1. Electromagnetic Waves and Polarization Physics 102: Lecture 15 HE2: Monday March 15 Material covered: lectures through 12 NOT INCLUDING AC CIRCUITS! discussions through 6 homework through 6 labs through 5 Review: Sunday March 14, 3-4:30, room 141 Will go over Fall 2009 Exam

2. Today: Electromagnetic Waves Energy Intensity Polarization

3. x z y E B 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!!!) loop in xy plane loop in xz plane loop in yz plane 123123 7 Preflight 15.1, 15.2 “In order to find the loop that detects the electromagnetic wave, we should find the loop that has the greatest flux through the loop.” 32%26%42%

4. Propagation of EM Waves Changing B field creates E field Changing E field creates B field E = c B x z y 10 If you decrease E, you also decrease B! This is important !

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 17% 74% 8%

6. Energy in E field 13 Electric Fields Recall Capacitor Energy: U = ½ C V 2 Energy Density (U/Volume): u E = ½  0 E 2 d A E

7. Energy in B field 13 Magnetic Fields Recall Inductor Energy: U = ½ L I 2 Energy Density (U/Volume): u B = ½ B 2 /  0 A l

8. Energy in EM wave Light waves carry energy but how? 13 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

9. Energy Density Calculate the average electric and magnetic energy density of sunlight hitting the earth with E rms = 720 N/C Use 18

10. Energy in EM wave Light waves carry energy but how? 13 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

11. 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 ] = uc = c  0 E 2 rms 23 U = Energy u = Energy Density (Energy/Volume) A = Cross section Area of light L = Length of box 23

12. 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 25

13. Linear Polarizers Linear Polarizers absorb all electric fields perpendicular to their transmission axis. 30

14. 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 33 Always true for unpolarized light!

15. Linearly Polarized Light on Linear Polarizer (Law of Malus) E tranmitted = E incident cos(  ) S transmitted = S incident cos 2 (  ) TA   is the angle between the incoming light’s polarization, and the transmission axis  Transmission axis Incident E E Transmitted E absorbed =E incident cos(  ) 36

16. 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 1. zero 2. 1/2 what it was before 3. 1/4 what it was before 4. 1/3 what it was before 5. need more information 37

17. 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 38

18. Law of Malus – 2 Polarizers Cool Link 39 1) Intensity of unpolarized light incident on linear polarizer is reduced by ½. S 1 = ½ S 0 S = S 0 S1S1 S2S2 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

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

20. Law of Malus – 3 Polarizers 43 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

21.  TA S1S1 S2S2 S0S0  TA S1S1 S2S2 S0S0  ACT: Law of Malus AB 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 E0E0 E0E0 48

22. See You Monday!