For a plane wave with a given frequency w, the index of refraction in an anisotropic medium depends on the direction of propagation. For a plane wave incident with a wave vector pointing in the direction and phase velocity c/n there will, in general, exist two reflected waves with wave vectors pointing in the and directions and phase velocities c/n1 and c/n2, respectively, and two refracted waves with wave vector pointing in the and and phase velocities c/n1’ and c/n2’, respectively.
Maxwell’s equations require that the tangential components of E and H are continuous at any interface.
For waves of the form E(r,t) = E exp(i(k×r-wt)), H(r,t) = H exp(i(k×r-wt)) these boundary conditions require
kt×r = ki×r = kr×r
n(×r) = n1(×r) = n2(×r) =n1'(×r) =n2'(×r).
for any r lying in the interface.
For r = we find
n sinq = n1 sinq1 = n2 sinq2 = n1’sinq1’ = n2’ sinq2.
The simple law of reflection qi = qr will not generally hold since the indices of refraction are different for the incident and the reflected waves.
Total internal reflection at an interface between anisotropic media can occur in two ways. The incident wave vector makes the critical angle qc with the normal to the interface if either the phase velocity or the group velocity of the transmitted wave makes an angle of 90o with the normal to the interface.
In an anisotropic crystal the electric displacement vectors D associated with a wave-normal direction are parallel to the principal axes of the elliptical section through the index ellipsoid perpendicular to . In a biaxial crystal there is are special directions ' of when this section is circular. All directions of D perpendicular to ' are then permissible. Therefore there exist an infinity of directions of the electric field vector E (calculated for each D), and an infinity of directions for vg. The vectors lie on the surface of a cone. If a plate of a biaxial crystal is cut so that its parallel faces are perpendicular to ', and the plate is illuminated by a narrow beam of monochromatic light normal to one of the faces, then the energy will spread out in the plate in a hollow cone, the cone of internal conical refraction. Upon emerging on the other side it will form a hollow cylinder. On a screen parallel to the crystal face we see a bright circular ring.
Large scale experiments on conical refraction
External conical refraction
Assume an incident and a reflected TM wave exist in a lih medium. Ei(r,t) = Ei exp(-i(ki×r-wt)) denotes the electric field of the incident wave, Er(r,t) = Er exp(-i(kr×r-wt)) denotes the electric field of the reflected wave, E(r,t) = Ei(r,t) + Er(r,t).
The magnetic induction is given by
Hi = n1/(m1c)Ei(-), Hr = n1/(m1c)Er(-), H = Hi + Hr.
The wave impedance Z(z) is defined as Z(z) = Ex(z)/Hy(z). We have
Z(z) = (m1c/n1) cosqi(Hiy(z) - Hry(z))/(Hiy(z) + Hry(z)) = Z0(Hiy(z) - Hry(z))/(Hiy(z) + Hry(z)).
Z(z) = Z0(Hiy(0) - Hry(0)exp(2ikzz))/(Hiy(0) + Hry(0)exp(2ikzz)).
[Note: For a TM wave Z0 = (mc/n) cosq.]
With the reflection coefficient defined as r12p = Hry(0)/Hiy(0) we have
Z(z) = Z0(1 - r12pexp(2ikzz))/(1 + r12pexp(2ikzz)).
If we write G = -r12p,
then Z(z) = Z0(1 + Gexp(2ikzz))/(1
[Note: G is the ratio of the magnitudes of the reflected and incident electric field amplitudes at z = 0.]
For z = 0 we have Z(0) = Z0(1 + G)/(1 - G). We can therefore write
Z(z) = Z0(Z(0) - iZ0 tankzz)/(Z0 - iZ(0) tankzz).
This expression lets us find Z(z) at any position in the medium if we know it at one position in the medium.
If we define G(z) = G exp(2ikzz) as the ratio of the electric fields of the reflected and the incident waves at position z, then we can write
Z(z) = Z0(1 + G(z))/(1 - G(z)), and G(z) = (Z(z) - Z0)/(Z(z) + Z0).
All equations shown in red also hold for a TE wave as long as we define G = r12s, for a TE wave.
The wave impedance is a useful concept because Z(z) is always continuous and at any interface we can find the reflection coefficient from G(z) = (Z(z) - Z0)/(Z(z) + Z0).
Knowing Z(z) at any position in a medium, we can find the ratio of the electric fields of the reflected and the incident waves. If G(z) = 0 at an interface, then no reflected wave exists.
Example: an anti-reflection coating
Consider a TE wave incident on a dielectric layer on top of a
substrate as shown below. In the dielectric layer we have a transmitted
and a reflected wave. In the substrate we only have a transmitted
wave. The wave impedance at z = 0 is continuous across the interface,
Z(0)dielectric = Z(0)substrate = Z0substrate = [(m0/e0)1/2]/(n cosq).
[Note: For a TM wave Z0 = (mc/ncosq).]
At z = -d the impedance therefore is given by
Z(-d) = Z0(Z(0) + iZ0 tank1zd)/(Z0 + iZ(0) tank1zd).
If we choose 4d ® 2p/k1z,
then tankzd ® Ą, and Z(-d) will be real.
Z(-d) = (Z0dielectric)2/Z(0) = [(m0/e0)1/2](n cosq)/(n1 cosq1)2.
A dielectric of this thickness is called a quarter-wave layer.
[2p/k1z = l1z = l1 cosq1].
If we match Z(-d) to Z0air = [(m0/e0)1/2]/(cosq0),
then reflections in air at will be eliminated.
To match we need
(n1 cosq1)2 = n cosq cosq0.
This equation tells us what the index of refraction of a quarter wave layer needs to be to produce an anti-reflection coating. For normal incidence, q0 = 0, we need
n12 = n.
The matching condition is a function of angle, but if we match for q0 = 0, we greatly reduce the reflectance over a wide range of angles.
In practice, it may be difficult to find a dielectric material
with an index n1 that satisfies n12 = n.
Usually several quarter wave layers are applied.
One quarter wave layer with index n1 produces
Z(-d1) = (Z0dielectric_1)2/Z(0), Z(-d1) = Z(0)*(Z0dielectric_1)2/(Z(0))2
= [(m0/e0)1/2/(n cosq)] [ncosq/n1cosq1]2,
i.e. it multiplies Z(0) by a factor of [ncosq/n1cosq]2.
Assume you add another quarter wave layer with index n2.
Then Z(-d1-d2) = (Z0dielectric_2)2/(Z(-d1)) = Z(0) (Z0dielectric_2)2/(Z0dielectric_1)2
= [(m0/e0)1/2/(n cosq)] [n1cosq1/n2cosq2]2.
Z(0) is multiplied by a factor of [n1cosq1/n2cosq2]2.
If m pairs of quarter-wave layers layers with n1 and n2 are applied, then Z(0) is multiplied by a factor of [n1cosq1/n2cosq2]2m.
Z(after m pairs) = [(m0/e0)1/2/(n cosq)] [n1cosq1/n2cosq2]2m.
To match to Z0air = [(m0/e0)1/2]/(cosq0) we then need
[(m0/e0)1/2/(n cosq)] [n1cosq1/n2cosq2]2m = [(m0/e0)1/2]/(cosq0),
[n1cosq1/n2cosq2]2m = (n cosq)/(cosq0),
If q0 = 0, we need (n1/n2)2m = n
It may be possible to find a power m for which this relation is satisfied to a high degree of accuracy and thus produce an antireflection coating.
A different choice of dielectric materials with different
n1 and n2 may be used to produce a mirror with a very high
The reflection coefficient is therefore given by
G = (Z(surface) - Z0air)/(Z(surface) + Z0air).
We adjust m and therefore Z(surface) = Z(after m pairs) so that G approaches unity.