Electromagnetic Waves and Polarization Physics 102: Lecture 15 Electromagnetic Waves and Polarization 1

Today: Electromagnetic Waves Energy Intensity Polarization

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.” x z y E B loop in yz plane loop in xy plane loop in xz plane Note wavelength must be much large than loop size Radio waves: 3 meters Demo 158 1 2 3

Propagation of EM Waves x z y Changing B field creates E field Changing E field creates B field E = c B Note E=cB is only true for EM wave, not in general This is important ! If you decrease E, you also decrease B!

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

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 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

Energy Density Example Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C Example Again note energy density is same only for EM wave, not in general

Energy Density Example Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C Example Again note energy density is same only for EM wave, not in general Use

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 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 In EM waves, E field energy = B field energy! ( uE = uB ) utot = uE + uB = 2uE = e0E2 rms

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 P = U/t = uAc Intensity (I or S): S = P/A [W/m2] = uc = ce0E2rms L=ct U = Energy u = Energy Density (Energy/Volume) A = Cross section Area of light L = Length of box 23

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

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

Unpolarized Light on Linear Polarizer demo 324 Most light comes from electrons accelerating in random directions and is unpolarized. Averaging over all directions: Stransmitted= ½ Sincident Always true for unpolarized light!

Linearly Polarized Light on Linear Polarizer (Law of Malus) TA Etranmitted = Eincident cos(q) Stransmitted = Sincident cos2(q) q q is the angle between the incoming light’s polarization, and the transmission axis Transmission axis Incident E Eabsorbed q ETransmitted =Eincidentcos(q)

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 zero      1/2 what it was before      1/4 what it was before      1/3 what it was before      need more information

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

Law of Malus – 2 Polarizers Example S = S0 S1 S2 1) Intensity of unpolarized light incident on linear polarizer is reduced by ½ . S1 = ½ S0 Demo 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

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

Law of Malus – 3 Polarizers Example 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) Demo 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) = I0/8

ACT: Law of Malus A B Cool Link S2 S2 E0 E0 S0 S0 S1 S1 90 ° TA S1 S2 S0 60 TA S1 S2 S0 60 ° E0 E0 A B Cool Link S1= S0cos2(60) S1= S0cos2(60) S2= S1cos2(30) = S0 cos2(60) cos2(30) S2= S1cos2(60) = S0 cos4(60) 1) S2A > S2B 2) S2A = S2B 3) S2A < S2B