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UF PHY 2049 - Example

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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 fieldsIntensity 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 objectCan 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 -> pressureFrom mechanicsIf 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 10-6Unit? 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 examplesRadiation pressure, together with gas pressure, prevents stars from collapsing due to gravity.Comet tails—shapes and directions.PHY2049: Chapter 339PolarizationÎPolarized EM wavesE field oscillates along a fixed direction, as does the B fieldRadio waves, EM waves for TV broadcastingLaser lightÎUnpolarized EM wavesLight 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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UF PHY 2049 - Example

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