8 Antireflection coatings and thin-film optics

Run this chapter’s code in Google Colab (no installation).

Slides 148–154 (antireflection coating of a solar cell) and slides 267–276 (multiple-beam thin-film interference, minimum and maximum reflectance, transmission fringes of an a-Se film, incoherent multiple reflections in a plate).

8.1 Key relations

A film of index $n_2$ and thickness $d$ on a substrate $n_3$ reflects a sum of partial waves (slides 267–269). At normal incidence, with the round-trip phase $\phi=2\cdot\frac{2\pi}{\lambda}n_2d$:

\[ r=\frac{r_{12}+r_{23}e^{-j\phi}}{1+r_{12}r_{23}e^{-j\phi}},\qquad r_{ij}=\frac{n_i-n_j}{n_i+n_j}. \]

result	formula	slide
AR condition (phase)	$\phi=m\pi$, $m$ odd $\Rightarrow d=m\,\dfrac{\lambda}{4n_2}$	151–152, 270
AR condition (amplitude)	$n_2=\sqrt{n_1n_3}$ so that $r_{12}=r_{23}$	153, 270
extreme reflectances ($n_1<n_2<n_3$)	$R_\text{min}=\left(\dfrac{n_1n_3-n_2^2}{n_1n_3+n_2^2}\right)^2$ at $\phi=$ odd $\pi$; $\ R_\text{max}=\left(\dfrac{n_3-n_1}{n_3+n_1}\right)^2$ at $\phi=$ even $\pi$	272
incoherent plate	$T_\text{plate}=(1-R)^2(1+R^2+R^4+\dots)=\dfrac{1-R}{1+R}$	276

If $n_1<n_3<n_2$ the expressions for $R_\text{min}$ and $R_\text{max}$ are interchanged (slide 272).

8.1.1 Thin-film designer, live

Show the Python

viewof tf = Inputs.form({
  n2: Inputs.range([1, 4], {value: 1.9, step: 0.01, label: "film index n₂"}),
  n3: Inputs.range([1, 4], {value: 3.5, step: 0.01, label: "substrate index n₃"}),
  d:  Inputs.range([0, 600], {value: 92, step: 1, label: "film thickness d (nm)"})
})

With the defaults (Si$_3$N$_4$ on silicon, $d=92$ nm) you get the solar-cell example below.

8.2 Worked example: antireflection coating on a silicon solar cell (slides 148–154)

Show the Python

R_si = ((1 - 3.5)/(1 + 3.5))**2
print(f"bare Si (n = 3.5): R = {R_si:.1%}   → about 30 % of the light is lost (slide 149)")
print(f"ideal coating index √(1·3.5) = {np.sqrt(3.5):.2f}  (Si3N4, n ≈ 1.9, is a good choice)")
print(f"quarter-wave thickness at 700 nm: d = λ/(4 n2) = {700/(4*1.9):.1f} nm (or odd multiples)")

bare Si (n = 3.5): R = 30.9%   → about 30 % of the light is lost (slide 149)
ideal coating index √(1·3.5) = 1.87  (Si3N4, n ≈ 1.9, is a good choice)
quarter-wave thickness at 700 nm: d = λ/(4 n2) = 92.1 nm (or odd multiples)

A single quarter-wave layer is perfect only at one wavelength and only if $n_2=\sqrt{n_1n_3}$ (slide 154). SiNₓ ($n\approx2.0$) is the industrial compromise. A dispersion-free $n_\text{Si}=3.5$ is assumed here for clarity.

Show the Python

lam = np.linspace(nm(400), nm(1100), 700); l0, nSi = nm(700), 3.5
fig, ax = plt.subplots()
ax.plot(lam*1e9, ol.tmm([1, nSi], [], lam)[0], color=ol.INK2, label="bare Si")
for n2, lbl in ((np.sqrt(nSi), "ideal n₂ = √3.5"), (2.0, "SiNₓ n₂ = 2.0")):
    ax.plot(lam*1e9, ol.tmm([1, n2, nSi], [l0/(4*n2)], lam)[0], label=f"λ/4 {lbl}, d = {l0/(4*n2)*1e9:.0f} nm")
ax.plot(lam*1e9, ol.tmm([1, 1.38, 2.3, nSi], [l0/(4*1.38), l0/(4*2.3)], lam)[0], "--", label="double layer 1.38 / 2.3")
ax.set(xlabel="λ (nm)", ylabel="R", ylim=(0, 0.4)); ax.legend(fontsize=8); plt.show()

Figure 8.1: Reflectance of silicon with single- and double-layer AR coatings designed for 700 nm (non-dispersive model).

Show the Python

try:
    from refractiveindex import RefractiveIndexMaterial
    Si = RefractiveIndexMaterial(shelf="main", book="Si", page="Green-2008")
    lam_nm = np.linspace(400, 1100, 200)
    N_Si = np.array([Si.get_refractive_index(l) - 1j*Si.get_extinction_coefficient(l) for l in lam_nm]).squeeze()
    R = [ol.tmm([1, 2.0, Ni], [nm(700)/(4*2.0)], nm(l))[0][0] for Ni, l in zip(N_Si, lam_nm)]
    Rb = [ol.tmm([1, Ni], [], nm(l))[0][0] for Ni, l in zip(N_Si, lam_nm)]
    plt.plot(lam_nm, Rb, color=ol.INK2, label="bare c-Si"); plt.plot(lam_nm, R, label="λ/4 SiNₓ at 700 nm")
    plt.xlabel("λ (nm)"); plt.ylabel("R"); plt.legend(); plt.show()
except Exception as err:
    print("refractiveindex package or database not available here:", type(err).__name__)

Figure 8.2: Same SiNx coating on crystalline silicon with measured dispersion and absorption (refractiveindex.info, Green 2008), if the optional package is available.

8.3 Worked example: a dielectric film on a semiconductor (slides 273–275)

A semiconductor ($n_3=3.5$) is coated with a 160 nm transparent film ($n_2=2.5$) in air. Find the minimum and maximum reflectances and transmittances and their wavelengths in the visible.

Show the Python

n1, n2, n3, d = 1.0, 2.5, 3.5, 160e-9
Rmin = ((n1*n3 - n2**2)/(n1*n3 + n2**2))**2
Rmax = ((n3 - n1)/(n3 + n1))**2
print(f"R_min = {Rmin:.3f} → T_max = {1-Rmin:.0%};  R_max = {Rmax:.3f} → T_min = {1-Rmax:.0%}")
for m in (1, 3):
    print(f"R_min (φ = {m}π) at λ = 4 n2 d / {m} = {4*n2*d/m*1e9:.0f} nm")
for m in (1, 2, 3):
    print(f"R_max (φ = {2*m}π) at λ = 2 n2 d / {m} = {2*n2*d/m*1e9:.0f} nm")
lam = np.linspace(nm(350), nm(900), 1500)
R = ol.tmm([n1, n2, n3], [d], lam)[0]
fig, ax = plt.subplots(); ax.plot(lam*1e9, R, label="R"); ax.plot(lam*1e9, 1 - R, "--", label="T")
for l in (533.3e-9, 800e-9, 400e-9):
    ax.plot(l*1e9, ol.tmm([n1, n2, n3], [d], l)[0][0], "o", color=ol.INK)
ax.set(xlabel="λ (nm)", ylabel="R, T"); ax.legend(); plt.show()

R_min = 0.080 → T_max = 92%;  R_max = 0.309 → T_min = 69%
R_min (φ = 1π) at λ = 4 n2 d / 1 = 1600 nm
R_min (φ = 3π) at λ = 4 n2 d / 3 = 533 nm
R_max (φ = 2π) at λ = 2 n2 d / 1 = 800 nm
R_max (φ = 4π) at λ = 2 n2 d / 2 = 400 nm
R_max (φ = 6π) at λ = 2 n2 d / 3 = 267 nm

Figure 8.3: Reflectance of a 160 nm film (n = 2.5) on n = 3.5 (slide 275). The dots mark the visible extrema.

In the visible, $R_\text{min}$ (8 %) occurs at 533 nm and $R_\text{max}$ (31 %) at 400 nm, which gives $T_\text{max}=92\,\%$ and $T_\text{min}=69\,\%$ as on slide 275.

8.4 Coherent and incoherent reflections (slides 274, 276)

A thin film gives interference fringes (slide 274: transmission of an a-Se film on glass). A thick plate does not, because the source’s coherence length (Chapter 9) is shorter than the round trip. Then intensities, not amplitudes, add:

Show the Python

Rg = ((1.5 - 1)/(1.5 + 1))**2
print(f"glass plate, incoherent multiple reflections: T = (1−R)/(1+R) = {(1-Rg)/(1+Rg):.4f}  "
      f"(single surface 1−R = {1-Rg:.4f}, two surfaces without multiple reflections (1−R)² = {(1-Rg)**2:.4f})")

glass plate, incoherent multiple reflections: T = (1−R)/(1+R) = 0.9231  (single surface 1−R = 0.9600, two surfaces without multiple reflections (1−R)² = 0.9216)

Going deeper

The double-layer design above is illustrative. Optimise both thicknesses to minimise the solar-spectrum-weighted reflectance (AM1.5G spectrum, scipy.optimize). What is gained versus texturing the surface?
Graded-index and moth-eye surfaces: why does a gradual index transition suppress reflection? Relate it to the limit of many thin layers.
Swanepoel’s method (slide 274) extracts $n(\lambda)$ and thickness of a film from the envelopes of its transmission fringes. Implement it on a simulated spectrum from ol.tmm.
Facet coatings of semiconductor lasers and SOAs: AR (below $10^{-4}$) for amplifiers, HR for lasers. Estimate the tolerance on thickness for a single-layer AR at 1550 nm.

Complements

Macleod, Thin-Film Optical Filters: design of AR, HR and band-pass coatings.
R. Swanepoel, “Determination of the thickness and optical constants of amorphous silicon”, J. Phys. E 16, 1214 (1983).
M. A. Green, Solar Cells: Operating Principles, Technology and System Applications.

8.5 Try it yourself

Use the live panel to design a single-layer AR for a glass lens ($n_3=1.52$) at 550 nm. Why is MgF₂ ($n=1.38$) used even though it is not the ideal index?
Why do soap bubbles show colours, and why does a soap film turn black just before it bursts?
Find the thinnest film that makes a silicon surface look blue (minimum reflection in the red, maximum in the blue).

# Antireflection coatings and thin-film optics {#sec-thinfilm} [![Open in Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/{{< var github_user >}}/{{< var repo >}}/blob/main/{{< var subdir >}}/notebooks/thinfilm.ipynb){.colab-link}  Run this chapter's code in Google Colab (no installation). ```{python} #| include: false import numpy as np import matplotlib.pyplot as plt import optolib as ol ol.use_style() c0, nm, um = ol.c0, ol.nm, ol.um ``` *Slides 148–154 (antireflection coating of a solar cell) and slides 267–276 (multiple-beam thin-film interference, minimum and maximum reflectance, transmission fringes of an a-Se film, incoherent multiple reflections in a plate).* ## Key relations A film of index $n_2$ and thickness $d$ on a substrate $n_3$ reflects a sum of partial waves (slides 267–269). At normal incidence, with the round-trip phase $\phi=2\cdot\frac{2\pi}{\lambda}n_2d$: $$ r=\frac{r_{12}+r_{23}e^{-j\phi}}{1+r_{12}r_{23}e^{-j\phi}},\qquad r_{ij}=\frac{n_i-n_j}{n_i+n_j}. $$ | result | formula | slide | |---|---|---| | AR condition (phase) | $\phi=m\pi$, $m$ odd $\Rightarrow d=m\,\dfrac{\lambda}{4n_2}$ | 151–152, 270 | | AR condition (amplitude) | $n_2=\sqrt{n_1n_3}$ so that $r_{12}=r_{23}$ | 153, 270 | | extreme reflectances ($n_1<n_2<n_3$) | $R_\text{min}=\left(\dfrac{n_1n_3-n_2^2}{n_1n_3+n_2^2}\right)^2$ at $\phi=$ odd $\pi$; $\ R_\text{max}=\left(\dfrac{n_3-n_1}{n_3+n_1}\right)^2$ at $\phi=$ even $\pi$ | 272 | | incoherent plate | $T_\text{plate}=(1-R)^2(1+R^2+R^4+\dots)=\dfrac{1-R}{1+R}$ | 276 | If $n_1<n_3<n_2$ the expressions for $R_\text{min}$ and $R_\text{max}$ are interchanged (slide 272). ### Thin-film designer, live ```{ojs} //| echo: false viewof tf = Inputs.form({ n2: Inputs.range([1, 4], {value: 1.9, step: 0.01, label: "film index n₂"}), n3: Inputs.range([1, 4], {value: 3.5, step: 0.01, label: "substrate index n₃"}), d: Inputs.range([0, 600], {value: 92, step: 1, label: "film thickness d (nm)"}) }) ``` ```{ojs} //| echo: false tfData = { const n1 = 1, r12 = (n1 - tf.n2) / (n1 + tf.n2), r23 = (tf.n2 - tf.n3) / (tf.n2 + tf.n3); const out = []; for (let l = 300; l <= 1200; l += 1) { const phi = 4 * Math.PI * tf.n2 * tf.d / l, c = Math.cos(phi), s = Math.sin(phi); // r = (r12 + r23 e^{-jφ}) / (1 + r12 r23 e^{-jφ}) const nr = r12 + r23 * c, ni = -r23 * s, dr = 1 + r12 * r23 * c, di = -r12 * r23 * s; const R = (nr*nr + ni*ni) / (dr*dr + di*di); out.push({l, R, q: "with film"}, {l, R: ((n1 - tf.n3)/(n1 + tf.n3))**2, q: "bare substrate"}); } return out; } Plot.plot({ width: Math.min(width, 1050), height: 320, x: {label: "λ (nm)"}, y: {label: "reflectance R", domain: [0, Math.max(0.45, d3.max(tfData, d => d.R))], grid: true}, color: {legend: true, domain: ["with film", "bare substrate"], range: ["#2a78d6", "#b9b8b2"]}, marks: [Plot.rectX([{x1: 380, x2: 750}], {x1: "x1", x2: "x2", fill: "#eda100", fillOpacity: 0.07}), Plot.line(tfData, {x: "l", y: "R", stroke: "q", strokeWidth: 2}), Plot.tip(tfData.filter(d => d.q === "with film"), Plot.pointerX({x: "l", y: "R", title: d => `λ = ${d.l} nm\nR = ${(100*d.R).toFixed(2)} %`}))] }) ``` ```{ojs} //| echo: false { const n1 = 1, nideal = Math.sqrt(n1 * tf.n3); const Rmin = ((n1*tf.n3 - tf.n2**2) / (n1*tf.n3 + tf.n2**2))**2; const lmin = [1, 3, 5].map(m => 4 * tf.n2 * tf.d / m).filter(l => l >= 300 && l <= 1200); return html`<div class="result-box">ideal AR index √(n₁n₃) = ${nideal.toFixed(3)} · quarter-wave minima (R = ${(100*Rmin).toFixed(2)} %) at ${lmin.length ? lmin.map(l => l.toFixed(0) + " nm").join(", ") : "none in range"} · bare substrate R = ${(100*((n1-tf.n3)/(n1+tf.n3))**2).toFixed(1)} % · shaded: visible</div>` } ``` With the defaults (Si$_3$N$_4$ on silicon, $d=92$ nm) you get the solar-cell example below. ## Worked example: antireflection coating on a silicon solar cell (slides 148–154) ```{python} R_si = ((1 - 3.5)/(1 + 3.5))**2 print(f"bare Si (n = 3.5): R = {R_si:.1%} → about 30 % of the light is lost (slide 149)") print(f"ideal coating index √(1·3.5) = {np.sqrt(3.5):.2f} (Si3N4, n ≈ 1.9, is a good choice)") print(f"quarter-wave thickness at 700 nm: d = λ/(4 n2) = {700/(4*1.9):.1f} nm (or odd multiples)") ``` A single quarter-wave layer is perfect only at one wavelength and only if $n_2=\sqrt{n_1n_3}$ (slide 154). SiNₓ ($n\approx2.0$) is the industrial compromise. A dispersion-free $n_\text{Si}=3.5$ is assumed here for clarity. ```{python} #| label: fig-ar-si #| fig-cap: "Reflectance of silicon with single- and double-layer AR coatings designed for 700 nm (non-dispersive model)." lam = np.linspace(nm(400), nm(1100), 700); l0, nSi = nm(700), 3.5 fig, ax = plt.subplots() ax.plot(lam*1e9, ol.tmm([1, nSi], [], lam)[0], color=ol.INK2, label="bare Si") for n2, lbl in ((np.sqrt(nSi), "ideal n₂ = √3.5"), (2.0, "SiNₓ n₂ = 2.0")): ax.plot(lam*1e9, ol.tmm([1, n2, nSi], [l0/(4*n2)], lam)[0], label=f"λ/4 {lbl}, d = {l0/(4*n2)*1e9:.0f} nm") ax.plot(lam*1e9, ol.tmm([1, 1.38, 2.3, nSi], [l0/(4*1.38), l0/(4*2.3)], lam)[0], "--", label="double layer 1.38 / 2.3") ax.set(xlabel="λ (nm)", ylabel="R", ylim=(0, 0.4)); ax.legend(fontsize=8); plt.show() ``` ```{python} #| label: fig-ar-real #| fig-cap: "Same SiNx coating on crystalline silicon with measured dispersion and absorption (refractiveindex.info, Green 2008), if the optional package is available." try: from refractiveindex import RefractiveIndexMaterial Si = RefractiveIndexMaterial(shelf="main", book="Si", page="Green-2008") lam_nm = np.linspace(400, 1100, 200) N_Si = np.array([Si.get_refractive_index(l) - 1j*Si.get_extinction_coefficient(l) for l in lam_nm]).squeeze() R = [ol.tmm([1, 2.0, Ni], [nm(700)/(4*2.0)], nm(l))[0][0] for Ni, l in zip(N_Si, lam_nm)] Rb = [ol.tmm([1, Ni], [], nm(l))[0][0] for Ni, l in zip(N_Si, lam_nm)] plt.plot(lam_nm, Rb, color=ol.INK2, label="bare c-Si"); plt.plot(lam_nm, R, label="λ/4 SiNₓ at 700 nm") plt.xlabel("λ (nm)"); plt.ylabel("R"); plt.legend(); plt.show() except Exception as err: print("refractiveindex package or database not available here:", type(err).__name__) ``` ## Worked example: a dielectric film on a semiconductor (slides 273–275) > A semiconductor ($n_3=3.5$) is coated with a 160 nm transparent film ($n_2=2.5$) in air. Find the minimum and maximum > reflectances and transmittances and their wavelengths in the visible. ```{python} #| label: fig-film #| fig-cap: "Reflectance of a 160 nm film (n = 2.5) on n = 3.5 (slide 275). The dots mark the visible extrema." n1, n2, n3, d = 1.0, 2.5, 3.5, 160e-9 Rmin = ((n1*n3 - n2**2)/(n1*n3 + n2**2))**2 Rmax = ((n3 - n1)/(n3 + n1))**2 print(f"R_min = {Rmin:.3f} → T_max = {1-Rmin:.0%}; R_max = {Rmax:.3f} → T_min = {1-Rmax:.0%}") for m in (1, 3): print(f"R_min (φ = {m}π) at λ = 4 n2 d / {m} = {4*n2*d/m*1e9:.0f} nm") for m in (1, 2, 3): print(f"R_max (φ = {2*m}π) at λ = 2 n2 d / {m} = {2*n2*d/m*1e9:.0f} nm") lam = np.linspace(nm(350), nm(900), 1500) R = ol.tmm([n1, n2, n3], [d], lam)[0] fig, ax = plt.subplots(); ax.plot(lam*1e9, R, label="R"); ax.plot(lam*1e9, 1 - R, "--", label="T") for l in (533.3e-9, 800e-9, 400e-9): ax.plot(l*1e9, ol.tmm([n1, n2, n3], [d], l)[0][0], "o", color=ol.INK) ax.set(xlabel="λ (nm)", ylabel="R, T"); ax.legend(); plt.show() ``` In the visible, $R_\text{min}$ (8 %) occurs at 533 nm and $R_\text{max}$ (31 %) at 400 nm, which gives $T_\text{max}=92\,\%$ and $T_\text{min}=69\,\%$ as on slide 275. ## Coherent and incoherent reflections (slides 274, 276) A thin film gives interference fringes (slide 274: transmission of an a-Se film on glass). A thick plate does not, because the source's coherence length (@sec-coherence) is shorter than the round trip. Then intensities, not amplitudes, add: ```{python} Rg = ((1.5 - 1)/(1.5 + 1))**2 print(f"glass plate, incoherent multiple reflections: T = (1−R)/(1+R) = {(1-Rg)/(1+Rg):.4f} " f"(single surface 1−R = {1-Rg:.4f}, two surfaces without multiple reflections (1−R)² = {(1-Rg)**2:.4f})") ``` ::: {.callout-tip collapse="true"} ## Going deeper - The double-layer design above is illustrative. Optimise both thicknesses to minimise the solar-spectrum-weighted reflectance (AM1.5G spectrum, `scipy.optimize`). What is gained versus texturing the surface? - Graded-index and moth-eye surfaces: why does a gradual index transition suppress reflection? Relate it to the limit of many thin layers. - Swanepoel's method (slide 274) extracts $n(\lambda)$ and thickness of a film from the envelopes of its transmission fringes. Implement it on a simulated spectrum from `ol.tmm`. - Facet coatings of semiconductor lasers and SOAs: AR (below $10^{-4}$) for amplifiers, HR for lasers. Estimate the tolerance on thickness for a single-layer AR at 1550 nm. ::: ::: {.callout-note collapse="true"} ## Complements - Macleod, *Thin-Film Optical Filters*: design of AR, HR and band-pass coatings. - R. Swanepoel, "Determination of the thickness and optical constants of amorphous silicon", *J. Phys. E* 16, 1214 (1983). - M. A. Green, *Solar Cells: Operating Principles, Technology and System Applications*. ::: ## Try it yourself 1. Use the live panel to design a single-layer AR for a glass lens ($n_3=1.52$) at 550 nm. Why is MgF₂ ($n=1.38$) used even though it is not the ideal index? 2. Why do soap bubbles show colours, and why does a soap film turn black just before it bursts? 3. Find the thinnest film that makes a silicon surface look blue (minimum reflection in the red, maximum in the blue).