Chapter 8: Guided-Wave Optics ============================= Source: Saleh and Teich, *Fundamentals of Photonics*, second edition, Chapter 8. These solutions retain the intermediate algebra so that every boundary condition, mode count, and numerical result can be checked. Shared notation and conventions ------------------------------- The phasor convention is :math:`e^{j\omega t}` with guided-wave factor :math:`e^{-j\beta z}`. The vacuum quantities are :math:`\lambda_0` and :math:`k_0=2\pi/\lambda_0`; in a core of index :math:`n_1`, :math:`\lambda=\lambda_0/n_1`. The bounce angle :math:`\theta` is measured from the guide axis, so .. math:: :label: fop-ch08-critical-complement \bar\theta_c=\cos^{-1}\!\left(\frac{n_2}{n_1}\right), \qquad \mathrm{NA}=n_1\sin\bar\theta_c =\sqrt{n_1^2-n_2^2}. For a symmetric slab of full width :math:`d`, define .. math:: :label: fop-ch08-normalized-variables k_y=n_1k_0\sin\theta,\quad \gamma=\sqrt{\beta^2-n_2^2k_0^2},\quad u=\frac{k_yd}{2},\quad w=\frac{\gamma d}{2},\quad V=\frac{k_0d}{2}\mathrm{NA},\qquad u^2+w^2=V^2. Here scalar :math:`u` is a dimensionless transverse phase, whereas :math:`u_m(y)` is the book's normalized mode function, :math:`\int|u_m(y)|^2dy=1`. In-text exercises ----------------- Exercise 8.1-1 — Modal power ^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Step 1 — Definitions and setup.** The TE modal field is .. math:: :label: fop-exercise-8-1-1-field \boldsymbol E_m=\hat{\boldsymbol x}\,a_mu_m(y)e^{-j\beta_mz}, \qquad \int|u_m(y)|^2dy=1, where :math:`\beta_m=n(\omega/c_0)\cos\theta_m` and :math:`\eta=\sqrt{\mu/\epsilon}=\eta_0/n`. Power is per unit length in the invariant :math:`x` direction. .. _fop-exercise-8-1-1-illustration: .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/exercise_illustrations/exercise_08_01_01.svg :alt: Illustrated calculation map for Exercise 8.1-1, Modal power :align: center :width: 95% **Figure 56 — Exercise 8.1-1: Modal power.** The normalized transverse field and axial propagation constant determine the integrated Poynting flux. **Step 2 — Mathematical formulas used.** We use :ref:`vector-calculus identities `, :ref:`complex phasors `, and :ref:`integration identities `. **Step 3 — Derive the associated magnetic field.** From :math:`\nabla\times\boldsymbol E=-j\omega\mu\boldsymbol H`, .. math:: :label: fop-exercise-8-1-1-hy H_{y,m}=\frac{\beta_m}{\omega\mu} a_mu_m(y)e^{-j\beta_mz}. The sign gives positive axial flux because :math:`\hat{\boldsymbol x}\times\hat{\boldsymbol y} =\hat{\boldsymbol z}`. **Step 4 — Integrate the time-averaged Poynting vector.** .. math:: :label: fop-exercise-8-1-1-poynting \begin{aligned} P_{z,m} &=\frac12\operatorname{Re}\int E_{x,m}H_{y,m}^*dy\\ &=\frac12\frac{\beta_m}{\omega\mu}|a_m|^2 \int|u_m(y)|^2dy\\ &=\frac{|a_m|^2}{2}\frac{n/c_0}{\mu}\cos\theta_m. \end{aligned} Since :math:`(n/c_0)/\mu=\sqrt{\epsilon/\mu}=1/\eta`, .. math:: :label: fop-exercise-8-1-1-result \boxed{P_{z,m}=\frac{|a_m|^2}{2\eta}\cos\theta_m}. **Step 5 — Check.** At :math:`\theta_m=0` this is the plane-wave value; as :math:`\theta_m\to90^\circ`, the axial power tends to zero. The result has units :math:`|a_m|^2/\eta=\mathrm W` under the modal normalization. Exercise 8.1-2 — Multimode power ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Step 1 — Definitions and setup.** .. math:: :label: fop-exercise-8-1-2-fields E_x=\sum_m a_mu_m(y)e^{-j\beta_mz},\qquad H_y=\sum_n\frac{\beta_n}{\omega\mu} a_nu_n(y)e^{-j\beta_nz}, and the real mirror-guide modes obey :math:`\int u_m(y)u_n(y)dy=\delta_{mn}`. .. _fop-exercise-8-1-2-illustration: .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/exercise_illustrations/exercise_08_01_02.svg :alt: Illustrated calculation map for Exercise 8.1-2, Multimode power :align: center :width: 95% **Figure 57 — Exercise 8.1-2: Multimode power.** Orthogonal modes add in field amplitude, while their integrated powers add without cross terms. **Step 2 — Mathematical formulas used.** We use :ref:`integration identities ` and :ref:`complex phasors `. **Step 3 — Expand before using orthogonality.** .. math:: :label: fop-exercise-8-1-2-double-sum \begin{aligned} P_z&=\frac12\operatorname{Re}\int E_xH_y^*dy\\ &=\frac12\operatorname{Re}\sum_m\sum_n \frac{\beta_n}{\omega\mu}a_ma_n^* e^{-j(\beta_m-\beta_n)z}\int u_m(y)u_n(y)dy. \end{aligned} The Kronecker delta removes every :math:`m\ne n` cross term, and the phase of each surviving diagonal term is unity: .. math:: :label: fop-exercise-8-1-2-diagonal P_z=\frac12\sum_m\frac{\beta_m}{\omega\mu}|a_m|^2. **Step 4 — Substitute the single-mode result.** .. math:: :label: fop-exercise-8-1-2-result \boxed{P_z=\sum_m\frac{|a_m|^2}{2\eta}\cos\theta_m}. **Step 5 — Check.** With only :math:`a_q\ne0`, this reduces to Exercise 8.1-1. It is independent of :math:`z`, as required in a lossless guide. Exercise 8.2-1 — Slab confinement ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Step 1 — Definitions and setup.** Use the unnormalized symmetric-slab profile .. math:: :label: fop-exercise-8-2-1-profile f_m(y)= \begin{cases} \cos(k_yy),&|y|\le d/2,\ m\ \text{even},\\ \sin(k_yy),&|y|\le d/2,\ m\ \text{odd},\\ f_m(d/2)e^{-\gamma(y-d/2)},&y>d/2, \end{cases} with even or odd continuation below. Normalization cancels from :math:`\Gamma_m=P_{\rm core}/P_{\rm total}`. .. _fop-exercise-8-2-1-illustration: .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/exercise_illustrations/exercise_08_02_01.svg :alt: Illustrated calculation map for Exercise 8.2-1, Slab confinement :align: center :width: 95% **Figure 58 — Exercise 8.2-1: Slab confinement.** The harmonic core power is compared with both evanescent tails. **Step 2 — Mathematical formulas used.** We use :ref:`integration identities `, :ref:`trigonometric identities `, and :ref:`exponential identities `. **Step 3 — Integrate the core.** Let :math:`\sigma_m=(-1)^m` and :math:`Q=k_yd`. The even and odd cases combine as .. math:: :label: fop-exercise-8-2-1-core-integral I_{\rm core}=\frac d2+\sigma_m\frac{\sin Q}{2k_y} =\frac d2\left(1+\sigma_m\frac{\sin Q}{Q}\right), while the boundary intensity is .. math:: :label: fop-exercise-8-2-1-boundary-field |f_m(d/2)|^2=\frac{1+\sigma_m\cos Q}{2}. **Step 4 — Integrate both tails.** .. math:: :label: fop-exercise-8-2-1-cladding-integral I_{\rm clad}=2|f_m(d/2)|^2\int_0^\infty e^{-2\gamma s}ds =\frac{|f_m(d/2)|^2}{\gamma}. Express the transverse phase and decay only through the requested variables: .. math:: :label: fop-exercise-8-2-1-angle-variables Q=2\pi\frac d\lambda\sin\theta_m,\qquad G=\gamma d=2\pi\frac d\lambda \sqrt{\sin^2\bar\theta_c-\sin^2\theta_m}. Therefore .. math:: :label: fop-exercise-8-2-1-result \boxed{\Gamma_m=\left[ 1+\frac{1+\sigma_m\cos Q} {G(1+\sigma_m\sin Q/Q)}\right]^{-1}}, \qquad \sigma_m=(-1)^m. **Step 5 — Check.** Increasing :math:`m` increases :math:`\theta_m`, decreases :math:`\gamma`, and increases the tail length :math:`1/\gamma`, which diverges at cutoff. Thus the nodeless :math:`m=0` mode has the greatest confinement. The limit :math:`\gamma\to0^+` correctly gives :math:`\Gamma_m\to0` near cutoff. Exercise 8.2-2 — Asymmetric slab ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Step 1 — Definitions and setup.** The core, substrate, and cover indices are :math:`n_1`, :math:`n_2`, and :math:`n_3`, with :math:`n_3` and :ref:`algebraic rearrangement `. **Step 3 — Part (a): maximum angle and NA.** Since :math:`\cos\bar\theta_{cj}=n_j/n_1` and :math:`n_2>n_3`, the substrate has the smaller complementary critical angle: .. math:: :label: fop-exercise-8-2-2-part-a \boxed{\theta_{\max}=\cos^{-1}(n_2/n_1)}, \qquad \boxed{\mathrm{NA}=n_1\sin\theta_{\max} =\sqrt{n_1^2-n_2^2}}. **Step 4 — Part (b): self-consistency.** For TE polarization define .. math:: :label: fop-exercise-8-2-2-reflection-phases \tan\frac{\phi_{1j}}2 =\sqrt{\frac{\sin^2\bar\theta_{cj}}{\sin^2\theta}-1}, \qquad j=2,3. One round trip contains one reflection at each boundary, so .. math:: :label: fop-exercise-8-2-2-result \boxed{2k_yd-\phi_{12}-\phi_{13}=2\pi m}, \qquad k_y=\frac{2\pi}{\lambda}\sin\theta. When :math:`n_3=n_2`, the two phases become equal and the symmetric-slab condition is recovered. **Step 5 — Part (c): many-mode limit.** Reflection phase changes shift the endpoints only by order unity. With approximate spacing :math:`\Delta(\sin\theta)=\lambda/(2d)`, .. math:: :label: fop-exercise-8-2-2-mode-count \boxed{M\simeq\frac{2d}{\lambda}\sin\theta_{\max} =\frac{2d}{\lambda_0}\sqrt{n_1^2-n_2^2}},\qquad M\gg1. The cover still changes individual propagation constants, but the higher-index substrate controls the leading mode count. **Check.** Setting :math:`n_3=n_2` recovers both the symmetric phase condition and its symmetric large-mode count. End-of-chapter problems ----------------------- Problem 8.1-3 — Mirror-guide field ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** The perfect mirrors are at :math:`y=\pm d/2`; both boundary values must vanish for every :math:`z`. **Mathematical formulas used.** We use :ref:`complex-exponential identities ` and :ref:`algebraic rearrangement `. **Worked derivation.** **Part (a): test one plane wave.** At the upper mirror, .. math:: :label: fop-problem-8-1-3-single-boundary E_x(d/2,z)=Ae^{-jk_yd/2}e^{-j\beta z}=0. The exponentials never vanish, so :math:`A=0`. A nonzero single traveling wave cannot form nodes at both mirrors. **Part (b): impose both boundaries on two waves.** Set :math:`k_{y1}=q`, :math:`k_{y2}=-q`, and :math:`\beta_1=\beta_2=\beta`. Using :ref:`complex-exponential identities `, .. math:: :label: fop-problem-8-1-3-two-wave-field E_x=e^{-j\beta z}\left(A_1e^{-jqy}+A_2e^{jqy}\right). The equations at :math:`y=+d/2` and :math:`y=-d/2` are .. math:: :label: fop-problem-8-1-3-boundary-system A_1e^{-jqd/2}+A_2e^{jqd/2}=0,\qquad A_1e^{jqd/2}+A_2e^{-jqd/2}=0. A nonzero solution requires the determinant to vanish: .. math:: :label: fop-problem-8-1-3-determinant e^{-jqd}-e^{jqd}=-2j\sin(qd)=0 \quad\Longrightarrow\quad \boxed{q=k_y=\frac{m\pi}{d}},\quad m=1,2,\ldots. The first boundary equation then fixes the relative sign: .. math:: :label: fop-problem-8-1-3-result \boxed{\frac{A_2}{A_1}=-e^{-jqd}=(-1)^{m+1}}. Odd :math:`m` therefore gives :math:`A_2=A_1` and a cosine mode; even :math:`m` gives :math:`A_2=-A_1` and a sine mode. .. note:: The printed problem says the sum “does not satisfy” the boundaries under these conditions. Direct substitution shows that an arbitrary :math:`\pm` sign fails, but the parity-matched sign above satisfies both mirrors. This appears to be a wording error in the printed question. **Check.** :math:`\cos(m\pi y/d)` vanishes at both mirrors for odd :math:`m`, while :math:`\sin(m\pi y/d)` vanishes there for even :math:`m`. Problem 8.1-4 — Mirror-guide dispersion ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** :math:`\lambda_0=0.633\ \mu\mathrm m`, :math:`d=10\ \mu\mathrm m`, and :math:`n=1`, hence :math:`\lambda=\lambda_0/n=0.633\ \mu\mathrm m`. **Mathematical formulas used.** We use the :ref:`chain rule ` and :ref:`algebraic and dimensional checks `. **Worked derivation.** **Mode count.** The guide permits :math:`m=1,\ldots,M` with :math:`m\lambda/(2d)<1`. Thus .. math:: :label: fop-problem-8-1-4-count \frac{2d}{\lambda}=\frac{20}{0.633}=31.5956 \quad\Longrightarrow\quad \boxed{M_{\rm TE}=31,\qquad M_{\rm TM}=31}. There are 62 polarization-resolved modes. The book's mirror-guide sequence starts at :math:`m=1`, not at a TEM :math:`m=0` endpoint. **Derive the group velocity.** With nondispersive :math:`n`, .. math:: :label: fop-problem-8-1-4-dispersion \beta_m^2=\left(\frac{n\omega}{c_0}\right)^2 -\left(\frac{m\pi}{d}\right)^2. Differentiate using the :ref:`chain rule `: .. math:: :label: fop-problem-8-1-4-group-velocity v_{g,m}=\left(\frac{d\beta_m}{d\omega}\right)^{-1} =\frac{c_0}{n}\sqrt{1-\left(\frac{m\lambda}{2d}\right)^2}. The fastest and slowest modes are :math:`m=1` and :math:`m=31`: .. math:: :label: fop-problem-8-1-4-velocities \begin{aligned} v_{g,1}&=c_0\sqrt{1-(0.633/20)^2} =\boxed{2.99642\times10^8\ \mathrm{m\,s^{-1}}},\\ v_{g,31}&=c_0\sqrt{1-(31\times0.633/20)^2} =\boxed{5.79342\times10^7\ \mathrm{m\,s^{-1}}}. \end{aligned} **Pulse spread over 1 m.** .. math:: :label: fop-problem-8-1-4-result \begin{aligned} \Delta t &=L\left(\frac1{v_{g,31}}-\frac1{v_{g,1}}\right)\\ &=1.726096\times10^{-8}-3.337313\times10^{-9}\ \mathrm s\\ &=\boxed{13.924\ \mathrm{ns}}. \end{aligned} The high-order zigzag ray follows the longer path, so the positive delay is also the expected physical sign. **Check.** Both velocities lie between zero and :math:`c_0`, and direct substitution of them in :math:`L/v_g` reproduces the stated delay. Problem 8.2-3 — Film in index-1.4 cladding ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** :math:`\lambda_0=0.87\ \mu\mathrm m`, :math:`d=2.00\ \mu\mathrm m`, :math:`n_1=1.60`, and :math:`n_2=1.40`. **Mathematical formulas used.** We use :ref:`trigonometric identities `, the :ref:`chain rule `, and :ref:`algebraic rearrangement `. **Worked derivation.** **Part (a): angles and aperture.** .. math:: :label: fop-problem-8-2-3-part-a \begin{aligned} \theta_c&=\sin^{-1}(1.4/1.6)=\boxed{61.045^\circ},\\ \bar\theta_c&=90^\circ-\theta_c=\boxed{28.955^\circ},\\ \mathrm{NA}&=\sqrt{1.6^2-1.4^2}=\boxed{0.774597}. \end{aligned} At an air entrance face, Snell's law gives :math:`\sin\theta_{a,\max}=\mathrm{NA}`, hence .. math:: :label: fop-problem-8-2-3-acceptance \boxed{\theta_{a,\max}=\sin^{-1}(0.774597)=50.768^\circ}. **Part (b): TE mode count.** The chapter uses the smallest integer greater than :math:`2d\,\mathrm{NA}/\lambda_0`: .. math:: :label: fop-problem-8-2-3-count \frac{2d\,\mathrm{NA}}{\lambda_0}=3.56136 \quad\Longrightarrow\quad \boxed{M_{\rm TE}=4}. **Part (c): TE0 eigenvalue and angle.** .. math:: :label: fop-problem-8-2-3-root V=\frac{\pi d}{\lambda_0}\mathrm{NA}=5.594177,\qquad u\tan u=\sqrt{V^2-u^2}. Solving on :math:`0` gives .. math:: :label: fop-problem-8-2-3-beta-derivative \frac{d\beta}{dk_0}= \frac{n_1^2k_0-(2u\,\mathrm{NA}/d)(du/dV)}{\beta}. Since :math:`\omega=c_0k_0`, .. math:: :label: fop-problem-8-2-3-result \boxed{v_g=\frac{c_0}{d\beta/dk_0} =1.86508\times10^8\ \mathrm{m\,s^{-1}}}. The simpler ray estimate :math:`(c_0/n_1)\cos\theta_0 =1.86124\times10^8\ \mathrm{m/s}` differs by 0.21%; the exact derivative includes the frequency-dependent reflection-phase delay. **Check.** The root obeys :math:`u\tan u=w` to the shown precision and :math:`\theta_0<\bar\theta_c`. The exact group velocity is close to the ray estimate and may lie slightly below :math:`c_0/n_1`, as expected from the chapter's waveguide-dispersion curve. Problem 8.2-4 — Film suspended in air ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Keep the preceding :math:`\lambda_0,d,n_1`, but set :math:`n_2=1`. **Mathematical formulas used.** We use :ref:`trigonometric identities `, the implicit :ref:`chain rule `, and :ref:`algebraic rearrangement `. **Worked derivation.** **Part (a).** .. math:: :label: fop-problem-8-2-4-part-a \theta_c=\boxed{38.682^\circ},\qquad \bar\theta_c=\boxed{51.318^\circ},\qquad \mathrm{NA}=\boxed{1.24900}. The formal entrance relation :math:`\sin\theta_a=1.249>1` means the guide accepts the entire propagating angular range available in air: :math:`\boxed{\theta_{a,\max}=90^\circ}` in the ideal end-face model. **Part (b).** .. math:: :label: fop-problem-8-2-4-count \frac{2d\,\mathrm{NA}}{\lambda_0}=5.74253 \quad\Longrightarrow\quad \boxed{M_{\rm TE}=6}. **Part (c).** The even-mode equation gives .. math:: :label: fop-problem-8-2-4-root V=9.020340,\qquad u=1.4134518,\qquad w=8.9089107,\qquad \boxed{\theta_0=7.02606^\circ}. Repeating :eq:`fop-problem-8-2-3-du-dv`-- :eq:`fop-problem-8-2-3-beta-derivative` with :math:`du/dV=0.0158137` yields .. math:: :label: fop-problem-8-2-4-result \boxed{v_g=1.86244\times10^8\ \mathrm{m\,s^{-1}}}. The ray estimate is :math:`1.85963\times10^8\ \mathrm{m/s}`. Relative to Problem 8.2-3, air cladding increases the mode count from 4 to 6 and raises the TE0 confinement from 99.12% to 99.75%, while changing its angle and group velocity only slightly. **Check.** The lower cladding index increases NA and therefore cannot decrease the mode count; all calculated angles remain within the TIR limit. Problem 8.2-5 — TE0 field and confinement ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** :math:`n_1=1.48`, :math:`n_2=1.46`, :math:`d=0.500\ \mu\mathrm m`, and :math:`\lambda_0=0.850\ \mu\mathrm m`. **Mathematical formulas used.** We use :ref:`integration identities `, :ref:`exponential identities `, and :ref:`trigonometric identities `. **Worked derivation.** **Ratio of proportionality constants.** Matching the book's :math:`A\cos(k_yy+\varphi)` core field to :math:`Be^{-\gamma y}` at :math:`y=d/2` gives .. math:: :label: fop-problem-8-2-5-ratio-general A\cos(k_yd/2+\varphi)=Be^{-\gamma d/2} \quad\Longrightarrow\quad \boxed{\frac BA=e^{\gamma d/2}\cos(k_yd/2+\varphi)}. For even TE0, :math:`\varphi=0`, so :math:`B/A=e^w\cos u`. **Solve the eigenvalue.** .. math:: :label: fop-problem-8-2-5-root \mathrm{NA}=0.242487,\quad V=0.448115,\quad u\tan u=\sqrt{V^2-u^2} \Longrightarrow u=0.4108277,\ w=0.1789630. Thus :math:`k_y=1.643311\ \mu\mathrm m^{-1}`, :math:`\gamma=0.715852\ \mu\mathrm m^{-1}`, and :math:`B/A=1.096460`. A continuous normalized profile is .. math:: :label: fop-problem-8-2-5-normalized-profile u_0(y)=N\begin{cases} \cos(k_yy),&|y|\le d/2,\\ \cos u\,e^{-\gamma(|y|-d/2)},&|y|>d/2. \end{cases} Using :ref:`integration identities `, .. math:: :label: fop-problem-8-2-5-normalization \begin{aligned} N^{-2}&=\left[\frac d2+\frac{\sin(2u)}{2k_y}\right] +\frac{\cos^2u}{\gamma}\\ &=0.4728045+1.1741324=1.6469369\ \mu\mathrm m, \end{aligned} so :math:`\boxed{N=0.779223\ \mu\mathrm m^{-1/2}}`. .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/ch08/problem_08_02_05_te0_field.svg :alt: Normalized TE0 field and evanescent cladding tails :align: center :width: 88% **Figure 60 — Problem 8.2-5: normalized TE0 field.** The shaded core is :math:`-0.25\le y\le0.25\ \mu\mathrm m`; field and slope are continuous at both interfaces. **Confinement.** The factor :math:`N^2` cancels, so .. math:: :label: fop-problem-8-2-5-result \boxed{\Gamma_0=\frac{0.4728045}{0.4728045+1.1741324} =0.287081=28.71\%}. This is plausible because :math:`V=0.448\ll1` and the intensity decay length :math:`1/(2\gamma)=0.6985\ \mu\mathrm m` exceeds the core half-width. **Check.** Substitution of :math:`N` makes the total field integral unity, and the core and cladding fractions sum to one. Problem 8.2-6 — Maxwell derivation ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** The TE field has one electric-field component; the media are nonmagnetic and the interfaces are parallel to the guide axis. **Mathematical formulas used.** The curl, Helmholtz equation, exponential derivative, and electromagnetic boundary conditions are used below. **Worked derivation.** Let :math:`\boldsymbol E=\hat{\boldsymbol x}u(y)e^{-j\beta z}`. With the :math:`e^{j\omega t}` convention and :ref:`vector-calculus identities `, .. math:: :label: fop-problem-8-2-6-curl \nabla\times\boldsymbol E =(0,-j\beta u,-u')e^{-j\beta z} =-j\omega\mu\boldsymbol H. Therefore .. math:: :label: fop-problem-8-2-6-h-fields \boxed{H_y=\frac{\beta}{\omega\mu}u(y)e^{-j\beta z}}, \qquad \boxed{H_z=-\frac{j}{\omega\mu}u'(y)e^{-j\beta z}}. The :ref:`constant-coefficient ODE solutions ` applied to the Helmholtz equation require .. math:: :label: fop-problem-8-2-6-dispersion-pieces k_y^2+\beta^2=n_1^2k_0^2,\qquad -\gamma^2+\beta^2=n_2^2k_0^2. Subtracting them and multiplying by :math:`d^2/4` verifies :math:`u^2+w^2=V^2`. **Apply both boundary conditions.** Equal permeabilities make continuity of tangential :math:`E_x,H_z` equivalent to continuity of :math:`u,u'`. For an even core field at :math:`y=d/2`, .. math:: :label: fop-problem-8-2-6-even-boundary A\cos(k_yd/2)=Be^{-\gamma d/2},\qquad -Ak_y\sin(k_yd/2)=-\gamma Be^{-\gamma d/2}. Dividing the equations gives .. math:: :label: fop-problem-8-2-6-even-eigenvalue \boxed{k_y\tan(k_yd/2)=\gamma} \quad\Longleftrightarrow\quad \boxed{u\tan u=w}. For odd :math:`A\sin(k_yy)`, the same operation yields .. math:: :label: fop-problem-8-2-6-odd-eigenvalue \boxed{-k_y\cot(k_yd/2)=\gamma} \quad\Longleftrightarrow\quad \boxed{-u\cot u=w}. Combine the parity cases and substitute the angle definitions: .. math:: :label: fop-problem-8-2-6-unified \tan\left(\frac{k_yd}{2}-\frac{m\pi}{2}\right)=\frac{\gamma}{k_y}, \qquad \frac{\gamma}{k_y}=\sqrt{ \frac{\sin^2\bar\theta_c}{\sin^2\theta}-1}. Because :math:`k_yd/2=\pi(d/\lambda)\sin\theta`, this becomes .. math:: :label: fop-problem-8-2-6-result \boxed{\tan\left(\pi\frac d\lambda\sin\theta_m-\frac{m\pi}{2}\right) =\sqrt{\frac{\sin^2\bar\theta_c}{\sin^2\theta_m}-1}}, which is Eq. (8.2-4). Derivative continuity is the step that turns a merely continuous trial field into a permitted eigenmode. **Check.** Either parity profile recovers the two regional dispersion relations, and both tangential field components are continuous. Problem 8.2-7 — Single-mode thickness ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** The guide is symmetric and material dispersion is not included. **Mathematical formulas used.** The mode-count inequality and algebraic rearrangement are used below. **Worked derivation.** For :math:`n_1=1.50` and :math:`n_2=1.46`, .. math:: :label: fop-problem-8-2-7-na \mathrm{NA}=\sqrt{1.50^2-1.46^2}=0.344093. The :math:`m=1` branch reaches cutoff at :math:`2d\,\mathrm{NA}/\lambda_0=1`. At equality it is marginal and not evanescent outside, so the largest thickness retaining only confined TE0 is .. math:: :label: fop-problem-8-2-7-thickness \boxed{d_{\max}=\frac{1.30}{2(0.344093)} =1.88902\ \mu\mathrm m}. At :math:`\lambda_0'=0.85\ \mu\mathrm m`, keeping this thickness, .. math:: :label: fop-problem-8-2-7-result \frac{2d_{\max}\mathrm{NA}}{\lambda_0'} =\frac{1.30}{0.85}=1.52941 \quad\Longrightarrow\quad \boxed{M_{\rm TE}=2}. The inverse-wavelength scaling confirms that the shorter wavelength must support more modes. **Check.** Substitution of the limiting thickness returns the first higher-order-mode cutoff, with dimensions of length. Problem 8.2-8 — Cutoff approximation ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** The index difference is small and the mode is a higher-order TE mode. **Mathematical formulas used.** The cutoff phase condition, difference of squares, and weak-guidance approximation are used below. **Worked derivation.** Let :math:`\Delta n=n_1-n_2` and :math:`m>0`. At cutoff, :math:`\gamma\to0`, the reflection phase tends to zero, and :math:`\theta_m\to\bar\theta_c`. The phase condition reduces to .. math:: :label: fop-problem-8-2-8-cutoff-phase \frac{2\pi}{\lambda}(2d\sin\bar\theta_c)=2\pi m \quad\Longrightarrow\quad \frac{2d}{\lambda_0}\sqrt{n_1^2-n_2^2}=m. Solving and factoring the difference of squares gives .. math:: :label: fop-problem-8-2-8-exact \lambda_{0,c}^2 =\frac{4d^2}{m^2}(n_1-n_2)(n_1+n_2). For weak guidance :math:`n_2\simeq n_1`, hence :math:`n_1+n_2\simeq2n_1` and .. math:: :label: fop-problem-8-2-8-result \boxed{\lambda_{0,c}^2\simeq \frac{8n_1\Delta n\,d^2}{m^2}}. Dimensions are length squared on both sides. Increasing :math:`d` or index contrast correctly moves cutoff to a longer wavelength. **Check.** Replacing the index sum by twice the core index in the exact result reproduces the approximation without changing dimensions. Problem 8.2-9 — TM modes ^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** The bounce angle is measured from the guide axis; the requested plot uses the sine of that angle. **Mathematical formulas used.** The TM total-internal-reflection phase, trigonometric identities, and algebraic rearrangement are used below. **Worked derivation.** For total internal reflection of a TM wave, .. math:: :label: fop-problem-8-2-9-tm-phase \tan\frac{\phi_{\rm TM}}2 =\frac{n_1^2}{n_2^2} \sqrt{\frac{\sin^2\bar\theta_c}{\sin^2\theta}-1}. Substitute this in :math:`2k_yd-2\phi_{\rm TM}=2\pi m` and take the tangent of the half-phase equation: .. math:: :label: fop-problem-8-2-9-result \boxed{\tan\left(\pi\frac d\lambda\sin\theta_m-\frac{m\pi}{2}\right) =\frac{n_1^2}{n_2^2} \sqrt{\frac{\sin^2\bar\theta_c}{\sin^2\theta_m}-1}}. The index-squared factor is the difference from TE polarization. For :math:`s=\sin\theta`, :math:`s_c=0.3`, and :math:`\lambda/(2d)=0.1`, one has :math:`d/\lambda=5` and :math:`n_1^2/n_2^2=1/(1-s_c^2)=1.098901`. The numerical equation is .. math:: :label: fop-problem-8-2-9-numeric-equation \tan(5\pi s-m\pi/2)=1.098901\sqrt{0.09/s^2-1}. .. list-table:: TM branch intersections :header-rows: 1 :widths: 15 28 28 29 * - :math:`m` - :math:`\sin\theta_m` - :math:`\theta_m` - status * - 0 - 0.0835713 - :math:`4.79387^\circ` - guided * - 1 - 0.165506 - :math:`9.52662^\circ` - guided * - 2 - 0.242842 - :math:`14.0544^\circ` - guided * - 3 - 0.300000 - :math:`17.4576^\circ` - cutoff boundary .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/ch08/problem_08_02_09_tm_modes.svg :alt: Graphical solution for the TM slab modes :align: center :width: 88% **Figure 61 — Problem 8.2-9: graphical TM-mode solution.** Positive branches of the left side intersect the reflection-phase curve at the listed roots; the fourth point lies exactly at cutoff. The book's count is the smallest integer strictly greater than :math:`s_c/[\lambda/(2d)]=3`, so .. math:: :label: fop-problem-8-2-9-count \boxed{M_{\rm TM}=4}. Only the first three have :math:`\gamma>0`; :math:`m=3` is marginal at the specified exact cutoff. A plot that omits its endpoint will appear to show only three intersections. **Check.** Removing the TM index-squared factor recovers the TE equation; every listed root is on its proper branch and does not exceed cutoff. Problem 8.3-1 — Rectangular-guide mode count ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** The guide is square, and the requested result counts only the TE polarization family. **Mathematical formulas used.** The area mode-count approximation, wavelength-frequency relation, and dimensional checks are used below. **Worked derivation.** The square area is :math:`A=d^2=10^{-2}\ \mathrm{mm^2}=10^{-8}\ \mathrm{m^2}`, so :math:`d=10^{-4}\ \mathrm m`; :math:`\mathrm{NA}=0.1`. For one polarization, Eq. (8.3-3) is .. math:: :label: fop-problem-8-3-1-start M_{\rm TE}\simeq\frac\pi4 \left(\frac{2d}{\lambda_0}\right)^2\mathrm{NA}^2, \qquad \lambda_0=\frac{c_0}{\nu}. Substitution gives .. math:: :label: fop-problem-8-3-1-result \boxed{M_{\rm TE}(\nu)= \pi A\left(\frac{\mathrm{NA}\,\nu}{c_0}\right)^2 =(3.49549\times10^{-27}\ \mathrm{Hz^{-2}})\nu^2}. .. list-table:: Approximate TE mode count :header-rows: 1 :widths: 50 50 * - :math:`\nu` (THz) - :math:`M_{\rm TE}` * - 50 - 8.739 * - 100 - 34.955 * - 200 - 139.820 * - 300 - 314.594 * - 400 - 559.279 .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/ch08/problem_08_03_01_mode_count.svg :alt: Rectangular dielectric guide mode count versus frequency :align: center :width: 88% **Figure 62 — Problem 8.3-1: TE count versus frequency.** A square guide has quadratic growth; the same-width slab comparison is linear. Exact counts would be staircases around these large-mode approximations. Indeed, :math:`M_{\rm slab}\simeq2d\,\mathrm{NA}\,\nu/c_0`, so .. math:: :label: fop-problem-8-3-1-comparison M_{\rm rectangular}\simeq\frac\pi4M_{\rm slab}^2. This is the two-dimensional analogue of Fig. 8.2-4. The area estimate is not an exact integer count at small :math:`M`. **Check.** The coefficient of frequency squared has inverse-hertz-squared units, and substitution at 100 THz gives the tabulated value 34.955. Problem 8.4-1 — Two-slab coupler ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** The identical isolated TE0 fields are normalized to unity and weakly overlap across the gap, so they are phase matched. **Mathematical formulas used.** Field normalization, exponential and trigonometric integration, and coupled-mode power exchange are used below. **Worked derivation.** Each slab has :math:`d=0.500\ \mu\mathrm m`, :math:`n_s=1.48`; the medium has :math:`n=1.46`; the inner-edge gap is :math:`2a=1.00\ \mu\mathrm m`, so :math:`a=0.500\ \mu\mathrm m`; and :math:`\lambda_0=0.850\ \mu\mathrm m`. Put guide 1 in :math:`[-a-d,-a]` and guide 2 in :math:`[a,a+d]`. **Part (a): evaluate Eq. (8.5-6).** Problem 8.2-5 supplies .. math:: :label: fop-problem-8-4-1-mode-data u=0.4108277,\ k_y=1.643311\ \mu\mathrm m^{-1},\quad \gamma=0.715852\ \mu\mathrm m^{-1},\quad \beta=10.816010\ \mu\mathrm m^{-1},\quad N=0.779223\ \mu\mathrm m^{-1/2}. For a guide centered at :math:`y_c`, use .. math:: :label: fop-problem-8-4-1-shifted-profile U(y-y_c)=N\begin{cases} \cos[k_y(y-y_c)],&|y-y_c|\le d/2,\\ \cos u\,e^{-\gamma(|y-y_c|-d/2)},&|y-y_c|>d/2. \end{cases} Then :math:`u_1(y)=U[y+(a+d/2)]` and :math:`u_2(y)=U[y-(a+d/2)]`. For identical guides, .. math:: :label: fop-problem-8-4-1-coupling-equation \kappa=\frac12(n_s^2-n^2)\frac{k_0^2}{\beta} \int_a^{a+d}u_1(y)u_2(y)dy. Inside guide 2, guide 1 supplies its exponential tail. With :math:`x=y-a`, .. math:: :label: fop-problem-8-4-1-overlap \begin{aligned} I&=N^2\cos u\,e^{-2\gamma a} \int_0^d e^{-\gamma x}\cos[k_y(x-d/2)]dx\\ &=0.1111544. \end{aligned} Now :math:`k_0=2\pi/0.85=7.391983\ \mu\mathrm m^{-1}` and :math:`n_s^2-n^2=0.0588000`, so .. math:: :label: fop-problem-8-4-1-result-a \boxed{\kappa=\tfrac12(0.0588) \frac{(7.391983)^2}{10.816010}(0.1111544) =0.0165093\ \mu\mathrm m^{-1} =1.65093\times10^4\ \mathrm m^{-1}}. The normalized overlap is dimensionless, leaving the correct inverse-length unit from :math:`k_0^2/\beta`. **Part (b): 3-dB length.** Identical, phase-matched guides obey .. math:: :label: fop-problem-8-4-1-power-transfer P_1(z)=P_1(0)\cos^2(\kappa z),\qquad P_2(z)=P_1(0)\sin^2(\kappa z). Equal powers require :math:`\kappa L_{3\rm dB}=\pi/4`, hence .. math:: :label: fop-problem-8-4-1-result \boxed{L_{3\rm dB}=\frac{\pi}{4\kappa}=47.573\ \mu\mathrm m}. Complete transfer occurs at twice this length, :math:`L_0=95.146\ \mu\mathrm m`. Increasing the gap decreases the overlap as :math:`e^{-2\gamma a}` and must increase both lengths, providing an independent trend check. **Check.** At the 3-dB length both squared trigonometric factors are one half; at twice that length all power transfers to guide 2.