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

2. Physics 102: Lecture 15, Slide 2 x z y E B loop in xy plane loop in xz plane loop in yz plane 123123 Which orientation will have the largest induced emf? Hint: Loops use B not E. Preflight 15.1, 15.2

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 Which orientation will have the largest induced emf? Preflight 15.1, 15.2

4. Physics 102: Lecture 15, Slide 4 Propagation of EM Waves Changing B field creates E field Changing E field creates B field E = c B x z y If you decrease E, you also decrease B! This is important !

5. Physics 102: Lecture 15, Slide 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

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

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

10. Physics 102: Lecture 15, Slide 10 Energy in EM wave Electric and magnetic fields carry equal amounts of energy. Electric Fields Average Energy Density: u E = ½ (½  0 E 0 2 ) = ½  0 E 2 rms Magnetic Fields Average Energy Density: u B = ½ (½ B 0 2 /  0 ) = ½ B 2 rms /  0 = ½ E 2 rms /(c 2  0 ) = ½  0 E 2 rms In EM waves, E field energy = B field energy! ( u E = u B )

11. Physics 102: Lecture 15, Slide 11 Intensity (I or S) = Power/Area Energy (U) in box: U = u x Volume = u (AL) Power (P): L A L=ct P = U/t = U (c/L) = u A c Intensity (I or S): S = P/A = 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. Physics 102: Lecture 15, Slide 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

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

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 Always true for unpolarized light!

15. Physics 102: Lecture 15, Slide 15 Linearly Polarized Light on Linear Polarizer (Law of Malus) E tranmitted = S transmitted = 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(  )

16. Physics 102: Lecture 15, Slide 16 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(  )

17. Physics 102: Lecture 15, Slide 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 1.zero 2.½ what it was before 3.¼ what it was before 4. ⅓ what it was before 5.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 1.zero 2.½ what it was before 3.¼ what it was before 4. ⅓ what it was before 5.Need more information

18. Physics 102: Lecture 15, Slide 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 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

19. Physics 102: Lecture 15, Slide 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

20. Physics 102: Lecture 15, Slide 20 Law of Malus – 2 Polarizers S 1 = S 2 = S = S 0 S1S1 S2S2

21. Physics 102: Lecture 15, Slide 21 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 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

22. Physics 102: Lecture 15, Slide 22 Law of Malus – 3 Polarizers I 2 = I 3 = I1 =I1 =

23. 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 1 = ½ I 0

24. Physics 102: Lecture 15, Slide 24  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 E0E0 E0E0

25. Physics 102: Lecture 15, Slide 25  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

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