Multiple Parallel Interfaces
6.9 Repeated Multilayer Stacks
(high reflector, p-polarized) (6.80) The matrices for a high and a low refractive index layer are multiplied together in the usual manner. Each layer pair takes the form
"
6.9 Repeated Multilayer Stacks 155 and using (6.82) we can compute A(p)
A(p) = 1
This stack of q periods can achieve extraordinarily high reflectivity. In the limit of q → ∞, we have tp→ 0 and rp → −1 from (6.71) and (6.72), giving 100% reflection.
Sometimes multilayer coatings are made with repeated stacks of layers. In general, if the same series of layers in (6.82) is repeated many times, say q times, the following formula known as Sylvester’s theorem (see appendix 0.4) comes in handy:
A B
This formula relies on the condition AD − BC = 1, which is true for matrices of the form (6.67) and (6.75) or any product of them. Here, A, B, C, and D represent the elements of a matrix composed of a block of matrices corresponding to a repeated pattern within the stack.
Many different types of multilayer coatings are possible. For example, a Brewster’s-angle polarizer has a coating designed to transmit with high efficiency p-polarized light while simultaneously reflecting s-polarized light with high efficiency. The backside of the substrate is left uncoated where p-polarized light passes with 100% efficiency at Brewster’s angle.
Exercises
6.2 Double Boundary Problem Solved Using Fresnel Coefficients
P6.1 You have a 1 micron thick coating of dielectric material (n = 2) on a piece of glass (n = 1.5). Use a computer to plot the magnitude of the Fresnel coefficient (6.10) from air into the glass at normal incidence. Plot as a function of wavelength for wavelengths between 200 nm and 800 nm (assume the index remains constant over this range).
6.3 Double Boundary Problem at Sub Critical Angles
P6.2 A light wave impinges at normal incidence on a thin glass plate with index n and thickness d.
(a) Show that the transmittance through the plate as a function of wavelength is
Ttot= 1
1 +(n24n−1)2 2 sin2
2πnd λvac
HINT: Find
r12 = r10 = −r01= n − 1 n + 1 and then use
Ti→m= 1 − R01 T12= 1 − R12
(b) If n = 1.5, what is the maximum and minimum transmittance through the plate?
(c) If the plate thickness is d = 150 µm, what wavelengths transmit with maxi-mum efficiency?
HINT: Give a formula involving an integer N .
P6.3 Consider the “beam splitter” introduced in Example 6.2. Show that the maximum reflectance possible from the single coating at the first surface is 46%. Find the smallest possible d1 that accomplishes this for light with wavelength λvac = 633 nm.
6.4 Beyond Critical Angle: Tunneling of Evanescent Waves
P6.4 Re-compute (6.32) in the case of s-polarized light. Write the result in the same form as the last expression in (6.32). HINT: You need to redo (6.28)–(6.30).
Exercises 157 L6.5 Consider s-polarized microwaves (λvac= 3 cm) encountering an air gap separating two paraffin wax prisms (n = 1.5). The 45◦ right-angle prisms are arranged with the geometry shown in Fig. 6.3. The presence of the second prism frustrates the total internal reflection that would have occurred if the first prism were by itself.
This occurs because “feedback” from the second surface disrupts the evanescent waves.
Figure 6.15
(a) Use a computer to plot the transmittance through the gap as a function of separation d (normal to gap surface). Do not consider reflections from other surfaces of the prisms.
HINT: Plot the result of P 6.4.
(b) Measure the transmittance of the microwaves through the prisms as function of spacing d (normal to the surface) and superimpose the results on the graph of part (a).
RESULT: See the graph below. Presumably experimental error causes some dis-crepancy, but the trend is clear.
Figure 6.16
6.7 Distinguishing Nearby Wavelengths in a Fabry-Perot Instrument
P6.6 A Fabry-Perot interferometer has silver-coated plates each with reflectance R = 0.9, transmittance T = 0.05, and absorbance A = 0.05. The plate separation
is d = 0.5 cm with interior index n1 = 1. Suppose that the wavelength being observed near normal incidence is 587 nm.
(a) What is the maximum and minimum transmittance through the interferome-ter?
(b) What are the free spectral range ∆λFSR and the fringe width ∆λFWHM? (c) What is the resolving power?
P6.7 Generate a plot like Fig. 6.10(a), showing the fringes you get in a Fabry-Perot etalon when θ1 is varied. Let Tmax = 1, F = 10, λ = 500 nm, d = 1 cm, and n1 = 1.
(a) Plot T vs. θ1 over the angular range used in Fig. 6.10(a).
(c) Suppose d was slightly different, say 1.00002 cm. Make a plot of T vs θ1 for this situation.
P6.8 Consider the configuration depicted in Fig. 6.9, where the center of the diverging light beam λvac = 633 nm approaches the plates at normal incidence. Suppose that the spacing of the plates (near d = 0.5 cm) is just right to cause a bright fringe to occur at the center. Let n1 = 1. Find the angle for the mth circular bright fringe surrounding the central spot (the 0th fringe corresponding to the center). HINT: cos θ ∼= 1 − θ2/2. The answer has the form a√
m; find the value of a.
L6.9 Characterize a Fabry-Perot etalon in the laboratory using a HeNe laser (λvac = 633 nm). Assume that the bandwidth ∆λHeNe of the HeNe laser is very narrow compared to the fringe width of the etalon ∆λFWHM. Assume two identical reflec-tive surfaces separated by 5.00 mm. Deduce the free spectral range ∆λFSR, the fringe width ∆λFWHM, the resolving power, and the reflecting finesse (small f ).
Figure 6.17
L6.10 Use the same Fabry-Perot etalon to observe the Zeeman splitting of the yellow line λ = 587.4 nm emitted by a krypton lamp when a magnetic field is applied.
As the line splits and moves through half of the free spectral range, the peak of the decreasing wavelength and the peak of the increasing wavelength meet on the screen. When this happens, by how much has each wavelength shifted?
Exercises 159
Figure 6.18
6.8 Multilayer Coatings
P6.11 (a) Write (6.54) through (6.59) for s-polarized light.
(b) From these equations, derive (6.73)–(6.75).
P6.12 Beginning with (6.76) for a single layer between two materials (i.e. two interfaces), derive (6.21). WARNING: This is more work than it may appear at first.
6.9 Repeated Multilayer Stacks
P6.13 (a) What should be the thickness of the high and the low index layers in a periodic high-reflector mirror? Let the light be p-polarized and strike the mirror surface at 45◦. Take the indices of the layers be nH= 2.32 and nL= 1.38, deposited on a glass substrate with index n = 1.5. Let the wavelength be λvac= 633 nm.
(b) Find the reflectance R with 1, 2, 4, and 8 periods in the high-low stack.
P6.14 Find the high-reflector matrix for s-polarized light that corresponds to (6.82).
P6.15 Design an anti-reflection coating for use in air (assume the index of air is 1):
(a) Show that for normal incidence and λ/4 films (thickness= 14 the wavelength of light inside the material), the reflectance of a single layer (n1) coating on a glass is
R = ng− n21 ng+ n21
2
(b) Show that for a two coating setup (air-n1-n2-glass; n1 and n2 are each a λ/4 film), that
R = n22− ngn21 n22+ ngn21
2
(c) If ng = 1.5, and you have a choice of these common coating materials: ZnS (n = 2.32), CeF (n = 1.63) and MgF (n = 1.38), find the combination that gives you the lowest R for part (b). (Be sure to specify which material is n1 and which is n2.) What R does this combination give?
P6.16 Suppose you design a two-coating “anti-reflection optic” (each coating set for λ/4, as in the last problem) using n1 = 1.6 and n2 = 2.1. Assume you’ve got ng = 1.5 and normal incidence. If you design your coatings to be quarter-wave for λ = 550 nm (in the middle of the visible range) the R that you found in P 6.15(b) will be true only for that specific wavelength for two reasons: the index changes with λ, but more importantly, the thicknesses used in the coatings will not be λ/4 for other wavelengths. Let’s ignore the index change with λ and focus on the wavelength dependence. Use the matrix techniques and a computer to plot R(λair
for 400 to 700 nm (visible range). Do this for a single bilayer (one layer of each coating, two bilayers, four bilayers, and 25 bilayers.