Chapter 3: Theory of Optical Waveguides
=======================================

Source: Hunsperger, sixth edition (2009), Problems 3.1--3.10,
printed pages 48--49. The full core thickness is used throughout.

Problem 3.1 — Oscillatory and evanescent constants
--------------------------------------------------

**Working.** Substitute :math:`E_y(x,z,t)=F(x)e^{i\omega t-i\beta z}`
into the source-free Helmholtz equation in region :math:`j`:

.. math::

   F''+(k_0^2n_j^2-\beta^2)F=0.

In the core, :math:`\beta<k_0n_2`, so the coefficient is positive
and the solution is sinusoidal with
:math:`h^2=k_0^2n_2^2-\beta^2`. In the claddings,
:math:`\beta>k_0n_j`, giving decaying exponentials.

**Answer.**
:math:`h=\sqrt{k_0^2n_2^2-\beta^2}`,
:math:`q=\sqrt{\beta^2-k_0^2n_1^2}`, and
:math:`p=\sqrt{\beta^2-k_0^2n_3^2}`.

**Check.** All have units of inverse length. Select the exterior
exponential signs so the field decreases away from the core on both sides.

Problem 3.2 — Derive the asymmetric weak-step condition
-------------------------------------------------------

**Working.** At substrate cutoff, :math:`p=0`. For a strongly asymmetric
guide :math:`q/h\gg1`, so the remaining reflection half-phase tends
to :math:`\pi/2`. The resonance condition becomes

.. math::

   ht\simeq(m+\tfrac12)\pi,\qquad
   h^2=k_0^2(n_2^2-n_3^2)\simeq2k_0^2n_2\Delta n.

Squaring and using :math:`k_0=2\pi/\lambda_0` gives

.. math::

   \Delta n_c\simeq\frac{(2m+1)^2\lambda_0^2}{32n_2t^2}.

**Answer.** Propagation requires a contrast greater than this limiting
value within the stated approximation.

**Check.** This is not the exact cutoff for arbitrary cladding indices.
It assumes both strong asymmetry and :math:`\Delta n\ll n_2`.

Problem 3.3 — Rectangular electric fields
-----------------------------------------

**Working.** A separable interior approximation on
:math:`0<x<a,\ 0<y<b` is

.. math::

   E^y_{mn}\propto\sin(m\pi x/a)\sin(n\pi y/b).

Thus :math:`E^y_{12}` has one lobe across :math:`x` and two
opposite-sign lobes across :math:`y`. :math:`E^y_{22}` has four
lobes in a checkerboard of alternating signs. Continue the fields as
evanescent tails outside the dielectric core rather than literal zero walls.

.. figure:: /_static/knowledge_base/worked_exercises/integrated_optics_6e/rectangular_modes.svg
   :alt: Signed E12, E22, and E33 rectangular field patterns and their squared intensities.
   :width: 100%

   Separable schematic fields (top) and squared magnitudes (bottom).

**Answer.** The nodal-line counts are :math:`m-1` and :math:`n-1`
along the two transverse coordinates.

**Check.** This sketches the dominant component of a dielectric hybrid
mode, not an exact metallic-waveguide boundary solution.

Problem 3.4 — Rectangular optical power distributions
-----------------------------------------------------

**Working.** In the same separable approximation, power density is
proportional to :math:`|E^y_{mn}|^2`. Consequently
:math:`E^y_{33}` has a three-by-three array of nine positive intensity
lobes, with two dark nodal lines in each direction.
:math:`E^y_{22}` has a two-by-two array of four lobes.

**Answer.** Use the lower row of the preceding figure: negative field
lobes become positive intensity lobes, while field nodes stay dark.

**Check.** Intensity cannot be negative. For precise vector-mode
power use :math:`\frac12\operatorname{Re}(\mathbf E\times\mathbf H^*)`;
the dominant-field sketch is sufficient to identify the requested lobes.

Problem 3.5 — Symmetric slab counts and fundamental limit
---------------------------------------------------------

**Working.** For a symmetric slab, cutoff is
:math:`V=m\pi`, where
:math:`V=(2\pi t/\lambda_0)\sqrt{n_2^2-n_c^2}`.
Use :math:`t=3\,\mu{\rm m}`, :math:`n_2=1.5`, and
:math:`\lambda_0=0.6328\,\mu{\rm m}`.
Count integers :math:`m\ge0` strictly smaller than :math:`V/\pi`.

**Answer.** (a) Air claddings give 11 TE modes.
(b) Claddings of index 1.48 give 3 TE modes.
(c) The ideal symmetric dielectric slab supports TE0 for every
positive thickness; its limiting cutoff thickness is zero.

**Check.** Zero itself is not a finite guide. As thickness approaches
zero, confinement vanishes and the field extends arbitrarily far into
the cladding.

Problem 3.6 — Single-mode thickness and bounce angle
----------------------------------------------------

**Data.** Take the uncovered surface as air:
:math:`n_1=1,\ n_2=1.62,\ n_3=1.346,\ \lambda_0=0.6328\,\mu{\rm m}`.

**Working.** Applying the exact TE cutoff to orders zero and one gives

.. math::

   t_{c,0}=0.0877103\,\mu{\rm m},\qquad
   t_{c,1}=0.438685\,\mu{\rm m}.

For the supplied :math:`\beta=1.53\times10^5\,{\rm cm}^{-1}`,
convert wavelength to centimetres before forming
:math:`\cos\theta=\beta\lambda_0/(2\pi n_2)`.

**Answer.** (a) One TE order occurs for
:math:`0.0877103<t<0.438685\,\mu{\rm m}`.
(b) :math:`\theta=17.977^\circ` from the guide plane, or
72.023 degrees from its normal.

**Check.** If “one mode” means only one vector polarization as well,
TM0 begins at :math:`0.134793\,\mu{\rm m}`; the TE-only interval
ends there. A sketch must label which angle is being drawn.

Problem 3.7 — Exact asymmetric TE count
---------------------------------------

**Working.** Insert :math:`n_2=1.65,\ n_3=1.52,\ n_1=1` and
:math:`\lambda_0=0.63\,\mu{\rm m}` into
:eq:`integrated-slab-cutoff`. The cutoff sequence is an arithmetic
progression starting at :math:`t_{c,0}=0.165518\,\mu{\rm m}`
with spacing :math:`\lambda_0/(2\sqrt{n_2^2-n_3^2})`.
At :math:`t=1.18\,\mu{\rm m}` the first three, and only those,
are exceeded.

**Answer.** Three TE modes: TE0, TE1, and TE2.

**Check.** The dispersion roots are
:math:`N=1.63494,\ 1.59010,\ 1.52251`.
The highest order is only weakly above the substrate index 1.52.

Problem 3.8 — Why symmetry helps retain the fundamental
-------------------------------------------------------

**Working.** At symmetric cutoff both exterior decay constants vanish,
so the TE0 phase condition is :math:`ht=0`. The mode therefore has
no positive minimum thickness. In an asymmetric guide, the higher-index
substrate reaches cutoff while the cover still supplies a reflection
phase. A positive thickness is needed to balance that phase.

**Answer.** Symmetry removes the positive TE0 cutoff and avoids loss
of the fundamental through cover/substrate index imbalance.

**Check.** This does not mean every symmetric guide is single-mode.
Thickness must still stay below TE1 cutoff. At fixed contrast, compare
the actual two cutoff boundaries rather than assuming all fabrication
tolerances necessarily improve.

Problem 3.9 — Optical stripline and ray confinement
---------------------------------------------------

**Working.** A loading strip on an otherwise planar dielectric guide
changes the local vertical boundary conditions. The region beneath it
has a larger local effective index than the uncovered film.
The vertical slab first confines light normal to the wafer; the
effective-index step then provides lateral confinement.

.. figure:: /_static/knowledge_base/worked_exercises/integrated_optics_6e/stripline.svg
   :alt: Cross section and top view of a loaded slab; the loaded region has a higher effective index and a confined zigzag ray.
   :width: 100%

   The loading strip changes modal effective index; it need not exceed
   the core's material index.

**Answer.** Draw vertical reflections at the slab boundaries and a
lateral zigzag within the loaded strip region.

**Check.** Lateral ray confinement depends on local effective indices,
not merely on the bulk indices of strip and substrate.

Problem 3.10 — Phase velocity beneath a loading strip
-----------------------------------------------------

**Working.** At fixed optical frequency,
:math:`\beta=k_0N` and :math:`v_{\rm ph,z}=\omega/\beta=c/N`.
Loading raises the local effective index, so it increases :math:`\beta`
and lowers axial phase velocity.

**Answer.** The phase velocity is smaller beneath the strip.

**Check.** The core's bulk index need not change. The speed comparison
is for the guided mode's axial phase, not a claim about a group velocity
without dispersion data.