Inductive Heating

.

(real/imaginary parts of wave number) (9.70) Thus, we can rewrite the plane-wave solution as

E⁽⁺⁾(r) = E⁽⁺⁾₀ e^ikz= E⁽⁺⁾₀ e^−k−ze^ik+z. (9.71) (damped plane wave) As we saw for internal reflection, the plane wave breaks up into propagating and damping factors—the imaginary part of the refractive index leads to damping of the wave, while the real part represents the propagating aspect of the wave. We can again define a skin depth δ for the damped wave by

e^−k−z= e^−z/δ, (9.72)

so that

δ = 1

k₋ = 1

ω√µ0ǫ0nκ = λ0

2πnκ = λ0

2πIm[˜n]. (9.73)

(skin depth in conductor) As before, this is the length scale over which the wave damps exponentially away. Typically, for a good conductor, nκ is large, so that δ ≪ λ0. This is called the skin effect: for good conductors, an electromagnetic wave penetrates the surface by only a small fraction of the vacuum wavelength.

9.5.2 Inductive Heating

Since the transmitted wave gets damped, something must be taking up that dissipated energy. Of course, this goes into the metal as heat. While being undesirable from the point of view of making mirrors for high-power lasers, this effect can be useful in the machine shop or foundry for heat-treating or melting metals. An induction heater drives a high-voltage, radio-frequency signal through a coil that surrounds the conductive object to be heated. The electromagnetic wave generated by the coil couples into the conductive object, and generates heat as it is absorbed. Of course, the equivalent picture is that the radiation induces currents in the object, causing Joule heating—from our analysis above, we saw that it is precisely these induced currents that damp the electromagnetic wave after it penetrates the surface.

The following photographic sequence shows a quarter being heated and then melted by an inductive heater. Roughly speaking, you can think of the coil as being a solenoid, which ideally produces a uniform, oscillating magnetic field along the solenoid axis. The corresponding electric field is roughly a cylindrical wave collapsing onto the quarter, with the wave polarization being in the plane of the quarter. The wavelength for the 200 kHz radiation is quite large, around 1.5 km. However, the copper inside the quarter is an extremely good conductor; assuming a conductivity of 60 × 10⁶(m · Ω)⁻¹, the skin depth turns out to be only 0.15 mm.

Thus, due to the skin effect, the radiation is absorbed in a thin shell on the outside of the quarter. You can see this in the photo series, because the outside of the quarter is visibly much hotter than the middle, since

162 Chapter 9. Fresnel Relations

the middle is only being heated by thermal conduction.

Here is a movie of the heating process, where you can see the heat flowing in from the outside of the quarter to the inside.

9.5 Reflection at a Dielectric-Conductor Interface 163

9.5.3 Fresnel Relations

It is worth reiterating: with the conductor present, everything is the same as in the dielectric case, except that the refractive index is now complex. In particular, the Fresnel reflection coefficients still hold with the complex refractive index:

rS=n1cos θi− ˜n2cos θt

n1cos θi+ ˜n2cos θt

rP=n1cos θt− ˜n²cos θi

n1cos θt+ ˜n2cos θi

.

(9.74) (Fresnel relations with complex index)

Here, n1 is the refractive index of the dielectric (from which the wave is incident), and ˜n2 is the complex refractive index of the conductor. Note that this is even a bit more complicated than before, though, since from Snell’s law, θtis a complex number too! We won’t go into much detail regarding the wave in the medium beyond what we’ve already done, since we are usually only interested in this wave if it is then transmitted back into free space where it can be detected. This situation is better treated with the matrix formalism for multilayer films that we will get to shortly.

Let’s plug in some numbers for the simple example case of a polished silver mirror at normal incidence.

The reflectance is

R =    

n1− ˜n2

n1+ ˜n2

   

2

. (9.75)

For silver at 780 nm,² n = 0.27 + 4.47i, which give R = 95.0%. Thus silver is a good reflector, but some of˜ the incident energy is lost in the metal. This is because the wave penetrates a short distance into the metal.

Putting in numbers, we find that the skin depth is δ = λ0

2πnκ = λ0

28.1 = 27.8 nm (9.76)

2J. H. Weaver and H. P. R. Frederikse, “Optical Properties of Selected Elements,” in CRC Handbook of Chemistry and Physics, 82nd ed., David R. Lide, Ed. (CRC Press, Boca Raton, 2001), p. 12-133.

164 Chapter 9. Fresnel Relations

at 780 nm.

Here is a visualization of wave reflection from a dielectric-conductor interface. The specific indices used are vacuum (n1= 1) on the left and slightly conductive glass (˜n2= 1.52 + 0.152i) on the right. Note that the wave damping is similar to the case of internal reflection, but now the transmitted wave propagates away from the interface rather than along it.

9.6 Exercises 165

9.6 Exercises

Problem 9.1

We derived the Fresnel relations for S-polarized light giving the reflection coefficient rS= n1cos θi− n2cos θt

n1cos θi+ n2cos θt

(9.77) and transmission coefficient

tS= 2n1cos θi

n1cos θi+ n2cos θt (9.78)

of the electric field at a dielectric interface. Use the setup from the text to derive the Fresnel relations for the reflection coefficient,

rP= n1cos θt− n²cos θi

n1cos θt+ n2cos θi (9.79)

and transmission coefficient

tP= 2n1cos θi

n1cos θt+ n2cos θi

(9.80) for P-polarized light.

Problem 9.2

When we worked out the Fresnel relations for the reflection and coefficients at a planar, dielectric interface, we ignored two boundary conditions, namely that the components of D and B normal to the interface are continuous across the boundary. Write out both of these boundary conditions in terms of the Fresnel coefficients r and t for both polarizations.

Problem 9.3

One way of making a cheap but good polarizer (except for wavefront distortion) is to use a stack of microscope slides. Assuming an index of refraction n = 1.52 and that the slides are all oriented at Brewster’s angle with respect to a randomly polarized laser beam, how many slides does it take to attenuate one polarization by a factor of 10⁻⁴ in intensity compared to the other? Keep in mind that both sides of each slide are at Brewster’s angle. Ignore multiple reflections in your calculation.

Problem 9.4

Diode lasers in the near infrared (say, around 780 nm) are made from GaAlAs crystals. For low-power diodes (in the 5-15 mW range), the crystal ends are cleaved to form flat surfaces that act as the two flat reflectors of the laser cavity. Model a typical diode laser as a crystal (n = 3.5) of length 300 µm, surrounded by air (n = 1).

(a) For this resonator, calculate the round-trip time, the free spectral range, the finesse, the photon lifetime, and the frequency width (FWHM) of the modes. (Note that for higher power diodes, the output facet is typically antireflection coated to couple out more of the light in the cavity.)

(b) Assuming that the diode laser frequency is that of a single longitudinal mode, calculate the change in laser frequency per degree Celsius for a temperature change. Assume a nominal operating (vacuum) wavelength of 780 nm and a coefficient of thermal expansion (∆L/L) for GaAlAs of 7 × 10⁻⁶/^◦C. (In practice this tuning rate is only valid for small temperature changes; for larger changes, the change in the band gap is more important, causing much larger average tuning rates as the laser “hops” between longitudinal modes.)

Problem 9.5

Plot the intensity reflection coefficients as a function of incident angle for light incident from air (n = 1) onto crown glass (n = 1.52) for both S- and P-polarized light. Repeat for light incident from within the glass.

166 Chapter 9. Fresnel Relations

Problem 9.6

For total internal reflection from a glass–air interface (n1= 1.52 for glass, n2= 1 for air), plot the skin depth normalized to the vacuum wavelength (δ/λ0) as a function of incident angle.

Problem 9.7

Derive the wave equation for the magnetic field H in a homogeneous medium of permittivity ǫ and conductivity σ.

Problem 9.8

To describe the propagation of electromagnetic waves in a conductor, we derived a wave equation for E, coupled to the current density induced by the field in the conductor. An alternate approach is to consider the wave equation to be coupled to the polarization P of the medium.

As the first step in this method, recall that the electric displacement, electric field, and polarization density are related by

D = ǫ0E + P, (9.81)

and ignoring magnetic effects, we can assume

B = µ0H. (9.82)

Then use Maxwell’s equations in a source-free, dielectric medium,

∇ · D = 0

∇ · B = 0

∇ × E = −∂tB

∇ × H = ∂tD,

(9.83)

to derive a wave equation driven by P. You may assume the polarization field to be transverse, i.e.,

∇ · P = 0.

Problem 9.9

Compute the reflectance and skin depth for light at normal incidence from vacuum at wavelength 780 nm for polished surfaces of gold (˜n = 0.08 + 4.60i) and copper (˜n = 0.24 + 4.80i).

Problem 9.10

(a) Show that in the limit of low frequency, the skin depth for a good (nonmagnetic) conductor can be written

δ ≈

r 2

µ0ωσ. (9.84)

(b) Evaluate the skin depth at 200 kHz for titanium (σ = 2 × 10⁶ (m · Ω)⁻¹). Recall that µ0 = 4π × 10⁻⁷N/A².

(c) Suppose you have a disc of titanium the size and shape of a quarter. The disc is in the center of a solenoidal coil, which is driven at 200 kHz, and it lies in the plane orthogonal to the coil axis.

Assume that the solenoid length is much larger than all other length scales in this setup. What is the (qualitative) distribution of power absorption in the disc? Explain.

Problem 9.11

It is conventional to represent the mode functions of a compact region in space by normalized mode functions f^k,ζ(r), where k is the wave vector (which takes on discrete values), and ζ is an index marking the two possible polarizations. These mode functions are normalized in the sense that

Z

V

d³r |f^k,ζ(r)|²= 1, (9.85)

9.6 Exercises 167

where V is the volume over which the mode exists (i.e., the cavity size). More generally, the orthonor-mality relation is

Z

V

d³r f^k,ζ(r) · f^k∗′,ζ^′(r) = δk³,k^′δζ,ζ^′. (9.86) However, for mode functions defined over all space, the integration volume is unbounded and k takes on any value in R³, and this orthonormality relation generalizes to

Z

d³r f^k,ζ(r) · f^k∗′,ζ^′(r) = (2π)³δ³(k − k^′) δζ,ζ^′, (9.87) assuming

f (x) = 1 2π

Z _∞

−∞

f (k) e˜ ^ikxdk, f (k) =˜ Z _∞

−∞

f (x) e^−ikxdx, (9.88) as the Fourier-transform convention in each dimension.

In the presence of a vacuum–dielectric interface, the mode function for TE polarization can be written f^k,TE= ˆε^k,TEe^ik·rΘ(z) + rTEe^ikr·rΘ(z) + tTEe^ikt·rΘ(−z) , (9.89) where the dielectric (of permittivity ǫ, and index n) occupies the half-space z ≤ 0, rTEand tTEare the Fresnel reflection and transmission coefficients for TE (S) polarization, and Θ(z) is the Heaviside step function. Show that this mode function satisfies the orthonormality relation

Z

d³rǫ(r) ǫ0

fk^∗,TE(r) · f^k′,TE(r) = (2π)³δ³(k − k^′), (9.90) which reduces to the free-space relation as ǫ(r) −→ ǫ⁰. For simplicity, suppose that f^k,TE(r) and f^k′,TE(r) are both incident from the vacuum side.

Note: there is a bit of a trick to getting the delta function to come out right on the transmitting side. You might want to refer to the article that first proved this for help: C. K. Carniglia and L.

Mandel, “Quantization of Evanescent Electromagnetic Waves,” Physical Review D 3, 280 (1971) (doi:

10.1103/PhysRevD.3.280).

Chapter 10

In document Optics Notes (Page 161-169)