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Lect15 handout

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  • 1
  • Note wavelength must be much large than loop size Radio waves: 3 meters Demo 158
  • Note E=cB is only true for EM wave, not in general
  • Again note energy density is same only for EM wave, not in general
  • Again note energy density is same only for EM wave, not in general
  • demo 324
  • Demo
  • Demo
  • Transcript

    • 1. Electromagnetic Waves and Polarization Physics 102: Lecture 15
    • 2. Today: Electromagnetic Waves
      • Energy
      • Intensity
      • Polarization
    • 3. Preflight 15.1, 15.2 1 2 3 “ In order to find the loop that dectects the electromagnetic wave, we should find the loop that has the greatest flux through the loop.” x z y E B loop in xy plane loop in xz plane loop in yz plane
    • 4.  
    • 5. 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! x z y This is important !
    • 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
    • 7. Energy in EM wave
      • Light waves carry energy but how?
      • Electric Fields
      • Recall Capacitor Energy:
      • U = ½ C V 2
      • Energy Density (U/Volume): u E = ½  0 E 2
      • Average Energy Density:
        • u E = ½ (½  0 E 0 2 )
        • = ½  0 E 2 rms
      • Magnetic Fields
      • Recall Inductor Energy:
          • U = ½ L I 2
      • Energy Density (U/Volume):
      • u B = ½ B 2 /  0
      • Average Energy Density:
          • u B = ½ (½ B 0 2 /  0 )
          • = ½ B 2 rms /  0
    • 8.  
    • 9. Energy Density
      • Calculate the average electric and magnetic energy density of sunlight hitting the earth with E rms = 720 N/C
      Example
    • 10. Energy Density
      • Calculate the average electric and magnetic energy density of sunlight hitting the earth with E rms = 720 N/C
      Use Example
    • 11. Energy in EM wave
      • Light waves carry energy but how?
      • Electric Fields
      • Recall Capacitor Energy:
      • U = ½ C V 2
      • Energy Density (U/Volume): u E = ½  0 E 2
      • Average Energy Density:
        • u E = ½ (½  0 E 0 2 )
        • = ½  0 E 2 rms
      • Magnetic Fields
      • Recall Inductor Energy:
          • U = ½ L I 2
      • Energy Density (U/Volume):
      • u B = ½ B 2 /  0
      • Average Energy Density:
          • u B = ½ (½ B 0 2 /  0 )
          • = ½ B 2 rms /  0
      • In EM waves, E field energy = B field energy! ( u E = u B )
      • u tot = u E + u B = 2u E =  0 E 2 rms
    • 12.  
    • 13. 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 L=ct
        • P = U/t
        • = uAc
      • Intensity (I or S):
      • S = P/A [W/m 2 ]
        • = u c = c  0 E 2 rms
      23 U = Energy u = Energy Density (Energy/Volume) A = Cross section Area of light L = Length of box
    • 14. 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)
      x z y
    • 15. Linear Polarizers
      • Linear Polarizers absorb all electric fields perpendicular to their transmission axis.
    • 16.  
    • 17. Unpolarized Light on Linear Polarizer
      • Most light comes from electrons accelerating in random directions and is unpolarized.
      • Averaging over all directions: S transmitted = ½ S incident
      Always true for unpolarized light!
    • 18. Linearly Polarized Light on Linear Polarizer (Law of Malus)
      • E tranmitted = E incident cos(  )
      • S transmitted = S incident cos 2 (  )
        is the angle between the incoming light’s polarization, and the transmission axis  E Transmitted E absorbed =E incident cos(  ) TA Transmission axis Incident E
    • 19. 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
    • 20.  
    • 21. 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
    • 22. Law of Malus – 2 Polarizers Cool Link 1) Intensity of unpolarized light incident on linear polarizer is reduced by ½ . S 1 = ½ S 0 S = S 0 S 1 S 2 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is  =90º. S 2 = S 1 cos 2 (90º) = 0 Example
    • 23. How do polaroid sunglasses work? incident light unpolarized reflected light partially polarized the sunglasses reduce the glare from reflected light
    • 24.  
    • 25. Law of Malus – 3 Polarizers 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is  =45º. I 2 = I 1 cos 2 (45º) = ½ I 0 cos 2 (45º) 3) Light transmitted through second polarizer is polarized 45º from vertical. Angle between it and third polarizer is  =45º. I 3 = I 2 cos 2 (45º) I 2 = I 1 cos 2 (45) = ½ I 0 cos 4 (45º) = I 0 /8 I 1 = ½ I 0 Example
    • 26. ACT: Law of Malus A B 1) S 2 A > S 2 B 2) S 2 A = S 2 B 3) S 2 A < S 2 B S 1 = S 0 cos 2 (60) S 2 = S 1 cos 2 (30) = S 0 cos 2 (60) cos 2 (30) S 1 = S 0 cos 2 (60) S 2 = S 1 cos 2 (60) = S 0 cos 4 (60) Cool Link E 0 E 0   TA TA S 1 S 2 S 0    TA TA S 1 S 2 S 0   
    • 27.