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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
• ### Lect15 handout

1. 1. Electromagnetic Waves and Polarization Physics 102: Lecture 15
2. 2. Today: Electromagnetic Waves <ul><li>Energy </li></ul><ul><li>Intensity </li></ul><ul><li>Polarization </li></ul>
3. 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 <ul><li>Changing B field creates E field </li></ul><ul><li>Changing E field creates B field </li></ul><ul><li>E = c B </li></ul>If you decrease E, you also decrease B! x z y This is important !
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
6. 7. Energy in EM wave <ul><li>Light waves carry energy but how? </li></ul><ul><li>Electric Fields </li></ul><ul><li>Recall Capacitor Energy: </li></ul><ul><li>U = ½ C V 2 </li></ul><ul><li>Energy Density (U/Volume): u E = ½  0 E 2 </li></ul><ul><li>Average Energy Density: </li></ul><ul><ul><li> u E = ½ (½  0 E 0 2 ) </li></ul></ul><ul><ul><li> = ½  0 E 2 rms </li></ul></ul><ul><li>Magnetic Fields </li></ul><ul><li>Recall Inductor Energy: </li></ul><ul><ul><ul><li>U = ½ L I 2 </li></ul></ul></ul><ul><li>Energy Density (U/Volume): </li></ul><ul><li>u B = ½ B 2 /  0 </li></ul><ul><li>Average Energy Density: </li></ul><ul><ul><ul><li>u B = ½ (½ B 0 2 /  0 ) </li></ul></ul></ul><ul><ul><ul><li>= ½ B 2 rms /  0 </li></ul></ul></ul>
7. 9. Energy Density <ul><li>Calculate the average electric and magnetic energy density of sunlight hitting the earth with E rms = 720 N/C </li></ul>Example
8. 10. Energy Density <ul><li>Calculate the average electric and magnetic energy density of sunlight hitting the earth with E rms = 720 N/C </li></ul>Use Example
9. 11. Energy in EM wave <ul><li>Light waves carry energy but how? </li></ul><ul><li>Electric Fields </li></ul><ul><li>Recall Capacitor Energy: </li></ul><ul><li>U = ½ C V 2 </li></ul><ul><li>Energy Density (U/Volume): u E = ½  0 E 2 </li></ul><ul><li>Average Energy Density: </li></ul><ul><ul><li> u E = ½ (½  0 E 0 2 ) </li></ul></ul><ul><ul><li> = ½  0 E 2 rms </li></ul></ul><ul><li>Magnetic Fields </li></ul><ul><li>Recall Inductor Energy: </li></ul><ul><ul><ul><li>U = ½ L I 2 </li></ul></ul></ul><ul><li>Energy Density (U/Volume): </li></ul><ul><li>u B = ½ B 2 /  0 </li></ul><ul><li>Average Energy Density: </li></ul><ul><ul><ul><li>u B = ½ (½ B 0 2 /  0 ) </li></ul></ul></ul><ul><ul><ul><li>= ½ B 2 rms /  0 </li></ul></ul></ul><ul><li>In EM waves, E field energy = B field energy! ( u E = u B ) </li></ul><ul><li>u tot = u E + u B = 2u E =  0 E 2 rms </li></ul>
10. 13. Intensity (I or S) = Power/Area <ul><li>Energy (U) hitting flat surface in time t </li></ul><ul><ul><ul><li>= Energy U in red cylinder: </li></ul></ul></ul><ul><ul><li>U = u x Volume </li></ul></ul><ul><ul><li> = u (AL) = uAct </li></ul></ul><ul><li>Power (P): </li></ul>A L=ct <ul><ul><li>P = U/t </li></ul></ul><ul><ul><li> = uAc </li></ul></ul><ul><li>Intensity (I or S): </li></ul><ul><li> S = P/A [W/m 2 ] </li></ul><ul><ul><li>= u c = c  0 E 2 rms </li></ul></ul>23 U = Energy u = Energy Density (Energy/Volume) A = Cross section Area of light L = Length of box
11. 14. Polarization <ul><li>Transverse waves have a polarization </li></ul><ul><ul><li>(Direction of oscillation of E field for light) </li></ul></ul><ul><li>Types of Polarization </li></ul><ul><ul><li>Linear (Direction of E is constant) </li></ul></ul><ul><ul><li>Circular (Direction of E rotates with time) </li></ul></ul><ul><ul><li>Unpolarized (Direction of E changes randomly) </li></ul></ul>x z y
12. 15. Linear Polarizers <ul><li>Linear Polarizers absorb all electric fields perpendicular to their transmission axis. </li></ul>
13. 17. Unpolarized Light on Linear Polarizer <ul><li>Most light comes from electrons accelerating in random directions and is unpolarized. </li></ul><ul><li>Averaging over all directions: S transmitted = ½ S incident </li></ul>Always true for unpolarized light!
14. 18. Linearly Polarized Light on Linear Polarizer (Law of Malus) <ul><li>E tranmitted = E incident cos(  ) </li></ul><ul><li>S transmitted = S incident cos 2 (  ) </li></ul>  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
15. 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 <ul><li>zero </li></ul><ul><li>      1/2 what it was before </li></ul><ul><li>      1/4 what it was before </li></ul><ul><li>      1/3 what it was before </li></ul><ul><li>      need more information </li></ul>
16. 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 <ul><li>zero </li></ul><ul><li>      1/2 what it was before </li></ul><ul><li>      1/4 what it was before </li></ul><ul><li>      1/3 what it was before </li></ul><ul><li>      need more information </li></ul>
17. 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
18. 23. How do polaroid sunglasses work? incident light unpolarized reflected light partially polarized the sunglasses reduce the glare from reflected light
19. 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
20. 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   