Lecture 32Chapter 34Electromagnetic WavesReview• EM Waves –– Wavelengths of 108to 10-16meters (10-1024Hz)– Traveling wave of both Eand B fields– E field is ⊥⊥⊥⊥ B field– Wave moves in direction ⊥⊥⊥⊥to both E and B fields – E and B vary sinusoidally with same frequency– At large distances fields are in phase BErr×)sin( tkxEEmω−=)sin( tkxBBmω−=Review • EM waves move at the speed of light, c in free space (vacuum or air)• Relate velocity of wave by• Using definition of ω and wave number k • Find velocity of wave is• Also defined askvω=fπω2=λπ2=ksmc /1038×=λfcv ==mmBEc =001εµ=cReview• Poynting vector, S – rate of energy transported per unit area• Instantaneous energy flow rate • Defined intensity I to be time averaged value of S201rmsavgEcSIµ==aveaveavgareapowerareatimeenergySI===/BESrrr×=01µEBS01µ=EM Waves (12) • Problem – Isotropic point light source as power of 250 W. You are 1.8 meters away. Calculate the rms values of the E and B fields.• To find Ermsneed• Find intensity I from 201rmsEcIµ=24 rPIsπ=2004 rcPIcEsrmsπµµ==mVErms/1.48)8.1)(4()1026.1)(103)(250(288=××=−πEM Waves (13) • Problem – Isotropic point light source as power of 250 W. You are 1.8 meters away. Calculate the rms values of the E and B fields.• To find BrmsneedTsmmVBrms78106.1/103/1.48−×=×=rmsrmsBEc =cEBrmsrms=EM Waves (14) • Look at sizes of Ermsand Brms• This is why most instruments measure E• Does not mean that E component is stronger than B component in EM wave – Can’t compare different units• Average energies are equal for E and BTBrms7106.1−×=mVErms/1.48=EM Waves (15) • The energy density of electric field, uEis equal to energy density of magnetic field, uBBcE=2021EuEε=220212021)( BccBuEεε==001εµ=c0220002121µεµεBBuE==022µBuB=BEuu =EM Waves (16) • EM waves linear have momentum momentum as well as energy • Light shining on object exerts a pressure – radiation pressure• Object’s change in momentum is related to its change in energy • If object absorbs all radiation from EM wave (total absorption)• If object reflects all radiation back in original direction (total reflection)cUp∆=∆cUp∆=∆2EM Waves (17) • Just defined intensity, I as power per unit area A so power is • Change in energy is amount of power P in time t• Want force of radiation on object• For total absorption • Find force is IAP=tIAtPU ∆=∆=∆tpF∆∆=cUp∆=∆cIAtctIAtcUtpF =∆∆=∆∆=∆∆=EM Waves (18)• For total reflection back along original path• Express in terms of radiation pressure prwhich is force/area• SI unit is N/m2called pascal PacUp∆=∆2cIAtctIAtcUtpF222=∆∆=∆∆=∆∆=AFpr=cIpr=cIpr2=• Total absorption • Total reflectionEM Waves (19)• Source emits EM waves with E field always in same plane wave is polarized– Example, television station• Indicate a wave is polarized by drawing double arrow• Plane containing the E field is called plane of oscillationEM Waves (20)• Source emits EM waves with random planes of oscillation (E field changes direction) is unpolarized– Example, light bulb or Sun• Resolve E field into components • Draw unpolarized light as superposition of 2 polarized waves with E fields ⊥⊥⊥⊥ to each otherEM Waves (21)• Transform unpolarized light into polarized by using a polarizing sheet• Sheet contains long molecules embedded in plastic which was stretched to align the molecules in rows• E field component || to polarizing direction of sheet is passed (transmitted), but ⊥⊥⊥⊥component is absorbedEM Waves (22)• What is the intensity, Iof the light transmitted by polarizing sheet?• For unpolarized light, separate E field into components• Sum of 2 components are equal but only light || to polarizer is transmitted• One-half rule: Intensity of unpolarizedwave after a polarizer is half of original021II =EM Waves (23)• For polarized light, resolve E into components• Transmitted || component is• Use definition of intensity• Cosine-squared rule: Intensity of polarized wave changes as cos2θθθθθ20cosII =θcosEEy=θθµµ2022020coscos11IEcEcI ===EM Waves (24)• Have 2 polarizing sheets – First one called polarizer– Second one called analyzer• Intensity of unpolarized light going through polarizer is • Light is now polarized and intensity of light after analyzer is given by θ20cosII =021II =EM Waves (25)• Checkpoint #4 – Unpolarized light hits a polarizer and then an analyzer. The polarizing direction of each sheet is indicated by dashed line. Rank pairs according to fraction of initial intensity which is passed, greatest first.EM Waves (26)• Look at relative orientation of polarization direction between the 2 sheets.• What is the intensity if the sheets are…– Polarized || – all light passes– Polarized ⊥⊥⊥⊥ to each other – no light passes – For angles in between – get more light if closer to
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