Question 6 of 8: Power reflected at normal incidence on a dielectric
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2015 — 07-Elec-A7 Electromagnetics. Three hours, closed book; one of two approved calculators permitted. Eight questions, all of equal value (20 marks each). The rubric states that any five questions constitute a complete paper and that only the first five presented are marked — all eight are worked here, because this set is a study resource rather than a sat examination. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, from which $\eta_0=\sqrt{\mu_0/\varepsilon_0} = 376.8\ \Omega$ and $c = 1/\sqrt{\mu_0\varepsilon_0} = 3.00\times10^{8}\ \text{m/s}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, matching stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (plane waves, normal incidence); F. T. Ulaby, Fundamentals of Applied Electromagnetics, 8th ed. (polarisation, interference, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short current elements).
Question 6: Power reflected at normal incidence on a dielectric (20 marks)
Given. A monochromatic uniform plane wave in free space strikes the flat face of a lossless non-magnetic dielectric at normal incidence.
Given data
Quantity
Symbol
Value
Medium 1
$\varepsilon_{r1}$
$1$ (free space)
Medium 2
$\varepsilon_{r2}$
$1.69$
Magnetic character
$\mu_r$
$1$ in both media
Incidence
—
normal to a plane boundary
Find. The fraction of the incident power that is reflected.
Figure Q6. Normal incidence on a lossless dielectric half-space. The impedance step, not the permittivity step, sets the reflection.
Approach. Convert each medium's permittivity into an intrinsic impedance, form the field reflection coefficient from the impedance step, and square it to reach the power fraction.
Find the refractive index and the second impedance. For a non-magnetic medium $$n = \sqrt{\varepsilon_r} = \sqrt{1.69} = 1.30, \qquad \eta_2 = \frac{\eta_0}{\sqrt{\varepsilon_r}} = \frac{376.8}{1.30} = 289.9\ \Omega$$ against $\eta_1 = \eta_0 = 376.8\ \Omega$ in free space. The wave sees a step down in wave impedance, which is why the reflected electric field will come out with a reversed sign.
Apply the boundary conditions. Continuity of the tangential $E$ and $H$ at the interface gives the field reflection coefficient $$\Gamma = \frac{\eta_2-\eta_1}{\eta_2+\eta_1} = \frac{289.9-376.8}{289.9+376.8} = -0.1304$$ The negative sign is a 180° phase reversal on reflection from the optically denser medium; it does not affect the power, which depends only on the magnitude.
Square to obtain the power fraction. Since the incident and reflected waves share the same medium and hence the same impedance, the reflected power density is simply $|\Gamma|^{2}$ times the incident one: $$R = |\Gamma|^{2} = (0.1304)^{2} = \boxed{0.01701 \;\;(1.70\ \%)}$$
Check by the index form and by conservation. Written directly in terms of the index, $$R = \left(\frac{n-1}{n+1}\right)^{2} = \left(\frac{0.30}{2.30}\right)^{2} = 0.01701$$ identically, and the transmitted fraction is $T = 1-R = 0.98299$, so 98.3 % of the incident power crosses into the dielectric. Because the medium is lossless and the incidence normal, $R+T = 1$ exactly, which is the check to perform before quoting any answer of this type.
The number is small because the impedance step is small: a permittivity of 1.69 is barely denser than air optically, and reflection at a single dielectric interface grows only as the square of the relative impedance mismatch. This is precisely why a quarter-wave anti-reflection coating with $n = \sqrt{n_1 n_2}$ can drive even this modest reflection to zero.
