Thin Films
10.4 Gradient-Index Layers
10.4.7 Asymptotic Forms
10.4.7.2 Small-Reflection Limit
In the limit of large d, which smooths the transition between the vacuum and dielectric, we expect a small reflection coefficient. In this limit, we can use the asymptotic forms for real x for the Airy functions
Ai(−x) ∼ 1
190 Chapter 10. Thin Films
Then examining the components of the refection coefficient (10.121), we note that both terms in each coefficient A±0, B0±, A−, and B− scale as d−1/6, if we note that ˜k ˜d and ˜k ˜d0 scale as d2/3, and ˜α scales as d−1/3.
In the reflection coefficient (10.121),
rS= A−0B−− B0−A−
A+0B−− B0+A−, (10.128)
we will begin by evaluating the first term in the numerator:
A−0B− = αiαsAi(−˜k ˜d0)Bi(−˜k ˜d) − i˜αh
αiAi(−˜k ˜d0)Bi′(−˜k ˜d) + αsAi′(−˜k ˜d0)Bi(−˜k ˜d)i
− ˜α2Ai′(−˜k ˜d0)Bi′(−˜k ˜d).
(10.129) Comparing the first and last terms here in the limit of large d, we can see from the asymptotic forms (10.126) and (10.127) that the dependence on trigonometric functions will be the same. To compare them further, we can see that where we used Eqs. (10.121) to reduce the ratio. Thus, the first and last terms in Eq. (10.129) cancel in the large-d limit, and we have
A−0B− = −i˜αh
αiAi(−˜k ˜d0)Bi′(−˜k ˜d) + αsAi′(−˜k ˜d0)Bi(−˜k ˜d)i
. (10.131)
Proceeding to the second term in the numerator of Eq. (10.128), we see that this is the same as the first term (10.131) if we interchange d and d0, and then αiwith αs. Then from the last two equations in (10.121),
and writing out the numerator of (10.121),
A−0B−− B0−A−= −iαiα˜
we can put in the asymptotic forms (10.126) and (10.127) to see that this vanishes.
In the denominator, we have different signs, and so the first cancellation of the first and last terms in Eq. (10.129) does not occur, because the sign of the ˜α2term changes (as does the sign of the ˜ααsterm), and so we have
A+0B−= 2αiαsAi(−˜k ˜d0)Bi(−˜k ˜d) − i˜αh
αiAi(−˜k ˜d0)Bi′(−˜k ˜d) − αsAi′(−˜k ˜d0)Bi(−˜k ˜d)i
. (10.134)
Again, the other term in the denominator has the same form, but with ˜d0 and ˜d interchanged, and with αi−→ −αs and αs−→ −αi:
A−0B+= 2αiαsAi(−˜k ˜d)Bi(−˜k ˜d0) + i ˜αh
αsAi(−˜k ˜d)Bi′(−˜k ˜d0) − αiAi′(−˜k ˜d)Bi(−˜k ˜d0)i
. (10.135)
10.4 Gradient-Index Layers 191
From the asymptotic forms (10.126) and (10.127), along similar lines to Eq. (10.133), we see that the imaginary terms are equal in magnitude but opposite in sign, so that the denominator becomes
A+0B−− A−0B+= 2αiαs which is nonvanishing. Since we have a vanishing numerator and a nonvanishing denominator, we have established that rS−→ 0 as d −→ ∞.
Note that the denominator scales as d−1/3 for large d. The terms of the same order ended up canceling in the numerator. The next-order terms in the the asymptotic forms (10.126) and (10.127) are smaller by factors that scale as x−3/2 for large x. Since x here refers to ˜k ˜d and ˜k ˜d0, both of which scale as d2/3 the next-order part of the numerator should scale as 1/d compared to the denominator. Thus, for large d, the reflection coefficient scales as 1/d for large d, or the reflectance scales as R ∼ 1/d2.
To illustrate all this, below is plotted the reflectance at normal incidence for a vacuum–glass interface (assuming n = 1.52 for glass), connected by a linear permittivity gradient.
GRIN film thickness d/l0
0 1 2 3 4 5
Here, we can see that the reflectance starts at 4.3% at d = 0, which is what we expect for a vacuum–glass interface. We also see oscillations in the reflectance, and in the large-d regime, the envelope of the oscillations shows the expected 1/d2scaling. To gain a bit more insight into the oscillations, note that asymptotically, the Airy functions conspire to form trigonometric functions as in Eq. (10.136), with argument
2
where to keep things simple we are considering only normal incidence. This has the form of the optical path length of the GRIN film,
Z d
multiplied by k0, and so the oscillations here are similar to what we expect in for a thin, homogeneous film, provided we interpret the result in terms of the more general path length.
192 Chapter 10. Thin Films
10.4.8 P-Polarization
So far we have calculated the reflection coefficient in the case of S-polarization. But recall that we can adapt the calculation to P-polarization by moving the angle cosines from the numerators to the denominators of the α coefficients. Beginning with Eq. (10.61),
Fj≈
to write the alternate transfer matrix
Fj ≈
Passing over to the continuous limit, the differential system for the fields (10.70) becomes
d
where the definition (10.69) of nθ(x) still holds. Again E and H refer only to the components of the fields parallel to the film surface (now Ey and Hz in our coordinate system, for P-polarization). Written as two coupled equations, we have
Then proceeding with the effective impedance ˜η(x) defined in Eq. (10.73), we can eliminate the derivatives in Eq. (10.76) to find in place of Eq. (10.77) That is, part of the space dependence nθ2(x)/n2(x) of the variable index moves from the nonlinear term to the constant term. The procedure is then the same as before for calculating the reflection coefficient: first, use Eq. (10.75) as the initial condition for ˜η(d) (with suitably redefined αs), then use this differential equation to integrate backwards to ˜η(0), and finally, use Eq. (10.74),
r = αiη(0) − 1˜
αiη(0) + 1˜ , (10.147)
10.4 Gradient-Index Layers 193
for the reflection coefficient, with the modified αi. 10.4.8.1 Example: Linear Permittivity Gradient
Now we return to the example of the linear permittivity gradient from Section 10.4.6, and work out the reflection coefficient for P-polarization.
The Ricatti coefficients in Eq. (10.79) in this case become
a = −ik0n2(x) η0
, b = 0, c = ik0η0
nθ2(x)
n2(x), (10.148)
where Eq. (10.98) still holds for nθ(x) in this case. We should then solve the equation
z′′− a′ a
z′+ acz = 0, (10.149)
which becomes
z′′−
χ/d
1 + (χ/d)x
z′+ k02
cos2θi+χ dx
z = 0, (10.150)
which is the same as Eq. (10.101), except that the cosine in the coefficient of z′ is no longer there. Unfor-tunately, the variable-transformation tricks we used in the S-polarization case do not carry over here, and this equation does not have an obvious solution—except, that is, at normal incidence, when this case is equivalent to the S-polarization case.
Nevertheless, it is possible to solve Eq. (10.145) numerically to obtain the reflection coefficient. Doing this for the vacuum–glass (n = 1.52) interface for normal incidence, and 45◦incidence for both polarizations leads to the plot below.
GRIN film thickness d/l0
0 1 2 3 4 5
reflectance
10-1
10-6 10-2
10-3
10-4
10-5
normal incidence 45Ä, S
45Ä, P
In the d = 0 limit, S-polarization reflects more than does P, because 45◦ is not far from Brewster’s angle.
This behavior persists for all d, with similar oscillations in each case (but with longer period in the angled-incidence cases).
Note, however, that while something analogous to the polarizing effect at Brewster’s angle, the suppres-sion is not as complete as for a simple interface, as shown in the plot below, which shows both polarizations for three film thicknesses (with the d = 0 case corresponding to the usual vacuum–glass interface).
194 Chapter 10. Thin Films
angle of incidence
0Ä 30Ä 60Ä 90Ä
reflectance
100
10-7 10-1 10-2 10-3 10-4 10-5 10-6
S, do=o0 P, do=o0
S, do=ol0 P, do=ol0
S, do=o10l0
P, do=o10l0
However, notice that the reflectance is suppressed over a wide range of angles by the GRIN film.
10.5 Exercises 195
10.5 Exercises
Problem 10.1
Consider a planar glass window of refractive index n and thickness d, used as an etalon for light of wavelength λ incident at angle θ with respect to the window surface. Derive an expression for the changes in resonance wavelength δλ and frequency δν, given a small change δθi in the incident angle.
(That is, linearize the solution in δθi.) Problem 10.2
Consider a variable beam splitter consisting of two 45◦-45◦-90◦prisms5of refractive index n, separated by an air (n = 1) gap ∆ and operated as shown with light of (vacuum) wavelength λ0. Assume that n is large enough that frustrated internal reflection occurs at the gap at the incident angle of 45◦.
vary D D
(a) Write down an expression for the intensity reflection coefficient due to the air gap (ignore reflections from all other prism surfaces). Everything you write down should be defined in terms of parameters given in this problem, and make sure to treat both possible polarizations.
(b) Obtain the limits of your expression for small and large ∆, and comment on whether or not they make sense.
(c) Plot the reflectance of the beamsplitter as a function of ∆/λ0, assuming n = 1.52 for glass. Also include in your plot for comparison the simple estimate 1 − e−2∆/δ, where δ is the skin depth for internal reflection, which is based on the normalized intensity of the evanescent wave where it enters the second prism (hence, this fraction of the intensity transmits through the second prism).
Problem 10.3
Light of wavelength λ0is incident from air (n = 1) onto a single dielectric thin film (of index nf, and thickness λ/4, where λ is the wavelength inside the film), which covers a glass substrate (index ng).
no=o1 nf ng
E0i (+)
(a) Write down an expression for the film reflectance, assuming the light is at normal incidence, using the results of the reflection-summation formalism.
(b) Derive the value of nf that makes a perfect anti-reflection coating.
5for a realization of this beam splitter, see D. Bertani, M. Cetica, and R. Polloni, “A simple variable-ratio beam splitter for holography,” Journal of Physics E: Scientific Instruments 16, 602 (1983) (doi:10.1088/0022-3735/16/7/007). For application to Q-switched laser cavities, see Ian N. Court and Frederick K. von Willisen, “Frustrated Total Internal Reflection and Application of Its Principle to Laser Cavity Design,” Applied Optics 3, 719 (1964) (doi: dx.doi.org/10.1364/AO.3.000719).
196 Chapter 10. Thin Films
Problem 10.4
Repeat the derivation of the thin-film matrix formalism for P-polarization, showing that the transfer-matrix formalism is the same as for the S-polarization case that we already derived, provided the αj
coefficients are suitably redefined. Compute the film reflection and transmission coefficients for this case.
Problem 10.5
Discuss the advantage of a two-layer antireflection coating over a single-layer coating. Assume that you are only concerned with suppressing reflection at one incident wavelength and normal incidence, and distinguish practical vs. in principle performance of the coating in your answer. (Two or three sentences should suffice here.)
Problem 10.6
Plot the intensity reflection coefficients as a function of angle for light incident from air onto crown glass with a single-layer antireflection coating of MgF2. (Make plots for both polarizations.) Assume n = 1.38 for MgF2, and a coating thickness of λ/4 (where λ is the wavelength inside the coating!), where the design wavelength is λ = 550 nm (in vacuum). How thick is the coating in nm?
Problem 10.7
Plot the intensity reflection coefficients as a function of λ0 for light incident from air onto crown glass with a double-layer antireflection coating. The thin-film stack consists of a λ/4 layer of ZrO2
(n = 2.1) directly on top of the crown glass, followed by a λ/4 layer of CeF3 (n = 1.65). Assume a design wavelength of 550 nm (in vacuum) and extend the plot over the visible spectrum (400-700 nm).
Consider only the case of normal incidence. How thick are the layers in nm?
Problem 10.8
Consider a “slab” (film) of glass of index n surrounded on both sides by vacuum. Use the transfer-matrix method to derive the reflection coefficient for the film, assuming it is very thin (d ≪ λ0, λ), Taylor expanding to linear order in the film thickness d (at every stage, to keep it simple), and at normal incidence.