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Electromagnetic Waves and Polarization

Physics 102: Lecture 15. Electromagnetic Waves and Polarization. Today: Electromagnetic Waves. Energy Intensity Polarization. E. B. loop in xy plane. loop in yz plane. loop in xz plane. y. x. z. Preflight 15.1, 15.2.

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Electromagnetic Waves and Polarization

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  1. Physics 102: Lecture 15 Electromagnetic Wavesand Polarization

  2. Today: Electromagnetic Waves • Energy • Intensity • Polarization

  3. E B loop in xy plane loop in yz plane loop in xz plane y x z Preflight 15.1, 15.2 “In order to find the loop that dectects the electromagnetic wave, we should find the loop that has the greatest flux through the loop.” 1 2 3

  4. y x z This is important ! Propagation of EM Waves • Changing B field creates E field • Changing E field creates B field E = c B If you decrease E, you also decrease B!

  5. Preflight 15.4 Suppose that the electric field of an electromagnetic wave decreases in magnitude. The magnetic field: 1 increases 2 decreases 3 remains the same E=cB

  6. Energy in EM wave Electric Fields • Recall Capacitor Energy: U = ½ C V2 • Energy Density (U/Volume): uE = ½ e0E2 • Average Energy Density: uE = ½ (½ e0E02) = ½ e0E2rms Magnetic Fields • Recall Inductor Energy: U = ½ L I2 • Energy Density(U/Volume): uB = ½ B2/m0 • Average Energy Density: uB = ½ (½ B02/m0) = ½ B2rms/m0 Light waves carry energy but how?

  7. Energy Density Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C Example

  8. Energy Density Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C Example Use

  9. Energy in EM wave Electric Fields • Recall Capacitor Energy: U = ½ C V2 • Energy Density (U/Volume): uE = ½ e0E2 • Average Energy Density: uE = ½ (½ e0E02) = ½ e0E2rms Magnetic Fields • Recall Inductor Energy: U = ½ L I2 • Energy Density(U/Volume): uB = ½ B2/m0 • Average Energy Density: uB = ½ (½ B02/m0) = ½ B2rms/m0 Light waves carry energy but how? In EM waves, E field energy = B field energy! ( uE = uB ) utot = uE + uB = 2uE = e0E2rms

  10. Intensity (I or S) = Power/Area • Energy (U) hitting flat surface in time t = Energy U in red cylinder: U = u x Volume = u (AL) = uAct • Power (P): A • P = U/t • = uAc • Intensity (I or S): • S = P/A [W/m2] • = uc = ce0E2rms L=ct U = Energy u = Energy Density (Energy/Volume) A = Cross section Area of light L = Length of box 23

  11. y x z Polarization • Transverse waves have a polarization • (Direction of oscillation of E field for light) • Types of Polarization • Linear (Direction of E is constant) • Circular (Direction of E rotates with time) • Unpolarized (Direction of E changes randomly)

  12. Linear Polarizers • Linear Polarizers absorb all electric fields perpendicular to their transmission axis.

  13. Always true for unpolarized light! Unpolarized Light on Linear Polarizer • Most light comes from electrons accelerating in random directions and is unpolarized. • Averaging over all directions: Stransmitted= ½ Sincident

  14. TA Transmission axis Incident E Linearly Polarized Light on Linear Polarizer (Law of Malus) Etranmitted = Eincident cos(q) Stransmitted = Sincident cos2(q) q q is the angle between the incoming light’s polarization, and the transmission axis Eabsorbed q ETransmitted =Eincidentcos(q)

  15. ACT/Preflight 15.6 Unpolarized light (like the light from the sun) passes through a polarizing sunglass (a linear polarizer). The intensity of the light when it emerges is • zero •      1/2 what it was before •      1/4 what it was before •      1/3 what it was before •      need more information

  16. ACT/Preflight 15.7 Now, horizontally polarized light passes through the same glasses (which are vertically polarized). The intensity of the light when it emerges is • zero •      1/2 what it was before •      1/4 what it was before •      1/3 what it was before •      need more information

  17. Example Law of Malus – 2 Polarizers S = S0 S1 S2 1) Intensity of unpolarized light incident on linear polarizer is reduced by ½ . S1 = ½ S0 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is q=90º. S2 = S1 cos2(90º) = 0 Cool Link

  18. incident light unpolarized reflected light partially polarized the sunglasses reduce the glare from reflected light How do polaroid sunglasses work?

  19. Example Law of Malus – 3 Polarizers I1= ½ I0 I2= I1cos2(45) 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is q=45º. I2 = I1 cos2 (45º) = ½ I0 cos2 (45º) 3) Light transmitted through second polarizer is polarized 45º from vertical. Angle between it and third polarizer is q=45º. I3 = I2 cos2 (45º) = ½ I0 cos4 (45º) = I0/8

  20. ACT: Law of Malus 60 ° ° 60 ° ° TA TA TA 90 ° TA S0 S0 S1 S1 S2 S2 E0 E0 A B Cool Link S1= S0cos2(60) S1= S0cos2(60) S2= S1cos2(60) S2= S1cos2(30) = S0 cos2(60)cos2(30) = S0 cos4(60) 1) S2A > S2B 2) S2A = S2B 3) S2A < S2B

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