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Electromagnetic Waves and Polarization

Physics 102: Lecture 15. Electromagnetic Waves and Polarization. Today’s lecture will cover Textbook Sections 22.7-8. 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 • Today’s lecture will cover Textbook Sections 22.7-8 Physics 102: Lecture 15, Slide 1

  2. E B loop in xy plane loop in yz plane loop in xz plane y x z Preflight 15.1, 15.2 Which orientation will have the largest induced emf? Hint: Loops use B not E. 1 2 3 Physics 102: Lecture 15, Slide 2

  3. E B loop in xy plane loop in yz plane loop in xz plane y x z Preflight 15.1, 15.2 Which orientation will have the largest induced emf? 1 2 3 Only the loop in the xy plane will have a magnetic flux through it as the wave passes. The flux will oscillate with time and induce an emf. (Faraday’s Law!)

  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! Physics 102: Lecture 15, Slide 4

  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 Physics 102: Lecture 15, Slide 5

  6. 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 Physics 102: Lecture 15, Slide 6

  7. 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? Physics 102: Lecture 15, Slide 7

  8. Energy Density Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C Example Note: This is true only for EM waves. Physics 102: Lecture 15, Slide 8

  9. Energy Density Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C Example Use Note: This is true only for EM waves.

  10. Electric Fields • Average Energy Density: uE = ½ (½ e0E02) = ½ e0E2rms Magnetic Fields • Average Energy Density: uB = ½ (½ B02/m0) = ½ B2rms/m0 = ½ E2rms/(c2m0) = ½ e0E2rms Energy in EM wave Electric and magnetic fields carry equal amounts of energy. In EM waves, E field energy = B field energy! ( uE = uB ) Physics 102: Lecture 15, Slide 10

  11. Intensity (I or S) = Power/Area • Energy (U) in box: U = u x Volume = u (AL) • Power (P): • P = U/t • = U (c/L) • = u A c • Intensity (I or S): • S = P/A • = uc = ce0E2rms A L L=ct U = Energy u = Energy Density (Energy/Volume) A = Cross section Area of light L = Length of box 23 Physics 102: Lecture 15, Slide 11 23

  12. 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) Physics 102: Lecture 15, Slide 12

  13. Linear Polarizers • Linear Polarizers absorb all electric fields perpendicular to their transmission axis. Molecular View (link) Physics 102: Lecture 15, Slide 13

  14. 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

  15. TA Transmission axis Incident E Linearly Polarized Light on Linear Polarizer (Law of Malus) Etranmitted = Stransmitted = q q is the angle between the incoming light’s polarization, and the transmission axis Eabsorbed q ETransmitted =Eincidentcos(q) Physics 102: Lecture 15, Slide 15

  16. 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) Physics 102: Lecture 15, Slide 16

  17. Preflight 15.6, 15.7 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 • ½what it was before • ¼what it was before • ⅓ what it was before • need more information Now, horizontally polarized light passes through the same glasses (which are vertically polarized). The intensity of the light when it emerges is • zero • ½ what it was before • ¼ what it was before • ⅓ what it was before • Need more information Physics 102: Lecture 15, Slide 17

  18. 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 Physics 102: Lecture 15, Slide 18

  19. 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 Physics 102: Lecture 15, Slide 19

  20. S = S0 S1 S2 Example Law of Malus – 2 Polarizers S1 = S2 = Physics 102: Lecture 15, Slide 20

  21. 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 Physics 102: Lecture 15, Slide 21

  22. Example Law of Malus – 3 Polarizers I1= I2 = I3 = Physics 102: Lecture 15, Slide 22

  23. 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)

  24. ACT: Law of Malus 60 ° ° 60 ° ° TA TA TA 90 ° TA S0 S0 S1 S1 S2 S2 E0 E0 A B 1) S2A > S2B 2) S2A = S2B 3) S2A < S2B Physics 102: Lecture 15, Slide 24

  25. 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 Physics 102: Lecture 15, Slide 25

  26. See next time! • Read Sections 23.1-2, 7-8 Physics 102: Lecture 15, Slide 26

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