final_exam_222A_ 2000

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

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

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

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

View Full Document
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
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern