|^{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**.

**B**_{i}(**r**,t) = -**j**_{i }exp(i(**k**_{i}**∙r**- ωt)),

B_{r}(**r**,t) = -**j**_{r }exp(i(**k**_{r}**∙r**- ωt)),

B_{t}(**r**,t) = -**j**_{t }exp(i(**k**_{t}**∙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.

E_{i }exp(i(**k**_{i}**∙r**))cos(θ_{i}) - E_{r }exp(i(**k**_{r}**∙r**))cos(θ_{r}) = E_{t }exp(i(**k**_{t}**∙r**))cos(θ_{t}) for all**r**in the z = 0 plane.

This is only possible if**k**_{i}**∙r**=**k**_{r}**∙r**=**k**_{t}**∙r**for all**r**in the z = 0 plane.

Let**r**=**i**, then k_{ix}= k_{rx}= k_{tx}, or k_{i}sin(θ_{i}) = k_{r}sin(θ_{r}) = k_{t}sin(θ_{t}).

k_{i}= k_{r}= (μ_{0}ε_{0})^{½}ω = ω/c. k_{t}= (μ_{0}ε')^{½}ω = n_{2}ω/c. Therefore

sin(θ_{i}) = sin(θ_{r}) = n_{2}sin(θ_{t}).

This implies θ_{i}= θ_{r}(law of reflection) and sin(θ_{r}) = n_{2}sin(θ_{t}) (Snell's law) for p-polarization.

The tangential component of**B**= μ**H**is continuous across the boundary.

The y-direction is tangential to the interface.

B_{i}+ B_{r}= B_{t}.

**B**=**k**×**E**/ω, B_{i}= (1/ω)E_{i}k_{i}, = E_{i}/c. B_{r}= (1/ω)E_{r}k_{r}= E_{r}/c, B_{t}= (1/ω)E_{t}k_{t}= n_{2}E_{t}/c.

We then have E_{i}+ E_{r}= n_{2}E_{t}.

Combining this equation with E_{i }cos(θ_{i}) - E_{r }cos(θ_{r}) = E_{t }cos(θ_{t}) and θ_{i}= θ_{r}ffom above yields

E_{r}/E_{i}= (n_{2}cosθ_{i}- cosθ_{t})/(n_{2}cosθ_{i}+ cosθ_{t})

= ((sinθ_{i}/sinθ_{t})cosθ_{i}- cosθ_{t})/((sinθ_{i}/sinθ_{t})cosθ_{i}+ cosθ_{t})

= (sinθ_{i}cosθ_{i}- cosθ_{t}sinθ_{t})/(sinθ_{i}cosθ_{i}+ cosθ_{t}sinθ_{t})

= tan(θ_{t}- θ_{i})/tan(θ_{t}+ θ_{i}).

R =|E_{r}/E_{i}|^{2}= |tan(θ_{t}- θ_{i})/tan(θ_{t}+ θ_{i})|^{2}.

[(sina cosa - cosb sinb)/(sina cosa + cosb sinb)

= (sina cosa(sin^{2}b + cos^{2}b) - cosb sinb(sin^{2}a + cos^{2}a))/(sina cosa(sin^{2}b + cos^{2}b) + cosb sinb(sin^{2}a + cos^{2}a))

= (sin(a-b)cos(a+b)/[sin(a+b)cos(a-b)] = tan(a-b)/tan(a+b)].

(c) When θ_{t}+ θ_{i}= п/2, the reflectance is zero for p-polarization.

This happens when θ_{i}= θ_{B}, n_{2}sinθ_{B}= n_{2}sin((п/2) - θ_{B}), tanθ_{B}= n_{2}/n_{1}.

θ_{B}is called the Brewster angle.

When θ_{i}= θ_{B}, then only light with s-polarization is reflected. For a wave incident with mixed polarization incident with θ_{i}= θ_{B}, the reflected light is polarized.

With n_{1}= 1, n_{2}= 1.5 we have θ_{B}= 56.31^{o}.

An electromagnetic wave passes through a boundary between
two media with n_{1 }= 1 and n_{2 }= 3 at near-normal incidence
of θ_{i }= 0.5 degree.

(a) Find the angle θ_{t }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: n_{1}sinθ_{1 }= n_{2}sinθ_{2}.

Small angle approximation: n_{1}θ_{1 }= n_{2}θ_{2}.

θ_{1 }= θ_{i }= 0.5^{o}= 8.727 mrad. θ_{2 }= θ_{t }= (1/3)*8.727 mrad = 0.1667^{o}.(b) Fresnel reflection coefficient for p-polarization: r

_{12p }= tan(θ_{1 }- θ_{2})/tan(θ_{1 }+ θ_{2}).

Fresnel reflection coefficient for s-polarization: r_{12s }= sin(θ_{1 }- θ_{2})/sin(θ_{1 }+ θ_{2}).Small angle approximation: r

_{12 }= (θ_{1 }- θ_{2})/(θ_{1 }+ θ_{2}), independent of polarization.

The reflectance R = |r_{12}|^{2}, and the transmittance T = 1 - R.

r_{12 }= 0.5, R = 0.25 = 25%, T = 0.75 = 75%.

A beam of light with wavelength 450 nm in vacuum is incident on a prism as shown
in the figure, and totally reflected through 90^{o}. 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 n_{2 }< n_{1}, then for θ_{i}> θ_{c}= sin^{-1}(n_{2}/n_{1})_{ }we have**total internal reflection**. The wave vector**k**_{t}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.**k**_{i}makes an angle of 45_{o}with the z-axis. The critical angle for total internal reflection is θ_{c}= sin^{-1}(n_{2}/n_{1}) = sin^{-1}(1/1.6) = 38.7^{o}, so we have total internal reflection.

n sinθ_{i}= sinθ_{t}= 1.1313

k_{tx}= k_{t}sinθ_{t}. k_{tz}= k_{t}cosθ_{t}.

cosθ_{t}= (1 - sin^{2}θ_{t})^{½}= i*0.5295.

k_{t}cosθ_{t}= i*0.5295*(2π/450 nm) = iβ.

E_{t}= E_{0t}exp(i(**k**_{t}**∙r**- ωt)) = E_{0t}e^{-βz}exp(i(k_{tx}x - 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.