PHY2049: Chapter 331ExampleÎAn observer stands 1.8 m from an isotropic point source of light with power Ps= 250 W. Calculate the rms values of the electric and magnetic fields of the light at the position of the observer.Strategy: source power -> intensity at distance r -> E and B fieldsIntensity I ?Electric field Erms?2rms01EµcI =2244power totalat areaat power averagerπPrπrrIs==≡PHY2049: Chapter 332(continued)B field is very weak. ()()()V/m 488.1)250(308.125041041034780rms==××==−ππrPπµcEsT 106.11034878rmsrms−×=×==cEBPHY2049: Chapter 333Energy of EM WaveÎE/B=c, yet E and B are depicted with the same amplitude. Why?Compare energy densities uEand uBof the E and B fields.uE=uBeverywhere along the EM wave! Energy is shared equally by the E and B fields in the EM wave.2021EεuE=2021BµuB=BEuuu 22emw==(J/m3)PHY2049: Chapter 334(continued)Rewrite in terms of E and B:Energy transported per unit time through a plane drawn perpendicular to the propagation direction of the EM wave (a good definition of intensity): This is why the intensity has been defined to be <S> (Eq. 33-23) and the Poynting vector has the meaning claimed by Eq. 33-20.(J/m3)BEuuu 22emw==EBcµBµu020emw11==>=<>< EBµcu0emw1(J/m3)(m/s)=(W/m2)PHY2049: Chapter 335Momentum of EM Wave & Radiation PressureÎEM waves (e.g. light) carry energy and travel at velocity c. It is natural to expect that they also carry momentum, although they have no mass.Can be shown that they do, but the derivation is above the level of intro physics.ÎMomentum received by an objectCan show (derivation is tough)For total reflection back along path vs total absorption (analogy to completely elastic head-on collision)cU∆p∆ =for total absorptionabsref2 p∆p∆=Eq. 33-29 is misleadingPHY2049: Chapter 336(continued)ÎMomentum received -> Force received -> pressureFrom mechanicsIf EM wave is totally absorbed Intensity Therefore For total reflection back along path cU∆p∆ =for total absorptionabsref2 p∆p∆=impulse=momentum received/givent∆p∆F =At∆U∆I =≡areapower averagecIAt∆cU∆t∆p∆F ===AFpr≡cIpr=cIpr2=(p for pressure, not momentum)PHY2049: Chapter 337ExampleArtist’s view of the Cosmos 1 satellite with the solar sail unfolded. But the launching rocket failed.ÎThe intensity of the solar radiation near the equator is 1370 W/m2at noon. What is the radiation pressure on a completely reflecting surface that directly faces the sun?pr= 2I/c = 2(1370)/3x108= 9.1 x 10-6Unit? 9.1 x 10-6 Pa (Only ~10-10of atmospheric pressure)It is feeble but can be used:Pam NmWs/mm/sW/m222===PHY2049: Chapter 338Radiation PressureÎMore examplesRadiation pressure, together with gas pressure, prevents stars from collapsing due to gravity.Comet tails—shapes and directions.PHY2049: Chapter 339PolarizationÎPolarized EM wavesE field oscillates along a fixed direction, as does the B fieldRadio waves, EM waves for TV broadcastingLaser lightÎUnpolarized EM wavesLight from ‘ordinary’ sources (the Sun, light bulbs, etc.)ÎPeople--most mammals for that matter--cannot detect light polarization, but many insects and cephalopods can, as can some fish and amphibians.PHY2049: Chapter 3310Polarizing filterÎIdeally, absorbs only E field oscillating along one direction, unaffecting E field oscillating along the perpendicular direction, “polarizing” direction.Initially polarized light with an E field amplitude E0. Let θbe the angle between the direction of the E field and the polarizing direction yof the filter. Going through the filter, only the component parallel to the polarizing direction survives.Since light intensity is proportional to E2by definition, after the filter(polarized along y) If the light is initially unpolarized, after the filter(polarized along y)θEE cos0=θII20cos=2cos020IθII
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