17-Phys-A7 Optics · December 2016

Question 4 of 6: TE/TM plane waves, interference fringes, and refraction of the fringe pattern

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: TE/TM plane waves, interference fringes, and refraction of the fringe pattern (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — TE and TM plane-wave fields.

Given. Plane of incidence $=y$-$z$ plane; propagation angle $\beta$ from the $z$ axis (so $\hat{\mathbf k}=\sin\beta\,\hat{\mathbf y}+\cos\beta\,\hat{\mathbf z}$); isotropic, homogeneous, non-magnetic medium of index $n$ ($k=n\omega/c$, wave impedance $\eta=\eta_0/n$ with $\eta_0\approx377\ \Omega$).

Find. $\mathbf E,\mathbf H$ for the TE (E out of the plane of incidence) and TM (H out of the plane of incidence) polarizations.

1. TE (E perpendicular to the plane of incidence, i.e. along $\hat{\mathbf x}$). $\mathbf E_{TE}=\hat{\mathbf x}\,E_0\cos(ky\sin\beta+kz\cos\beta-\omega t)$. Using $\mathbf H=(\hat{\mathbf k}\times\mathbf E)/\eta$: $\boxed{\mathbf H_{TE}=\dfrac{E_0}{\eta}\big(\cos\beta\,\hat{\mathbf y}-\sin\beta\,\hat{\mathbf z}\big)\cos(ky\sin\beta+kz\cos\beta-\omega t)}$.
2. TM (H perpendicular to the plane of incidence, i.e. along $\hat{\mathbf x}$). $\mathbf H_{TM}=\hat{\mathbf x}\,H_0\cos(ky\sin\beta+kz\cos\beta-\omega t)$. Using $\mathbf E=\eta(\mathbf H\times\hat{\mathbf k})$: $\boxed{\mathbf E_{TM}=\eta H_0\big(-\cos\beta\,\hat{\mathbf y}+\sin\beta\,\hat{\mathbf z}\big)\cos(ky\sin\beta+kz\cos\beta-\omega t)}$.

Part (b) — check by limiting case and by $\mathbf E\times\mathbf H$.

Approach. Two independent checks: (1) set $\beta=0$ (normal incidence along $z$) and confirm the fields reduce to the familiar transverse plane wave; (2) compute $\mathbf E\times\mathbf H$ in general and confirm it points along $\hat{\mathbf k}$ with the expected magnitude $E_0H_0$.

1. Limiting case $\beta=0$. TE reduces to $\mathbf E=\hat{\mathbf x}E_0\cos(kz-\omega t)$, $\mathbf H=\hat{\mathbf x}\times... =\hat{\mathbf y}(E_0/\eta)\cos(kz-\omega t)$ — exactly the standard normal-incidence plane wave ($\mathbf E\parallel\hat{\mathbf x}$, $\mathbf H\parallel\hat{\mathbf y}$, both $\perp\hat{\mathbf z}=\hat{\mathbf k}$), as required.
2. $\mathbf E\times\mathbf H$ for TE, general $\beta$. $\mathbf E_{TE}\times\mathbf H_{TE}=E_0\hat{\mathbf x}\times\dfrac{E_0}{\eta}(\cos\beta\,\hat{\mathbf y}-\sin\beta\,\hat{\mathbf z})\cos^2(\cdot)=\dfrac{E_0^2}{\eta}\big(\cos\beta\,\hat{\mathbf z}+\sin\beta\,\hat{\mathbf y}\big)\cos^2(\cdot)=\boxed{\dfrac{E_0^2}{\eta}\,\hat{\mathbf k}\cos^2(\cdot)}$ (using $\hat{\mathbf x}\times\hat{\mathbf y}=\hat{\mathbf z}$, $\hat{\mathbf x}\times\hat{\mathbf z}=-\hat{\mathbf y}$) — points exactly along the propagation direction, with the expected magnitude $E_0H_0=E_0^2/\eta$; the identical check on the TM fields returns $(E_0^2/\eta)\hat{\mathbf k}\cos^2(\cdot)$ as well, confirming both polarizations were assigned consistent, self-checking directions.

Part (c) — interference of waves at $+\beta$ and $-\beta$: time-averaged Poynting vector at $z=0$.

Given. Two coherent, equal-amplitude TE waves in the $y$-$z$ plane, propagating at $+\beta$ and $-\beta$ from $z$; observation plane $z=0$.

Find. A simplified $\langle\mathbf S\rangle$ from which the fringe period follows by inspection.

Approach. Superpose the two TE fields (phases $\phi_{1,2}=\pm ky\sin\beta+kz\cos\beta-\omega t$) using the sum-to-product identity, then compute $\mathbf S=\mathbf E\times\mathbf H$ from the summed fields and time-average over one period.

1. Sum the E fields. $\mathbf E=E_0\hat{\mathbf x}\big[\cos\phi_1+\cos\phi_2\big]=2E_0\hat{\mathbf x}\cos(ky\sin\beta)\cos(kz\cos\beta-\omega t)$ — a standing-wave pattern across $y$ (the fringes) riding on a wave travelling in $z$.
2. Sum the H fields and form $\mathbf S$. Carrying through the same superposition for $\mathbf H$ (each wave's $\mathbf H$ from part (a), with $\beta\to-\beta$ for wave 2) and computing $\mathbf S=\mathbf E\times\mathbf H$ at $z=0$, the cross terms that oscillate at $2\omega t$ average to zero over a cycle, leaving $\boxed{\langle\mathbf S\rangle=\dfrac{2E_0^2\cos\beta}{\eta}\cos^2(ky\sin\beta)\,\hat{\mathbf z}}$ — a pure standing pattern in $y$ modulating a steady power flow along $z$, with zero net flow along $y$ as symmetry requires.
3. Read off the fringe period. $\cos^2(x)$ has period $\pi$ in its argument, so $\cos^2(ky\sin\beta)$ repeats when $k\,\Delta y\,\sin\beta=\pi$: $\boxed{\Lambda=\dfrac{\pi}{k\sin\beta}=\dfrac{\lambda}{2\sin\beta}}$ (illustratively, at $\beta=30^\circ$, $\lambda=500$ nm this is $0.500\ \mu$m — a fine fringe spacing typical of two beams crossing at a wide angle).

Part (d) — the same fringes viewed across a planar interface into a higher-index material.

Given. A planar interface normal to $z$, starting at $z=10$, separates the original medium (index $n$) from a second medium of index $n_2>n$; the two-beam fringe pattern of part (c) is now observed inside the second medium.

Find. The fringe period there.

Approach. A boundary with its normal along $z$ can only bend each beam's $z$-component of $\mathbf k$; the tangential (here, $y$) component $k_y=k\sin\beta$ must stay continuous across the interface (this IS Snell's law, restated as phase-matching), so each beam's own $k_y$ — the exact quantity that sets the fringe period in part (c) — is unchanged by refraction.

1. Apply continuity of $k_y$. In medium 2, $k_2=n_2\omega/c$ and the refracted angle $\beta_2$ satisfies $n\sin\beta=n_2\sin\beta_2$ (Snell's law), so $k_2\sin\beta_2=\dfrac{n_2\omega}{c}\sin\beta_2=\dfrac{\omega}{c}(n_2\sin\beta_2)=\dfrac{\omega}{c}(n\sin\beta)=k\sin\beta$ — identical to medium 1.
2. Fringe period in medium 2. $\Lambda_2=\dfrac{\pi}{k_2\sin\beta_2}=\dfrac{\pi}{k\sin\beta}=\boxed{\Lambda_2=\Lambda_1=\dfrac{\lambda}{2\sin\beta}}$ (with $\lambda$ the wavelength in the ORIGINAL medium, index $n$) — the fringe spacing on any $z=\text{const}$ plane is exactly the same on both sides of the interface, independent of $n_2$; refraction changes each beam's angle and its wavelength along its own propagation direction, but leaves the transverse ($y$) periodicity of the two-beam interference pattern untouched, because that periodicity is set entirely by $k_y$, and $k_y$ is precisely the quantity Snell's law conserves.