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22-Elec-B9 Electromagnetic Field, Transmission Lines, Antennas, and Radiation · May 2013

Question 8 of 8: Phase Shift on Total Internal Reflection

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

Notes on this paper

Paper format. National Exams, May 2013 — 07-Elec-B9, Electromagnetic Field, Transmission Lines, Antennas, and Radiation. Three hours, closed book; an approved Casio or Sharp calculator is permitted. Eight questions are printed and five constitute a complete paper, each of equal value (20 marks); the first five appearing in the answer book are the ones marked. The cover page supplies only two aids, $\varepsilon_0 = 8.85\times 10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times 10^{-7}\ \text{H/m}$, plus a Smith chart on page 4. Candidates are urged to state any interpretation assumptions with their answers. All eight questions are solved below, because this set is a study resource rather than a three-hour sitting.

Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission lines, the Smith chart, waveguides and cavities); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. and W. H. Hayt & J. A. Buck, Engineering Electromagnetics, 9th ed. (plane waves, oblique incidence, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short elements, arrays, apertures); F. T. Ulaby & U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed. (radar equation, ground-plane imaging). SI units throughout; the paper's own $c = 3\times 10^{8}\ \text{m/s}$ and $Z_0 = 377\ \Omega$ are used so that the printed aids reproduce exactly.

Check: reading of the printed data. Some symbols are printed loosely. Question 3 prints “10GH” and “1KWs total energy”; these are read as a 10 GHz carrier and a total pulse energy of $1\ \text{kW}\cdot\text{s} = 1000\ \text{J}$. Question 6 prints “15 MH” (15 MHz) and the aid $R_{\text{rad}} = (2\pi/3)Z_0 (l/\lambda)^2$. Question 7 prints “10GH” (10 GHz) and the aid $\theta_0 = 1.2\lambda/d$. Question 8 prints “an 600 angle” ($60^\circ$).

Question 8: Phase Shift on Total Internal Reflection (20 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.

Given. A plane wave inside a dielectric striking the boundary with vacuum from the dense side.

Given data
QuantitySymbolValue
Vacuum wavelength$\lambda_0$0.6 µm
Relative permittivity of the medium$\varepsilon_r$ 2.25
Refractive index of the medium$n_1$1.5
Second medium$n_2$1.0 (vacuum)
Angle of incidence$\theta_1$$60^\circ$
Polarisation— E parallel to the interface (perpendicular / TE / s)

Find. The phase of the reflected wave relative to the incident wave at the interface.

incidentreflectedtheta = 60 degmedium, n = 1.5vacuum, n = 1evanescent, decay 115 nmbeyond the critical angle nothing propagates away: the reflection is total
Figure 8.1 — Incidence from the dense side beyond the critical angle. No power crosses the boundary; the field in the vacuum is evanescent and decays within a fraction of a wavelength.

Approach. Test the incidence angle against the critical angle, recognise that “E parallel to the interface” means perpendicular (TE) polarisation, evaluate the Fresnel coefficient with an imaginary transmitted cosine, and read off its argument.

  1. Test for total internal reflection. $$\theta_c = \arcsin\frac{n_2}{n_1} = \arcsin\frac{1}{1.5} = 41.81^\circ \lt 60^\circ ,$$ so the wave is totally internally reflected: $|r| = 1$ and the only question is the phase.
  2. Identify the polarisation. An electric field lying in the plane of the interface is perpendicular to the plane of incidence — the TE, s or “perpendicular” case — so the relevant Fresnel coefficient is $$r_{\perp} = \frac{n_1\cos\theta_1 - n_2\cos\theta_2} {n_1\cos\theta_1 + n_2\cos\theta_2}.$$
  3. Evaluate the transmitted cosine. Snell's law gives $\sin\theta_2 = (n_1/n_2)\sin\theta_1 = 1.5\sin 60^\circ = 1.299 \gt 1$, so $\cos\theta_2$ is imaginary: $$n_2\cos\theta_2 = j\sqrt{(n_1\sin\theta_1)^{2} - n_2^{2}} = j\sqrt{1.6875 - 1} = j\,0.8292 .$$ With $n_1\cos\theta_1 = 1.5(0.5) = 0.750$, the coefficient becomes a ratio of a complex number to its own conjugate.
  4. Extract magnitude and phase. $$r_{\perp} = \frac{0.750 - j0.8292}{0.750 + j0.8292} \quad\Longrightarrow\quad |r_{\perp}| = 1, \qquad \arg r_{\perp} = -2\arctan\!\left(\frac{0.8292}{0.750}\right).$$ Equivalently, in the standard closed form $$\tan\frac{\delta_{\perp}}{2} = \frac{\sqrt{\sin^{2}\theta_1 - (n_2/n_1)^{2}}}{\cos\theta_1} = \frac{0.5528}{0.500} = 1.1055 .$$
  5. Report the relative phase. $$\boxed{\ \delta_{\perp} = 2\arctan(1.1055) = 95.74^\circ = 1.671\ \text{rad}\ }$$ so the reflected wave lags the incident wave by $95.74^\circ$ at the interface, with its amplitude unchanged.
  6. Describe the field beyond the boundary. Although nothing propagates away, the field does not stop at the interface: it continues as an evanescent wave decaying as $e^{-\alpha z}$ with $$\alpha = \frac{2\pi}{\lambda_0}\sqrt{(n_1\sin\theta_1)^{2} - 1} \quad\Longrightarrow\quad \frac{1}{\alpha} = 115\ \text{nm},$$ about one fifth of the vacuum wavelength. Time-averaged power flow across the boundary is zero, which is the energy-conservation statement behind $|r| = 1$.

The phase shift is the whole physical content of total internal reflection beyond the trivial statement that all the power comes back. Because it depends on both the angle and the polarisation, a wave that is not purely TE or TM emerges elliptically polarised — the principle of the Fresnel rhomb, which converts linear to circular polarisation using two total reflections instead of a birefringent plate. The same phase shift, accumulated over many reflections, is what fixes the propagation constants of the guided modes in a step-index optical fibre.

Final results — Question 8
QuantityValue
Critical angle$\theta_c = 41.81^\circ$
Regime at $60^\circ$total internal reflection
Reflection coefficient magnitude$|r_{\perp}| = 1$
Relative phase of reflected wave lags by $95.74^\circ$ (1.671 rad)
Reflectance$|r|^{2} = 1$ (100 % reflected)
Evanescent decay length in the vacuum115 nm

Check: sign convention. The $e^{j\omega t}$ convention is used with the evanescent field written as $e^{-\alpha z}$, which fixes $n_2\cos\theta_2 = +j\,0.8292$ and makes the reflected wave lag. Texts using $e^{-i\omega t}$ obtain $+95.74^\circ$; the magnitude of the relative phase, $95.74^\circ$, is the convention-independent answer.

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