tailieunhanh - Electromagnetic Field Theory: A Problem Solving Approach Part 58

Electromagnetic Field Theory: A Problem Solving Approach Part 58. Electromagnetic field theory is often the least popular course in the electrical engineering curriculum. Heavy reliance on vector and integral calculus can obscure physical phenomena so that the student becomes bogged down in the mathematics and loses sight of the applications. This book instills problem solving confidence by teaching through the use of a large number of worked problems. To keep the subject exciting, many of these problems are based on physical processes, devices, and models. This text is an introductory treatment on the junior level for a two-semester electrical engineering. | Applications to Optics 545 Most optical materials like glass have a permeability of free space z0. Therefore a Brewster s angle of no reflection only exists if the H field is parallel to the boundary. At the critical angle which can only exist if light travels from a high index of refraction material low light velocity to one of low index high light velocity there is a transmitted field that decays with distance as a nonuniform plane wave. However there is no time-average power carried by this evanescent wave so that all the time-average power is reflected. This section briefly describes various applications of these special angles and the rules governing reflection and refraction. 7-10-1 Reflections from a Mirror A person has their eyes at height h above their feet and a height A i below the top of their head as in Figure 7-20. A mirror in front extends a distance Ay above the eyes and a distance y below. How large must y and Ay be so that the person sees their entire image The light reflected off the person into the mirror must be reflected again into the person s eyes. Since the angle of incidence equals the angle of reflection Figure 7-20 shows that Ay A i 2 and y i 2. 7-10-2 Lateral Displacement of a Light Ray A light ray is incident from free space upon a transparent medium with index of refraction n at angle as shown in Figure 7-21. The angle of the transmitted light is given by Snell s law sin 6t 1 n sin ffi 1 Figure 7-20 Because the angle of incidence equals the angle of reflection a person can see their entire image if the mirror extends half the distance of extent above and below the eyes. 546 Electrodynamics Fields and Waves Figure 7-21 A light ray incident upon a glass plate exits the plate into the original medium parallel to its original trajectory but laterally displaced. When this light hits the second interface the angle 0 is now the incident angle so that the transmitted angle 02 is again given by Snell s law sin 02 n sin 0 sin 0j 2 so that the .