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Explore the concepts of electromagnetic waves, induced emf, energy density, and polarization in this lecture. Learn about the relationship between electric and magnetic fields, energy propagation in EM waves, and polarization types.
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Physics 102: Lecture 15 Electromagnetic Wavesand Polarization • Today’s lecture will cover Textbook Sections 22.7-8 Physics 102: Lecture 15, Slide 1
E B loop in xy plane loop in yz plane loop in xz plane y x z Preflight 15.1, 15.2 Which orientation will have the largest induced emf? Hint: Loops use B not E. 1 2 3 Physics 102: Lecture 15, Slide 2
E B loop in xy plane loop in yz plane loop in xz plane y x z Preflight 15.1, 15.2 Which orientation will have the largest induced emf? 1 2 3 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!)
y x z This is important ! 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! Physics 102: Lecture 15, Slide 4
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 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 E=cB Physics 102: Lecture 15, Slide 6
Energy in EM wave Electric Fields • Recall Capacitor Energy: U = ½ C V2 • Energy Density (U/Volume): uE = ½ e0E2 • Average Energy Density: uE = ½ (½ e0E02) = ½ e0E2rms Magnetic Fields • Recall Inductor Energy: U = ½ L I2 • Energy Density(U/Volume): uB = ½ B2/m0 • Average Energy Density: uB = ½ (½ B02/m0) = ½ B2rms/m0 Light waves carry energy but how? Physics 102: Lecture 15, Slide 7
Energy Density Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C Example Note: This is true only for EM waves. Physics 102: Lecture 15, Slide 8
Energy Density Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C Example Use Note: This is true only for EM waves.
Electric Fields • Average Energy Density: uE = ½ (½ e0E02) = ½ e0E2rms Magnetic Fields • Average Energy Density: uB = ½ (½ B02/m0) = ½ B2rms/m0 = ½ E2rms/(c2m0) = ½ e0E2rms Energy in EM wave Electric and magnetic fields carry equal amounts of energy. In EM waves, E field energy = B field energy! ( uE = uB ) Physics 102: Lecture 15, Slide 10
Intensity (I or S) = Power/Area • Energy (U) in box: U = u x Volume = u (AL) • Power (P): • P = U/t • = U (c/L) • = u A c • Intensity (I or S): • S = P/A • = uc = ce0E2rms A L L=ct U = Energy u = Energy Density (Energy/Volume) A = Cross section Area of light L = Length of box 23 Physics 102: Lecture 15, Slide 11 23
y x z 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) Physics 102: Lecture 15, Slide 12
Linear Polarizers • Linear Polarizers absorb all electric fields perpendicular to their transmission axis. Molecular View (link) Physics 102: Lecture 15, Slide 13
Always true for unpolarized light! Unpolarized Light on Linear Polarizer • Most light comes from electrons accelerating in random directions and is unpolarized. • Averaging over all directions: Stransmitted= ½ Sincident
TA Transmission axis Incident E Linearly Polarized Light on Linear Polarizer (Law of Malus) Etranmitted = Stransmitted = q q is the angle between the incoming light’s polarization, and the transmission axis Eabsorbed q ETransmitted =Eincidentcos(q) Physics 102: Lecture 15, Slide 15
TA Transmission axis Incident E Linearly Polarized Light on Linear Polarizer (Law of Malus) Etranmitted = Eincident cos(q) Stransmitted = Sincident cos2(q) q q is the angle between the incoming light’s polarization, and the transmission axis Eabsorbed q ETransmitted =Eincidentcos(q) Physics 102: Lecture 15, Slide 16
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 • zero • ½what it was before • ¼what it was before • ⅓ what it was before • 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 • zero • ½ what it was before • ¼ what it was before • ⅓ what it was before • Need more information Physics 102: Lecture 15, Slide 17
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 Physics 102: Lecture 15, Slide 18
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 Physics 102: Lecture 15, Slide 19
S = S0 S1 S2 Example Law of Malus – 2 Polarizers S1 = S2 = Physics 102: Lecture 15, Slide 20
Example Law of Malus – 2 Polarizers S = S0 S1 S2 1) Intensity of unpolarized light incident on linear polarizer is reduced by ½ . S1 = ½ S0 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is q=90º. S2 = S1 cos2(90º) = 0 Cool Link Physics 102: Lecture 15, Slide 21
Example Law of Malus – 3 Polarizers I1= I2 = I3 = Physics 102: Lecture 15, Slide 22
Example Law of Malus – 3 Polarizers I1= ½ I0 I2= I1cos2(45) 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is q=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 q=45º. I3 = I2 cos2 (45º) = ½ I0 cos4 (45º)
ACT: Law of Malus 60 ° ° 60 ° ° TA TA TA 90 ° TA S0 S0 S1 S1 S2 S2 E0 E0 A B 1) S2A > S2B 2) S2A = S2B 3) S2A < S2B Physics 102: Lecture 15, Slide 24
ACT: Law of Malus 60 ° ° 60 ° ° TA TA TA 90 ° TA S0 S0 S1 S1 S2 S2 E0 E0 A B Cool Link S1= S0cos2(60) S1= S0cos2(60) S2= S1cos2(60) S2= S1cos2(30) = S0 cos2(60)cos2(30) = S0 cos4(60) 1) S2A > S2B 2) S2A = S2B 3) S2A < S2B Physics 102: Lecture 15, Slide 25
See next time! • Read Sections 23.1-2, 7-8 Physics 102: Lecture 15, Slide 26