Download EM Waves: Lecture Notes on Electromagnetic Waves and more Study notes Geology in PDF only on Docsity! Lecture 32 Chapter 34 Electromagnetic Waves Review EM Waves Wavelengths of 108 to 10- 16 meters (10-1024 Hz) Traveling wave of both E and 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 BE rr × )sin( tkxEE m ω−= )sin( tkxBB m ω−= 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 Erms need Find intensity I from 2 0 1 rmsEc I µ = 24 r PI s π = 2 0 0 4 r cPIcE srms π µµ == mVErms /1.48)8.1)(4( )1026.1)(103)(250( 2 88 =××= − π EM 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 Brms need T sm mVBrms 7 8 106.1/103 /1.48 −×= × = rms rms B Ec = c EB rmsrms = EM Waves (14) Look at sizes of Erms and Brms This is why most instruments measure E Does not mean that E component is stronger than B component in EM wave Cant compare different units Average energies are equal for E and B TBrms 7106.1 −×= mVErms /1.48= EM 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 ∆=∆=∆ t pF ∆ ∆= c Up ∆=∆ c IA tc tIA tc U t pF = ∆ ∆= ∆ ∆= ∆ ∆= EM Waves (18) For total reflection back along original path Express in terms of radiation pressure pr which is force/area SI unit is N/m2 called pascal Pa c Up ∆=∆ 2 c IA tc tIA tc U t pF 222 = ∆ ∆= ∆ ∆= ∆ ∆= A Fpr = c Ipr = c Ipr 2= Total absorption Total reflection EM 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 oscillation EM Waves (22) What is the intensity, I of 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 unpolarized wave after a polarizer is half of original 02 1 II = 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 cos2θ θ 2 0 cosII = θcosEEy = θθ µµ 2 0 22 0 2 0 coscos11 IE c E c I === 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 θ20 cosII = 02 1 II =
