final_exam_222A_ 2000

final_exam_222A_ 2000 - ECE 222A Final Exam Dec 8 2000 Name...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 222A Final Exam Dec 8, 2000 Name Instructions. This a is three hour, closed book, examination. Calculators are permitted, but not needed. 1. (10 points) An exotic material has the properties μ = , σ c = 0, and ² = 0. What boundary conditions does such a surface impose on ~ E , ~ H , ~ B , and ~ D in an adjacent free space region at nonzero frequencies? Hint: Do not allow fields with infinite magnitudes. Also, do not allow currents or charges other than those owing to conduction effects. Solution : First, we know that since μ = , it follows that ~ H = 0 inside the material. We can’t say anything about ~ B because μ ~ H = ∞ · 0 is indeterminant. We also have ~ D = 0 inside the material, a consequence of ² = 0 . The electric field inside the material is unrestricted. Finally, there are no currents (or conduction electrons) since σ c = 0 . The four boundary conditions (valid for all materials) are ˆ n · h ~ D 2 - ~ D 1 i = 0 ˆ n · h ~ B 2 - ~ B 1 i = 0 ˆ n × h ~ E 2 - ~ E 1 i = 0 ˆ n × h ~ H 2 - ~ H 1 i = 0 where region “1” refers to the exotic material and region“2” is the vacuum. If we make use of the specific material properties of region “2,” we can refine the boundary conditions to ˆ n · ~ D 2 = ˆ n · ~ E 2 = 0 ˆ n · h ~ B 2 - ~ B 1 i = 0 ˆ n × h ~ E 2 - ~ E 1 i = 0 ˆ n × ~ H 2 = ˆ n × ~ B 2 = 0 2. (15 points) A beam of light is propagating inside glass with an index of refraction n i = p ² g 0 = 2, where ² g and μ g are the permittivity and permeability of the glass, respectively, and where we assume that μ g = μ 0 . The beam is incident upon a planar interface with a vacuum ( n 0 = 1) at an angle θ that exceeds the critical angle of total reflection θ c (see Figure 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2). The resulting evanescent wave in the vacuum decays as exp [ - z/ Δ]. What is the smallest possible value for the penetration depth Δ? Your answer should express Δ as a function of the wavenumber β g p ω 2 ² g μ 0 Solution : For the incident wave in the glass, we know that β 2 g = ω 2 ² g μ 0 = β 2 x + β 2 z where β x = β g sin θ i , and β z = β g cos θ i , where the direction normal to the surface is z. We assume that the incidence angle exceeds the critical
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern