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

Physics 102: Lecture 15, Slide 2 Today: Electromagnetic Waves Energy Intensity Polarization

Physics 102: Lecture 15, Slide 3 x z y E B loop in xy plane loop in xz plane loop in yz plane Preflight 15.1, 15.2 “In order to find the loop that detects the electromagnetic wave, we should find the loop that has the greatest flux through the loop.”

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 !

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 E field Electric Fields Recall Capacitor Energy: U = ½ C V 2 Energy Density (U/Volume): u E = ½  0 E 2 d E A

Physics 102: Lecture 15, Slide 7 Energy in B field Magnetic Fields Recall Inductor Energy: U = ½ L I 2 Energy Density (U/Volume): u B = ½ B 2 /  0 A l

Physics 102: Lecture 15, Slide 8 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

Physics 102: Lecture 15, Slide 9 Energy Density Calculate the average electric and magnetic energy density of sunlight hitting the earth with E rms = 720 N/C Use

Physics 102: Lecture 15, Slide 10 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 In EM waves, E field energy = B field energy! ( u E = u B ) u tot = u E + u B = 2u E =  0 E 2 rms

Physics 102: Lecture 15, Slide 11 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/m 2 ] = uc = c  0 E 2 rms U = Energy u = Energy Density (Energy/Volume) A = Cross section Area of light L = Length of box

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

Physics 102: Lecture 15, Slide 13 Linear Polarizers Linear Polarizers absorb all electric fields perpendicular to their transmission axis (TA) TA

Physics 102: Lecture 15, Slide 14 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(  )

Physics 102: Lecture 15, Slide 15 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!

Physics 102: Lecture 15, Slide 16 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

Physics 102: Lecture 15, Slide 17 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 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

Physics 102: Lecture 15, Slide 18 Law of Malus – 2 Polarizers 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

Physics 102: Lecture 15, Slide 19 How do polaroid sunglasses work? incident light unpolarized reflected light partially polarized the sunglasses reduce the glare from reflected light

Physics 102: Lecture 15, Slide 20 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

Physics 102: Lecture 15, Slide 21  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