17-Phys-A7 Optics · May 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Examinations, May 2015, 98-Phys-A7 Optics. Three hours, closed book, one approved Casio or Sharp calculator. Question 1 is mandatory; the rubric asks for Question 1 plus any four of Questions 2–6 (only the first four questions appearing in the answer book are marked) — all six are worked here, because the set is a study resource.
Reference texts. E. Hecht, Optics, 5th ed.; F. L. Pedrotti, L. M. Pedrotti and L. S. Pedrotti, Introduction to Optics, 3rd ed.
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 $\mathbf k = k(\sin\theta,\ \cos\theta,\ 0)$, $k=n\omega/c=nk_0$, propagating at angle $\theta$ from the $y$ axis in the $xy$ plane (the coordinate system above). Non-magnetic medium, $\mu=\mu_0$, wave impedance $\eta=\eta_0/n=\sqrt{\mu_0/\varepsilon_0}/n$.
TE (s-polarized, $\mathbf E\perp$ plane of incidence, along $\hat z$):
$$\mathbf E_{\text{TE}} = E_0\,\hat z\,\cos(\mathbf k\cdot\mathbf r-\omega t), \qquad \mathbf H_{\text{TE}} = \frac{n E_0}{\eta_0}\big(\cos\theta\,\hat x-\sin\theta\,\hat y\big)\cos(\mathbf k\cdot\mathbf r-\omega t),$$ obtained from $\mathbf H=\dfrac{\mathbf k\times \mathbf E}{\omega\mu_0}$: with $\mathbf k=k(\sin\theta,\cos\theta,0)$ and $\mathbf E=E_0\hat z$, $\mathbf k\times\mathbf E = kE_0(\cos\theta,-\sin\theta,0)$, and $k/(\omega\mu_0)=n/(c\mu_0)=n/\eta_0$.TM (p-polarized, $\mathbf H\perp$ plane of incidence, along $\hat z$):
$$\mathbf H_{\text{TM}} = H_0\,\hat z\,\cos(\mathbf k\cdot\mathbf r-\omega t), \qquad \mathbf E_{\text{TM}} = \frac{\eta_0 H_0}{n}\big(-\cos\theta\,\hat x+\sin\theta\,\hat y\big)\cos(\mathbf k\cdot\mathbf r-\omega t),$$ from $\mathbf E=-\dfrac{\mathbf k\times \mathbf H}{\omega\varepsilon}$ with $\varepsilon=n^2\varepsilon_0$ and $k/(\omega\varepsilon)=n/(c\varepsilon)=\eta_0/n$.Part (b) — checks. Orthogonality: in both cases $\mathbf E\cdot\mathbf k=0$, $\mathbf H\cdot\mathbf k=0$ and $\mathbf E\cdot\mathbf H=0$ term by term, as required of a transverse plane wave. Limiting case $\theta=0$ (propagation along $+y$, normal incidence on a surface with normal $\hat y$): TE reduces to $\mathbf E=E_0\hat z$, $\mathbf H=(nE_0/\eta_0)\hat x$; TM reduces to $\mathbf H=H_0\hat z$, $\mathbf E=-(\eta_0H_0/n)\hat x$ — in both cases $\mathbf E$, $\mathbf H$, $\hat y$ form a right-handed triad, as they must. Poynting vector: $\mathbf S=\mathbf E\times\mathbf H$; for TE at $\theta=0$, $\mathbf S=E_0\hat z\times(nE_0/\eta_0)\hat x=(nE_0^2/\eta_0)\hat y$ — positive $\hat y$, i.e. along $\mathbf k$, exactly as energy transport must be. The same check at general $\theta$ gives $\mathbf S\parallel\mathbf k$ in both polarizations (direct expansion of the cross product using $\hat z\times(\cos\theta\,\hat x-\sin\theta\,\hat y)=\cos\theta\,\hat y+\sin\theta\,\hat x=$ 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. Take two equal-amplitude TE waves, one at $+\theta$ and one at $-\theta$ from $y$: $\mathbf k_{1,2}=k(\pm\sin\theta,\cos\theta,0)$. Their (real) superposition is
$$E_z = E_0\cos(kx\sin\theta+ky\cos\theta-\omega t)+E_0\cos(-kx\sin\theta+ky\cos\theta-\omega t).$$Using $\cos A+\cos B=2\cos\!\big(\tfrac{A+B}{2}\big)\cos\!\big(\tfrac{A-B}{2}\big)$ with $\tfrac{A+B}{2}=ky\cos\theta-\omega t$ and $\tfrac{A-B}{2}=kx\sin\theta$:
$$E_z(x,y,t) = 2E_0\cos(kx\sin\theta)\,\cos(ky\cos\theta-\omega t).$$The time-averaged Poynting vector is dominated by $\langle S_y\rangle\propto\langle E_z^2\rangle_t$; since $\langle\cos^2(ky\cos\theta-\omega t)\rangle_t=\tfrac12$ for every $y$ (including $y=0$), the fringe pattern in the plane $y=0$ is
$$\boxed{\langle S_y(x)\rangle \propto \cos^2(kx\sin\theta)}$$— independent of $y$, exactly as required, and "simplified" in the sense the question asks: by inspection, $\cos^2(u)$ repeats when $u$ advances by $\pi$, so $k\sin\theta\cdot\Lambda=\pi$, giving the fringe period
$$\Lambda = \frac{\pi}{k\sin\theta}=\boxed{\dfrac{\lambda}{2\sin\theta}},$$the standard two-plane-wave interference-fringe spacing (the same relation that sets fringe spacing in holography and in a Young's double-slit far field, $\S$6(b) below, with $2\theta$ the full angle between the beams).
Part (d) — fringes observed across an interface into a higher-index material. At the interface $y=10$ (normal along $y$), each beam refracts according to Snell's law applied to its own angle: $n\sin\theta = n'\sin\theta'$, so $\theta'=\arcsin\!\big(\tfrac{n}{n'}\sin\theta\big)$ (both beams refract symmetrically to $\pm\theta'$, by symmetry of the incident pair). In the new medium the wavelength is $\lambda'=\lambda_0/n'$, so by the same derivation as part (c), the fringe period there is $\Lambda'=\lambda'/(2\sin\theta')$. Substituting:
$$\Lambda' = \frac{\lambda_0/n'}{2\sin\theta'} = \frac{\lambda_0/n'}{2\cdot(n/n')\sin\theta} = \frac{\lambda_0}{2n\sin\theta} = \frac{\lambda}{2\sin\theta} = \Lambda.$$The refractive index $n'$ of the second medium cancels exactly, so $\boxed{\Lambda'=\Lambda}$: the fringe period is unchanged by crossing into the higher-index material. This is not a coincidence — the fringe period is set purely by the wavevector component tangential to the $y=10$ interface, $k_x=k_0 n\sin\theta$, and that tangential component is continuous across any planar interface for any angle of incidence (this continuity condition is precisely Snell's law itself). Since $\Lambda=\pi/k_x$ depends only on $k_x$, and $k_x$ is conserved at the boundary, the fringe spacing observed on either side of the interface must be identical, regardless of $n'$. The plane $y=10$ itself plays no role beyond marking where the (unchanged) pattern crosses into the new medium.
