Run this chapter’s code in Google Colab (no installation).
Slides 176–190 (dielectric mirror, quarter-wave stack, reflectance versus number of pairs and index contrast, stop band, penetration depth) and slides 289–294 (photonic crystals in 1D, 2D and 3D, band gaps, defects).
A dielectric (Bragg) mirror stacks layers of alternating high (\(n_1=n_H\)) and low (\(n_2=n_L\)) index (slide 176). The reflection coefficient changes sign at each interface, and with quarter-wave layers all reflected waves come back in phase (slide 177):
\[ n_1d_1+n_2d_2=\frac{\lambda}{2}\quad\Rightarrow\quad d_1=\frac{\lambda}{4n_1},\ d_2=\frac{\lambda}{4n_2}\qquad\text{(slides 181–182)}. \]
| result | formula | slide |
|---|---|---|
| peak reflectance, \(N\) pairs | \(R_N=\left[\dfrac{n_0n_2^{2N}-n_3n_1^{2N}}{n_0n_2^{2N}+n_3n_1^{2N}}\right]^2\) | 179, 186 |
| stop-band width (infinite stack) | \(\Delta\lambda\approx\dfrac{4\lambda_0}{\pi}\arcsin\!\left(\dfrac{n_1-n_2}{n_1+n_2}\right)\) | 179, 188 |
| penetration depth | \(L_\text{pen}=-\dfrac12\dfrac{d\phi}{dk}\) of the reflection phase, \(\approx\dfrac{\lambda}{4\Delta n}\) | 189 |
\(R\) grows quickly with the number of pairs, and the bandwidth is set mainly by the index contrast (slides 184, 190).
Quarter-wave layers of Ta₂O₅ (\(n_H=1.78\)) and SiO₂ (\(n_L=1.55\)) on Pyrex (\(n_s=1.47\)), in air, at 850 nm. Find \(R\) for \(N=4\) and 12. What if TiO₂ (\(n_H=2.49\)) replaces Ta₂O₅? What is the bandwidth, and what if the layers are interchanged? What about a Si substrate?
lam0 = nm(850)
for name, nH in (("Ta2O5", 1.78), ("TiO2", 2.49)):
for N in (4, 12):
R_formula = ol.bragg_R_max(1, nH, 1.55, 1.47, N)
R_tmm = ol.tmm(*ol.qw_stack(nH, 1.55, N, lam0), lam0)[0][0]
print(f"{name}/SiO2 N = {N:2d}: R = {100*R_formula:7.3f} % (formula) = {100*R_tmm:7.3f} % (transfer matrix)")
R_swap = ol.bragg_R_max(1, 1.55, nH, 1.47, 12)
R_si = ol.bragg_R_max(1, nH, 1.55, 3.5, 12)
print(f"{name}/SiO2 N = 12, L first: R = {100*R_swap:.2f} %; on Si: R = {100*R_si:.2f} %; "
f"stop band Δλ ≈ {ol.bragg_bandwidth(nH, 1.55, lam0)*1e9:.1f} nm\n")Ta2O5/SiO2 N = 4: R = 40.043 % (formula) = 40.043 % (transfer matrix)
Ta2O5/SiO2 N = 12: R = 90.634 % (formula) = 90.634 % (transfer matrix)
Ta2O5/SiO2 N = 12, L first: R = 80.84 %; on Si: R = 95.95 %; stop band Δλ ≈ 74.8 nm
TiO2/SiO2 N = 4: R = 94.049 % (formula) = 94.049 % (transfer matrix)
TiO2/SiO2 N = 12: R = 99.997 % (formula) = 99.997 % (transfer matrix)
TiO2/SiO2 N = 12, L first: R = 99.99 %; on Si: R = 100.00 %; stop band Δλ ≈ 254.1 nm
The slides give R₄ = 94.0 % and R₁₂ ≈ 100 % for TiO₂/SiO₂, 80.8 % for the interchanged Ta₂O₅ stack, and Δλ = 74.8 nm for Ta₂O₅/SiO₂. The Ta₂O₅ values for N = 4 and 12 (40.0 % and 90.6 %) are not printed on the slides but follow from the same formula. The contrast matters: 4 pairs of TiO₂/SiO₂ do better than 12 pairs of Ta₂O₅/SiO₂.
lam = np.linspace(nm(600), nm(1200), 1500)
fig, ax = plt.subplots()
for k, (nH, N) in enumerate(((1.78, 4), (1.78, 12), (2.49, 4))):
ax.plot(lam*1e9, ol.tmm(*ol.qw_stack(nH, 1.55, N, lam0), lam)[0], color=ol.PALETTE[k], label=f"n_H = {nH}, N = {N}")
ax.set(xlabel="λ (nm)", ylabel="R"); ax.legend(fontsize=8); plt.show()
A Bragg mirror reflects as if it were a perfect mirror placed at a distance \(L_\text{pen}\) inside the stack. This matters for the effective cavity length of VCSELs. We extract it from the slope of the reflection phase.
lam = np.linspace(nm(840), nm(860), 801)
_, _, r = ol.tmm(*ol.qw_stack(2.49, 1.55, 12, lam0), lam)
k = 2*np.pi/lam
phi = np.unwrap(np.angle(r))
Lpen = -0.5*np.gradient(phi, k)[400]
print(f"TiO2/SiO2 12 pairs: L_pen = {Lpen*1e9:.0f} nm (estimate λ/(4Δn) = {850/(4*(2.49-1.55)):.0f} nm)")TiO2/SiO2 12 pairs: L_pen = 226 nm (estimate λ/(4Δn) = 226 nm)For an infinite bilayer lattice of period \(\Lambda=d_1+d_2\), Bloch’s theorem gives the dispersion relation of the 1D photonic crystal (slide 292):
\[ \cos(K\Lambda)=\cos(k_1d_1)\cos(k_2d_2)-\tfrac12\left(\tfrac{n_1}{n_2}+\tfrac{n_2}{n_1}\right)\sin(k_1d_1)\sin(k_2d_2). \]
Where the right-hand side exceeds 1 in magnitude, \(K\) is complex: no propagating Bloch wave exists, and a finite stack reflects.
n1, n2, Lam = 2.49, 1.55, 1.0
d1, d2 = Lam*n2/(n1 + n2), Lam*n1/(n1 + n2) # quarter-wave condition n1 d1 = n2 d2
f = np.linspace(1e-3, 1.0, 3000) # normalised frequency Λ/λ
k0 = 2*np.pi*f/Lam
rhs = np.cos(k0*n1*d1)*np.cos(k0*n2*d2) - 0.5*(n1/n2 + n2/n1)*np.sin(k0*n1*d1)*np.sin(k0*n2*d2)
gap = np.abs(rhs) > 1
KL = np.where(gap, np.nan, np.arccos(np.clip(rhs, -1, 1))/np.pi)
R8 = ol.tmm([1.0] + [n1, n2]*8 + [1.0], [d1, d2]*8, Lam/f)[0]
fig, ax = plt.subplots(1, 2, figsize=(10, 4.2), sharey=True)
ax[0].plot(KL, f); ax[0].set(xlabel="Bloch wave vector KΛ/π", ylabel="normalised frequency Λ/λ", xlim=(0, 1))
ax[1].plot(R8, f); ax[1].set(xlabel="R (N = 8)", xlim=(0, 1.05))
idx = np.where(gap)[0]
for blk in np.split(idx, np.where(np.diff(idx) > 1)[0] + 1):
if blk.size:
for a in ax: a.axhspan(f[blk[0]], f[blk[-1]], color=ol.PALETTE[3], alpha=0.2, lw=0)
plt.tight_layout(); plt.show()
The stop band of the finite stack coincides with the photonic band gap of the infinite crystal, and the index contrast sets the gap width. The “interference of reflections” picture explains why a band gap appears; the Bloch picture gives its size. In 2D and 3D (slides 289–293), a full band gap needs a large contrast and the right lattice. Point defects then act as cavities and line defects as waveguides (slide 294).
A fiber Bragg grating (FBG) is a Bragg mirror with a tiny index contrast (\(\Delta n\sim10^{-4}\)) but tens of thousands of periods. The same transfer-matrix approach applies; Chapter 2 treats FBGs in detail (coupled-mode theory, apodisation, sensing).
def periodic_R(n0, na, nb, da, db, N, ns, lam):
def layer(n, d):
dl = 2*np.pi*n*d/lam
M = np.empty((lam.size, 2, 2), complex)
M[:, 0, 0] = np.cos(dl); M[:, 0, 1] = 1j*np.sin(dl)/n
M[:, 1, 0] = 1j*n*np.sin(dl); M[:, 1, 1] = np.cos(dl)
return M
M = np.linalg.matrix_power(layer(na, da) @ layer(nb, db), N)
B = M[:, 0, 0] + M[:, 0, 1]*ns; C = M[:, 1, 0] + M[:, 1, 1]*ns
return np.abs((n0*B - C)/(n0*B + C))**2
neff, lamB = 1.447, nm(1550)
lam = np.linspace(nm(1549.0), nm(1551.0), 3000)
fig, ax = plt.subplots()
for dn, Lg in ((1e-4, 5e-3), (1e-4, 20e-3), (5e-4, 10e-3)):
na, nb = neff, neff + dn
da, db = lamB/(4*na), lamB/(4*nb)
N = int(Lg/(da + db))
ax.plot((lam - lamB)*1e12, periodic_R(neff, na, nb, da, db, N, neff, lam),
label=f"Δn = {dn:.0e}, L = {Lg*1e3:.0f} mm ({N} periods)")
ax.set(xlabel="λ − λ_B (pm)", ylabel="R"); ax.legend(fontsize=8); plt.show()
ol.tmm at oblique incidence.ol.tmm. What appears at the centre of the stop band? (This is the cavity of Chapter 12.)