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Applications to Optics 545 Most optical materials, like glass, have a permeability of fr~ee space o0 . 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 Ah 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 = Ah/2 and y = h/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 0,, as shown in Figure 7-21. The angle of the transmitted light is given by Snell's law: sin Ah 0, = (1/n) sin Oi (1) Ah Ay=- 2 ror 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-Fieldsand Waves dsin(6i-6,) -' &A.- lI COSU0r 62 6 Bi -<--d-->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 0 2 is again given by Snell's law: sin 02 = n Sin 0, = sin Oi (2) so that the light exits at the original incident angle Oi. However, it is now shifted by the amount: d sin (0i - 0,) cos 0, If the plate is glass with refractive index n = 1.5 and thickness d = 1 mm with incident angle Oi = 30*, the angle 0, in the glass is sin 0,= 0.33= 0,= 19.50 (4) so that the lateral displacement is s = 0.19 mm. 7-10-3 Polirization By Reflection Unpolarized light is incident upon the piece of glass in Section 7-10-2 with index of refraction n = 1.5. Unpolarized light has both E and H parallel to the interface. We assume that the permeability of the glass equals that of free space and that the light is incident at the Brewster's angle OB for light polarized with H parallel to the interface. The incident and I 547 Applications to Optics transmitted angles are then tan Os = EIE =n 0 = 56. 3 ° (5) tan 0, = IEo/e = 1/n = 0, = 33.70 The Brewster's angle is also called the polarizing angle because it can be used to separate the two orthogonal polarizations. The polarization, whose H field is parallel to the interface, is entirely transmitted at the first interface with no reflection. The other polarization with electric field parallel to the interface is partially transmitted and reflected. At the second (glass-free space) interface the light is incident at angle 0,. From (5) we see that this angle is the Brewster's angle with H parallel to the interface for light incident from the glass side onto the glass-free space interface. Then again, the H parallel to the interface polarization is entirely transmitted while the E parallel to the interface polarization is partially reflected and partially transmitted. Thus, the reflected wave is entirely polarized with electric field parallel to the interface. The transmitted waves, although composed of both polarizations, have the larger amplitude with H S H . E Polarized ligi (E parallel to inte zed ed Ilel lel ce) :e) Unpolarized light (E and H parallel to interface) Figure 7-22 Unpolarized light incident upon glass with A = A-o can be polarized by reflection if it is incident at the Brewster's angle for the polarization with H parallel to the interface. The transmitted light becomes more polarized with H parallel to the interface by adding more parallel glass plates. 548 Electrodynamics-Fieldsand Waves parallel to the interface because it was entirely transmitted with no reflection at both interfaces. By passing the transmitted light through another parallel piece of glass, the polarization with electric field parallel to the interface becomes further diminished because it is partially reflected, while the other polarization is completely transmitted. With more glass elements, as in Figure 7-22, the transmitted light can be made essentially completely polarized with H field parallel to the interface. 7-10-4 Light Propagation In Water (a) Submerged Source A light source is a distance d below the surface of water with refractive index n = 1.33, as in Figure 7-23. The rays emanate from the source as a cone. Those rays at an angle from the normal greater than the critical angle, sin O,= 1/n > 0, = 48.80 (6) are not transmitted into the air but undergo total internal reflection. A circle of light with diameter D = 2d tan Oc - 2.28d (7) then forms on the water's surface due to the exiting light. (b) Fish Below a Boat A fish swims below a circular boat of diameter D, as in Figure 7-24. As we try to view the fish from the air above, the incident light ray is bent towards the normal. The region below the boat that we view from above is demarcated by the light rays at grazing incidence to the surface (0i = 1r/2) just entering the water (n = 1.33) at the sides of the boat. The transmitted angle of these light rays is given from Snell's law as sin O 1 sin 0, = sin 1 = = 48.8" (8) n n J Figure 7-23 Light rays emanating from a source within a high index of refraction medium are totally internally reflected from the surface for angles greater than the critical angle. Lesser angles of incidence are transmitted. I Applications to Optics 549 Z vtzk - ýD- PWWW))~;)~W) HWIYrYlur D 2 Y=- tanO 1 Figure 7-24 A fish cannot be seen from above if it swims below a circular boat within the cone bounded by light rays at grazing incidence entering the water at the side of the boat. These rays from all sides of the boat intersect at the point a distance y below the boat, where tan D 0t =-- 2y y= D 2 tan 0, 0.44D If the fish swims within the cone, with vertex at the point y below the boat, it cannot be viewed from above. 7-10-5 Totally Reflecting Prisms The glass isoceles right triangle in Figure 7-25 has an index of refraction of n = 1.5 so that the critical angle for total - - = [n( n+' I 2z[ 22 Figure 7-25 A totally reflecting prism. The index of refraction n must exceed that the light incident on the hypotenuse at 450 exceeds the critical angle. 2 so 550 Electrodynamics-Fieldsand Waves internal reflection is 1 1 =n 1.5 sin oc 0c = 41.80 (10) The light is normally incident on the vertical face of the prism. The transmission coefficient is then given in Section 7-6-1 as E, 2n =-=-= Ei i7+tjo T 2/n 1+1/n 2 n+1 0.8 (11) where because the permeability of the prism equals that of free space n = ve/Eo while 1/1o0 = VE••e = 1/n. The transmitted light is then incident upon the hypotenuse of the prism at an angle of 450, which exceeds the critical angle so that no power is transmitted and the light is totally reflected being turned through a right angle. The light is then normally incident upon the horizontal face with transmission coefficient: T2 E2 2/0 = ------. 0.8Ei 7 + o 2 l/n + 1 2n = 1.2 n+l (12) The resulting electric field amplitude is then P2 = TIT 2 E, = 0.96Ei (13) The ratio of transmitted to incident power density is 12/7o 21|Pi21 = 217o, |/ 24 22 -0.92 1tl 2 (24 o(25 (14) 1 This ratio can be increased to unity by applying a quarterwavelength-thick dielectric coating with index of refraction -nh, as developed in Example 7-1. This is not usually ncoating= done because the ratio in (14) is already large without the expense of a coating. 7-10-6 Fiber Optics (a) Straight Light Pipe Long chin fibers of transparent material can guide light along a straight path if the light within the pipe is incident upon the wall at an angle greater than the critical angle (sin 0, = 1/n): sin 02 = cos 0, - sin 0~ (15) The light rays are then totally internally reflected being confined to the pipe until they exit, as in Figure 7-26. The I Applications to Optics 551 no =1 Figure 7-26 The index of refraction of a straight light pipe must be greater than /2 for total internal reflections of incident light at any angle. incident angle is related to the transmitted angle from Snell's law, sin 0, = (1/n) sin Oi (16) so that (15) becomes cos 0 = = %11-(1/n 2 ) sin" -sin - 1/n (17) which when solved for n yields n 2 - 1 +sin2 0i (18) If this condition is met for grazing incidence (i0 = ar/2), all incident light will be passed by the pipe, which requires that n2>-2*n - r2 (19) Most types of glass have n - 1.5 so that this condition is easily met. (b) Bent Fibers Light can also be guided along a tortuous path if the fiber is bent, as in the semi-circular pipe shown in Figure 7-27. The minimum angle to the radial normal for the incident light shown is at the point A. This angle in terms of the radius of the bend and the light pipe width must exceed the critical angle sin OA R R+d =- sin 0c (20) +d +d Figure 7-27 Light can be guided along a Circularly bent fiber if R/d > 1/(n - 1) as then there is always total internal reflection each time the light is incident on the walls. 552 Electrodynamics-Fieldsand Waves so that R/d Rd R/d + 11n (21) which when solved for Rid requires R 1 -a-d n-I (22) PROBLEMS Section 7-1 1. For the following electric fields in a linear media of permittivity e and permeability Cj find the charge density, magnetic field, and current density. (a) E = Eo(xi. +yi,) sin wt (b) E = Eo(yi, -xi,) cos wt (c) E= Re[Eo e" related so that \-•--&)i,]. How must k,, k,, and o be J = 0? 2. An Ohmic conductor of arbitrary shape has an initial charge distribution po(r) at t = 0. (a) What is the charge distribution for all time? (b) The initial charge distribution is uniform and is confined between parallel plate electrodes of spacing d. What are the electric and magnetic fields when the electrodes are opened or short circuited? (c) Repeat (b) for coaxial cylindrical electrodes of inner radius a and outer radius b. (d) When does a time varying electric field not generate a magnetic field? 3. (a) For linear media of permittivity e and permeability /, use the magnetic vector potential A to rewrite Faraday's law as the curl of a function. (b) Can a scalar potential function V be defined? What is the electric field in terms of V and A? The choice of V is not unique so pick V so that under static conditions E = -V V. (c) Use the results of (a) and (b) in Ampere's law with Maxwell's displacement current correction to obtain a single equation in A and V. (Hint: Vx (Vx A) = V(V - A) -V 2A.) (d) Since we are free to specify V *A, what value should we pick to make (c) an equation just in A? This is called setting the gauge. (e) Use the results of (a)-(d) in Gauss's law for D to obtain a single equation in V. Problem 553 (f) Consider a sinusoidally varying point charge at r = 0, Se"t'. . Solve (e) for r > 0. Hint: 1 8 (ra r 82a =- (rV) - Define a new variable (rV). By symmetry, V only depends on r and waves can only propagate away from the charge and not towards it. As r - 0, the potential approaches the quasi-static Coulomb potential. Section 7-2 4. Poynting's theorem must be modified if we have a hysteretic material with a nonlinear and double-valued relationship between the polarization P and electric field E and the magnetization M and magnetic field H. M I A k r , (a) For these nonlinear constitutive laws put Poynting's theorem in the form 8w V S+-= at -Pd - Pp- PM where Pp and PM are the power densities necessary to polarize and magnetize the material. (b) Sinusoidal electric and magnetic fields E = E, cos at and H = H, cos at are applied. How much energy density is dissipated per cycle? 5. An electromagnetic field is present within a superconductor with constituent relation aJf = wE 8t (a) Show that Poynting's theorem can be written in the form 8w V.s+-=0 8t What is w? 554 Electrodynamics-Fieldsand Waves (b) What is the velocity of the charge carriers each with charge q in terms of the current density Jr? The number density of charge carriers is n. (c) What kind of energy does the superconductor add? (d) Rewrite Maxwell's equations with this constitutive law for fields that vary sinusoidally with time. (e) Derive the complex Poynting theorem in the form V.-[½(r)XH*(r) + 2j < w > = 0 What is ? 6. A paradoxical case of Poynting's theorem occurs when a static electric field is applied perpendicularly to a static magnetic field, as in the case of a pair of electrodes placed within a magnetic circuit. y y (a) What are E, H, and S? (b) What is the energy density stored in the system? (c) Verify Poynting's theorem. 7. The complex electric field imaginary parts amplitude has real and E(r) = E, +jEi Under what conditions are the following scalar and vector products zero: (a) E E 10 (b) E • - 0 (c) E xE* 0 (d) E x E* 1 0 Section 7.3 8. Consider a lossy medium of permittivity e, permeability ;., and Ohmic conductivity or. (a) Write down the field equations for an x-directed electric field.

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