Chapter 22 Electromagnetic Waves
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Chapter 22 Electromagnetic Waves22.1 Maxwell’s Rainbow1.James Clerk Maxwell’s crowning achievement was to showthat a beam of light is a traveling wave of electric and magnetic field-an electromagnetic wave-and thus that optics, the study of visible light, is a branch of electromagnetism. In Maxwell’s day, the visible, infrared, and ultraviolet forms of light were the only electromagnetic waves known. Spurred on by Maxwell’s work, however, Heinrich Hertz discovered what we now call radio waves and verified that they move through the laboratory at the same speed as visible light.2.As figure shows, we now know a wide spectrum ofelectromagnetic waves, referred to by one imaginative writer as “Maxwell’s rainbow”. In the wavelength scale in figure(and similarly the corresponding frequency scale ), each scale marker represents a change in wavelength (and correspondingly in frequency) by a factor of 10. See the arrangement of the wavelength or frequency for different electromagnetic waves in the figure.22.2 The Traveling Electromagnetic Waves1. In this section, we will restrict ourselves to discuss the electromagnetic waves to that region of the spectrum (wavelength about 1m) in which the source of the radiation(the emittedwaves)is bothmacroscopic and of manageable dimension . Figure shows, in broad outline, the generation of such waves. To see the production of the electromagnetic waves in figure.2. The properties of the electromagnetic waves are shown in figure. (1) The electric field, magnetic field and the direction of the travel of the wave are always perpendicular to each other, and the cross product B E always gives the direction of travel of the wave. Thus the wave is a transverse wave. (2)Both the electric field and the magnetic field always vary sinusoidally . Moreover, the fields vary with the same frequency and in phase (in step) with each other. (3) All electromagnetic waves, including visible light, have the same speed c in vacuum. (4)If we assume that theelectromagnetic wave istraveling in the positivedirection of an x axis,the electric field isoscillating parallel to they axis, and that themagnetic field oscillatesparallel to the z axis.Then we can write the electric and magnetic fields as sinusoidal function of position x and time t:)sin(t kx E E m ω-= and )sin(t kx B B m ω-=, in which m E and m B are theamplitudes of thefields. (5) The wavespeedand the magnitudeof the electric field andmagnetic field are related by c B E =/.3. Figure shows the electric and magnetic field vectors in a “snapshot ” of the wave at a certain instant.22.3 Energy Transport and the Poynting Vector1. The rate of energy transport per unit area in an electromagnetic wave is described by a vector, called the Poynting vector , as B E S ⨯=01μ and has the SI unit watt persquare meter (W/m 2). The direction of the Poynting vector of an electromagnetic wave at any point gives the wave ’s direction of travel and the direction of energy transport at that point .2. Because E and B are perpendicular to each other in anelectromagnetic wave, then the magnitude ofS is20011E c EB S μμ==.3. More useful in practice is the average energy transported over time. So we need to find the time-averaged value of S, written S and also called the intensity I of the wave. That is022μc E S I m == 02μc E rms =, where rms E is the root-mean-square valueof the electric field.4. How intensity varies with distance from a real source ofelectromagnetic radiation is often complex . (1) However, in some situation we can assume that the source is a point source that emits the light isotropically , that is, with equal intensity in all direction. (2) If the energy of the waves is conserved as they spread from this source. Thus the rate at which energy is transferred through a sphere centered on the source by the radiation must equal the rate at which energy is emitted by the source, that is, the powers P of the source. (3) The intensity at the sphere must then be24r P I s π=.22.4 Radiation Pressure1. Electromagnetic waves have linear momentum as well as energy. This means that we can exert a pressure-a radiation pressure -on an object by shining light on it.2. To find an expression for the pressure , let us shine a beam of electromagnetic radiation on an object for a time interval t ∆. (1) Let us assume that the object is free to move and that the radiation is entirely absorbed by the object. The magnitude p ∆ of the linear momentum change of the object is related to the energy change , which is equal to the energy that the object gains from the radiation, by c U p /∆=∆, where c is the speed of light. (2) If the radiation is entirely reflected backalong its original path instead of being absorbed, the magnitude of the momentum change of the object is twice that given above c U p /2∆=∆.3. From Newton ’s second law, we know that a change in momentum is related to a force by t p F ∆∆=/.4. To find expression for the force exerted by radiation in terms of the intensity of the radiation, suppose that a flat surface of area A, perpendicular to the path of the radiation, intercepts the radiation. In time intervalt ∆, the energy intercepted by the area A is t IA U ∆=∆. (1) If the energy is completelyabsorbed , we have c IA t c t IA t c U t p F =∆∆=∆∆=∆∆=. (2) Similarly, Ifthe radiation is totally reflected back along its original path, we have cIA F 2=. (3) If the radiation is partly absorbed and partly reflected, the magnitude on the area A is between above two values .5. The radiation pressure r p is the force per unit area on anobject due to radiation. So the radiation pressure for the case of total absorption isc I p r = and that of total reflection back along path isc I p r 2=.22.5 Polarization1. VHF (very high frequency) television antennas in England areoriented vertically, but those in North America are horizontal. The difference is due to the direction of oscillation of the electromagnetic waves carrying the TV signal. In England, the transmitting equipment is designed to produce waves that are polarized vertically; that is, their electric field oscillates vertically. So, for the electric field of the incident television waves to drive a current along an antenna, the antenna must be vertical. In North America, the waves are horizontally polarized.2.Figure shows anelectromagnetic wave withits electric field oscillatingparallel to the vertical y axis.The plane containing the Evector is called the plane of oscillation of the wave. We can represent the wave’s polarization by showing the extent of the electric field oscillation in “head on”view of the plane of oscillation, as in figure.3.The electromagnetic waves emitted by television station all have the same polarization, but the electromagnetic waves emitted by any common source of light are polarized randomly or unpolarized. That is, the electric field at anygiven point is always perpendicular to the direction of travel of waves but changes direction randomly.4.In principle, we canresolve each electricfield of figure (a) into yand z components andthen finding the netfields along the y axis and z axis separately, as shown in figure (b). In doing so, we mathematically change unpolarized light into the superposition of two polarized waves whose planes of oscillation are perpendicular to each other. The result is the double-arrow representation of above figure (b). Similarly we can represent light that is partially polarized by drawing one of the arrows of the double-arrow representation longer than the other arrow.5.We can actually transformunpolarized visible light intopolarized light by sending itthrough a polarizing sheet, as isshown in figure. When light is sent through the sheet, electric field components along one direction pass through the sheet, while components perpendicular to that direction are absorbedand disappear . We shall assign to the sheet a polarizing direction , along which electric field components are passed.6. The intensity of the transmitted light will follow the one-half rule 021I I =. 7. Suppose now that the light reaching apolarizing sheet is already polarized.We can resolve E into twocomponents relative to the polarizingdirection of the sheet, as shown infigure. Thus we have the cosine-square rule for the intensity θ20cos I I =.22.6 Polarization by Reflection1. Figure shows a ray ofunpolarized lightincident on a glasssurface. In general,the reflected light alsohas both components but with unequal magnitudes. However, when the light is incident at a particular angle, called the Brewster angle B θ, the reflected light has only perpendicularcomponents, as shown in figure.2. Brewster ’s law : o r B 90=+θθ and n B 1tan -=θ.。
