17-Phys-A7 Optics · May 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
98-Phys-A7, Optics — National Exams, May 2016. 3 hours; closed book (formula sheet supplied). Notes 1–7 on the cover page require Q1 and Q2 (mandatory) plus any two of Q3–Q6 for a complete paper; every question is solved below as a full study resource, including all four of Q3–Q6.
Reference texts. Hecht, Optics, 5th ed.; Pedrotti, Pedrotti & Pedrotti, Introduction to Optics, 3rd ed.; Griffiths, Introduction to Electrodynamics, 4th ed. (Ch. 9, EM waves in matter).
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 fields. Take $\mathbf k=k(\cos\theta,\ \sin\theta,\ 0)$, $k=n\omega/c=nk_0$, propagating at angle $\theta$ from the $x$ axis in the $xy$ plane (coordinate system above). Non-magnetic medium, $\mu=\mu_0$, wave impedance $\eta=\eta_0/n$.
TE (s-polarized, $\mathbf E\perp$ plane of incidence, along $\hat z$):
$$\mathbf E_{\text{TE}}=E_0\,\hat z\,\cos(\mathbf k\cdot\mathbf r-\omega t),\qquad \mathbf H_{\text{TE}}=\frac{nE_0}{\eta_0}\big({-}\sin\theta\,\hat x+\cos\theta\,\hat y\big)\cos(\mathbf k\cdot\mathbf r-\omega t),$$from $\mathbf H=(\mathbf k\times\mathbf E)/(\omega\mu_0)$: with $\mathbf k=k(\cos\theta,\sin\theta,0)$, $\mathbf E=E_0\hat z$, $\mathbf k\times\mathbf E=kE_0({-}\sin\theta,\cos\theta,0)$ and $k/(\omega\mu_0)=n/\eta_0$.
TM (p-polarized, $\mathbf H\perp$ plane of incidence, along $\hat z$):
$$\mathbf H_{\text{TM}}=H_0\,\hat z\,\cos(\mathbf k\cdot\mathbf r-\omega t),\qquad \mathbf E_{\text{TM}}=\frac{\eta_0H_0}{n}\big(\sin\theta\,\hat x-\cos\theta\,\hat y\big)\cos(\mathbf k\cdot\mathbf r-\omega t),$$from $\mathbf E=-(\mathbf k\times\mathbf H)/(\omega\varepsilon)$ with $\varepsilon=n^2\varepsilon_0$ and $k/(\omega\varepsilon)=\eta_0/n$.
Part (b) — checks. Orthogonality: in both cases $\mathbf E\cdot\mathbf k=0$, $\mathbf H\cdot\mathbf k=0$, $\mathbf E\cdot\mathbf H=0$ term by term, as required of a transverse wave. Limiting case $\theta=0$ (propagation along $+x$): TE reduces to $\mathbf E=E_0\hat z$, $\mathbf H=(nE_0/\eta_0)\hat y$; TM reduces to $\mathbf H=H_0\hat z$, $\mathbf E=-(\eta_0H_0/n)\hat y$ — in both cases $\mathbf E,\mathbf H,\hat x$ form a right-handed triad, as required. Poynting vector: $\mathbf S=\mathbf E\times\mathbf H$; for TE at $\theta=0$, $\mathbf S=E_0\hat z\times(nE_0/\eta_0)\hat y=-(nE_0^2/\eta_0)(\hat z\times\hat y)\cdot(-1)=(nE_0^2/\eta_0)\hat x$ (using $\hat z\times\hat y=-\hat x$), positive $\hat x$ — along $\mathbf k$, exactly as energy transport must be. The same expansion at general $\theta$ gives $\mathbf S\parallel\mathbf k$ for both polarizations (direct use of $\hat z\times({-}\sin\theta\,\hat x+\cos\theta\,\hat y)=\cos\theta\,\hat y+\sin\theta\,\hat x$, the unit vector along $\mathbf k$), confirming the field amplitudes and the direction of $\mathbf k$ used in part (a) are self-consistent.
Part (c) — two-beam interference, fringe period in the plane $x=0$. Take two equal-amplitude TE waves at $\pm\theta$ from $x$: $\mathbf k_{1,2}=k(\cos\theta,\pm\sin\theta,0)$. Their real superposition is
$$E_z=E_0\cos(kx\cos\theta+ky\sin\theta-\omega t)+E_0\cos(kx\cos\theta-ky\sin\theta-\omega t).$$Using $\cos A+\cos B=2\cos(\tfrac{A+B}2)\cos(\tfrac{A-B}2)$ with $\tfrac{A+B}2=kx\cos\theta-\omega t$, $\tfrac{A-B}2=ky\sin\theta$:
$$E_z(x,y,t)=2E_0\cos(ky\sin\theta)\,\cos(kx\cos\theta-\omega t).$$The time-averaged Poynting vector is dominated by $\langle S_x\rangle\propto\langle E_z^2\rangle_t$; since $\langle\cos^2(kx\cos\theta-\omega t)\rangle_t=\tfrac12$ for every $x$ (including $x=0$), the pattern in the plane $x=0$ is
$$\boxed{\langle S_x(y)\rangle\propto\cos^2(ky\sin\theta)}$$independent of $x$ — a genuinely "simplified" answer: by inspection $\cos^2(u)$ repeats when $u$ advances by $\pi$, so $k\sin\theta\cdot\Lambda=\pi$, giving the fringe period
$$\Lambda=\frac{\pi}{k\sin\theta}=\boxed{\frac{\lambda}{2\sin\theta}}$$— the standard two-plane-wave fringe spacing, with $2\theta$ the full angle between the beams. (Illustrative numeric check: $\lambda_0=632.8$ nm, $\theta=20^\circ$, $n=1.00$ gives $\Lambda\approx0.925\ \mu$m — a typical holographic-interferometer fringe spacing.)
Part (d) — fringes observed across the interface at $x=10$ into a higher-index medium. At the interface (normal along $x$), each beam refracts by its own Snell's law, $n\sin\theta=n'\sin\theta'$, so $\theta'=\arcsin\!\big(\tfrac{n}{n'}\sin\theta\big)$ (both beams refract symmetrically to $\pm\theta'$). In the new medium $\lambda'=\lambda_0/n'$, so by the same derivation as part (c) the fringe period there is $\Lambda'=\lambda'/(2\sin\theta')$:
$$\Lambda'=\frac{\lambda_0/n'}{2\sin\theta'}=\frac{\lambda_0/n'}{2\cdot(n/n')\sin\theta}=\frac{\lambda_0}{2n\sin\theta}=\frac{\lambda}{2\sin\theta}=\Lambda.$$The index $n'$ cancels exactly, so $\boxed{\Lambda'=\Lambda}$: the fringe period is unchanged by crossing into the denser material. This is not a coincidence: the fringe period is set purely by the wavevector component $k_y=k_0n\sin\theta$ tangential to the $x=10$ interface, and that tangential component is continuous across any planar interface at any angle — that continuity condition is exactly Snell's law. Since $\Lambda=\pi/k_y$ depends only on the conserved $k_y$, the fringe spacing on either side of the boundary must be identical, regardless of $n'$.
