![Physics 102: Lecture 15, Slide 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/m2
]
= uc = cε0E2
rms
23
U = Energy
u = Energy Density
(Energy/Volume)
A = Cross section Area of light
L = Length of box](https://image.slidesharecdn.com/lect15handout-111223052211-phpapp02/85/Lect15-handout-13-320.jpg)
Lect15 handout
- 1. Physics 102: Lecture 15, Slide 1 Electromagnetic Waves and Polarization Physics 102: Lecture 15
- 2. Physics 102: Lecture 15, Slide 2 Today: Electromagnetic Waves • Energy • Intensity • Polarization
- 3. Physics 102: Lecture 15, Slide 3 x z y E B loop in xy plane loop in xz plane loop in yz plane 1 2 3 Preflight 15.1, 15.2 “In order to find the loop that dectects the electromagnetic wave, we should find the loop that has the greatest flux through the loop.”
- 4. Physics 102: Lecture 15, Slide 4
- 5. Physics 102: Lecture 15, Slide 5 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 This is important !
- 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 V2 • Energy Density (U/Volume): uE = ½ ε0E2 • Average Energy Density: uE = ½ (½ ε0E0 2 ) = ½ ε0E2 rms Magnetic Fields • Recall Inductor Energy: U = ½ L I2 • Energy Density (U/Volume): uB = ½ B2 /µ0 • Average Energy Density: uB = ½ (½ B0 2 /µ0) = ½ B2 rms/µ0
- 8. Physics 102: Lecture 15, Slide 8
- 9. Physics 102: Lecture 15, Slide 9 Energy Density Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C
- 10. Physics 102: Lecture 15, Slide 10 Energy Density Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C 2 0 2 1 rmsE Eu ε= 22 12 2 1 C N 8.85 10 720 2 Nm C − = × ÷ ÷ 0 2 2 1 µ rms B B u = 2 0 2 2 1 c Erms µ = 00 1 µε =cUse ErmsB uEu == 2 0 2 1 ε 3 6 106.42 m J uuuu EBEtotal − ×==+= 6 3 J 2.3 10 m − = ×
- 11. Physics 102: Lecture 15, Slide 11 Energy in EM wave Light waves carry energy but how? Electric Fields • Recall Capacitor Energy: U = ½ C V2 • Energy Density (U/Volume): uE = ½ ε0E2 • Average Energy Density: uE = ½ (½ ε0E0 2 ) = ½ ε0E2 rms Magnetic Fields • Recall Inductor Energy: U = ½ L I2 • Energy Density (U/Volume): uB = ½ B2 /µ0 • Average Energy Density: uB = ½ (½ B0 2 /µ0) = ½ B2 rms/µ0 In EM waves, E field energy = B field energy! ( uE = uB ) utot = uE + uB = 2uE = ε0E2 rms
- 12. Physics 102: Lecture 15, Slide 12
- 13. Physics 102: Lecture 15, Slide 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/m2 ] = uc = cε0E2 rms 23 U = Energy u = Energy Density (Energy/Volume) A = Cross section Area of light L = Length of box
- 14. Physics 102: Lecture 15, Slide 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. Physics 102: Lecture 15, Slide 15 Linear Polarizers • Linear Polarizers absorb all electric fields perpendicular to their transmission axis.
- 16. Physics 102: Lecture 15, Slide 16
- 17. Physics 102: Lecture 15, Slide 17 Unpolarized Light on Linear Polarizer • Most light comes from electrons accelerating in random directions and is unpolarized. • Averaging over all directions: Stransmitted= ½ Sincident Always true for unpolarized light!
- 18. Physics 102: Lecture 15, Slide 18 Linearly Polarized Light on Linear Polarizer (Law of Malus) Etranmitted = Eincident cos(θ) Stransmitted = Sincident cos2 (θ) T A θ θ is the angle between the incoming light’s polarization, and the transmission axis θ Transmission axisIncident E ETransmitte d E absorbed =Eincidentcos(θ)
- 19. Physics 102: Lecture 15, Slide 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 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
- 20. Physics 102: Lecture 15, Slide 20
- 21. Physics 102: Lecture 15, Slide 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. Physics 102: Lecture 15, Slide 22 Law of Malus – 2 Polarizers Cool Link unpolarized light E1 45° I =I0 TA TA 90° TA E0 I3 B1 unpolarized light E1 45° I =I0 TA TA 90° TA E0 I3 B1 1) Intensity of unpolarized light incident on linear polarizer is reduced by ½ . S1 = ½ S0 S = S0 S1 S2 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is θ=90º. S2 = S1 cos2 (90º) = 0
- 23. Physics 102: Lecture 15, Slide 23 How do polaroid sunglasses work? incident light unpolarized reflected light partially polarized he sunglasses reduce the glare from reflected light
- 24. Physics 102: Lecture 15, Slide 24
- 25. Physics 102: Lecture 15, Slide 25 unpolarized light E1 45° I =I0 TA TA 90° TA E0 I3 B1 unpolarized light E1 45° I =I0 TA TA 90° TA E0 I3 B1 Law of Malus – 3 Polarizers 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is θ=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 θ=45º. I3 = I2 cos2 (45º) I2= I1cos2 (45) = ½ I0 cos4 (45º) = I0/8 I1= ½ I0
- 26. Physics 102: Lecture 15, Slide 26 90° TA TA S1 S2 S0 60°° TATA S1 S2 S0 60°° ACT: Law of Malus A B 1) S2 A > S2 B 2) S2 A = S2 B 3) S2 A < S2 B S1= S0cos2 (60) S2= S1cos2 (30)= S0 cos2 (60) cos2 (30) S1= S0cos2 (60) S2= S1cos2 (60)= S0 cos4 (60) Cool Link E0 E0
- 27. Physics 102: Lecture 15, Slide 27
Editor's Notes
- #2: &lt;number&gt;
- #4: Note wavelength must be much large than loop size Radio waves: 3 meters Demo 158
- #6: Note E=cB is only true for EM wave, not in general
- #10: Again note energy density is same only for EM wave, not in general
- #11: Again note energy density is same only for EM wave, not in general
- #18: demo 324
- #23: Demo
- #26: Demo