Phys-272 Lecture 23 Polarization Birefringence TA y 45 ο x E fast w UP slo Es λ/4 z Ef RCP Polarization Orientation of E field matters when the EM wave traverses matter 4/20/2011 3 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) y x z 4/20/2011 4 Natural Light is Unpolarized • Light from sun • We can polarize light using special material • Crystals, Polymers with aligned atoms … 4/20/2011 5 Linear Polarizers • Linear Polarizers absorb all electric fields perpendicular to their transmission axis. 4/20/2011 6 Unpolarized Light on Linear Polarizer • Most light comes from electrons accelerating in random directions and is unpolarized. • Averaging over all directions, intensity of transmitted light reduces due to reduction in E c ε E 2 I = 0 4/20/2011 2 7 Linearly Polarized Light on Linear Polarizer (Law of Malus) E = E cos( θ) tranmitted incident TA 2 Itransmitted = Iincident cos (θ) θθθ θθθ is the angle between E ab so the incoming light’s rb ed Transmission axis polarization, and the Incident E transmission axis ETransmitted θθθ =Eincident cos( θθθ) 4/20/2011 8 Law of Malus Example TA 45° E0 TA E1 90° unpolarized light TA B1 II == I 0 I1 I2 I3 1) Intensity of unpolarized light incident on linear polarizer is reduced by half . I 1= I 0 / 2 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is 2 θθθ=90º . I 2= I 1 cos (90º) = 0 4/20/2011 9 Law of Malus Example TA 45° E0 TA E1 90° unpolarized light TA B1 I = I 0 I1= 0.5 I 0 2 I2= I 1cos (45) I3 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is θθθ=45º. 2 I2= I 1 cos (45º)=0.5I 1=0.25 I 0 3) Light transmitted through second polarizer is polarized 45º from vertical. Angle between it and third polarizer is 2 θθθ=45º. I 3= I 2 cos (45º) =0.125 I 0 4/20/2011 10 Polarization by Scattering • Light can also be polarized in scattering processes – Light scatters of molecules in air – Rotate polarized sunglasses while looking at blue sky » Light gets cut off - blue sky turns brighter and dimmer as you rotate indicating polarization 4/20/2011 11 Polarization by Reflection • Light can by polarized by reflection This is why polarized sunglasses are effective in reducing “glare” 4/20/2011 12 Brewster’s angle • Partial polarization on reflection • Part of the light is refracted (also polarized) • Depends on refractive indices – Brewster’s law Complete polarization for Brewster' s angle n2 tan θb = n1 when reflected and refracted light paths are perpendicular to one another 4/20/2011 13 So far we have considered plane waves that look like this: From now on just draw E and remember that B is still there: LinearLinear PolarizationPolarization The intensity of a wave does not depend on its polarization: 2 I= ε 0 c E 2 2 =ε 0c Ex + E y 2 2 =ε 0c Ex + E y 22 22 =ε0cE 0 sin( kz −− ωφθθ t )(sin + cos ) 1 2 1 1 So I= ε cE 2 Just like before 2 0 0 PolarizersPolarizers The molecular structure of a polarizer causes the component of the EE field perpendicular to the Transmission Axis to be absorbed. Polarization summary Consider an EM plane wave. The E field is polarized in the Y-direction. This is called “linearly Polarized light”. E y = E 0 sin( kx − ωt) E B = 0 sin( kx − ωt) z c Most light sources are not polarized in a particular direction. They produce unpolarized light or radiation. x polaroid (sunglasses) Long molecules absorb E- TA field parallel to molecule. (transmission axis) With and without a polarizing filter Polarizers at 90 and 0 degrees Linear Polarizers/Malus’ Law TA nˆ UP TA E1 θ (unpolarized light) I=I 0 θ I =?? E2 LP 1 LP I2=?? • This set of two linear polarizers produces LP light. What is the final intensity? – First LP transmits 1/2 of the unpolarized light: I1 = 1/2 I0 – Second LP projects out the E-field component parallel to the TA: E2 = E1 cos θ 2 I2 = I1 cos θ Law of Malus I ∝ E 2 An example of Malus’ Law Parallel axes Perpendicular axes Clicker 3) An EM wave is passed through a linear polarizer. Which component of the E-field is absorbed? The component of the E-field which is absorbed is ____________ . a) perpendicular to the transmission axis b) parallel to the transmission axis c) both components are absorbed Clicker An EM wave polarized along the y- axis, is incident on two orthogonal polarizers. I) What percentage of the intensity gets through both polarizers? a) 50% b) 25% c) 0% Q) Is it possible to increase this percentage by inserting another polarizer between the original two? [not a clicker problem (yet)] Will any light be transmitted through the crossed polarizers ? Why or why not ? Multi-part clicker I TA • Light of intensity 0, polarized along the x x θ direction is incident on a set of 2 E0 linear polarizers as shown. I y – Assuming θθθ = 45 °, what is 454545 , I I TA 1A = 0 the intensity at the exit of the 2 polarizers, in terms of I ? 0 z I =? 1 1 45 (a) I = I (b) I = I (c) I 45 = 0 45 2 0 45 4 0 1B • What is the relation between I454545 and I303030 , the final intensities in the situation above when the angle θθθ = 45 °and 30 °, respectively? I45 > I30 (a) I45 < I30 (b) I45 = I30 (c) 30 TA x 45 TA x E0 E0 y TA y I=I0 I=I 0 TA z z I45 =? I30 =? Multi-part clicker nˆ 1 I TA • Light of intensity 0, polarized along x the x direction is incident on a set of θ E0 2 linear polarizers as shown. E1 y ˆ I n 2 – Assuming θ = 45 °, what is , I I TA 1A 454545 = 0 the intensity at the exit of the E2 I z 2 polarizers, in terms of 0? I =? 1 1 45 (a) I = I (b) I = I (c) I 45 = 0 45 2 0 45 4 0 • We proceed through each polarizer in turn. I • The intensity after the first polarizer is: I= I cos2 () 45 − 0 = 0 1 0 2 • The electric field after the first polarizer is LP at θθθ1 = 45. • The intensity after the second polarizer is: 2 I1 1 I45= I 1 cos() 90 − 45 = I= I 2 454 0 Multipart clicker nˆ 1 I TA • Light of intensity 0, polarized along the x x direction is incident on a set of 2 θ E0 linear polarizers as shown. E1 y ˆ – Assuming θ = 45 °, what is the n 2 I=I TA 1A relation between the I , the intensity 0 454545 E at the exit of the 2 polarizers, in 2 I z terms of 0? I =? 1 1 45 (a) I = I (b) I = I (c) I = 0 45 2 0 45 4 0 45 1B • What is the relation between I454545 and I303030 , the final intensities in the situation above when the angle θθθ = 45 °and 30 °, respectively? I > I (a) I45 < I30 (b) I45 =I30 (c) 45 30 • In general, the first polarizer reduces the intensity by cos 2θθθ, while the second polarizer reduces it by an additional factor of cos 2(90 – θθθ). • Thus, the final output intensity is given by: 2 2 I30 = I 0 cos (30) cos (90 – 30) = 0.1875 22 22 2 In general: Iout = I0cos(θ) cos( 90 −= θ) I 0 cos( θθ) sin( ) ∝ sin( 2 θ ) This has a maximum when θθθ = 45 °. Polarization by reflection y E0 E0y unpolarized E0x x Can always describe θθθi θθθi E in terms of na components in two nb arbitrary directions. The components are equal for unpolarized Medium b light. θθθr The reflected rays are partially polarized in the horizontal plane. The transmitted rays are also partially polarized. Polarization by reflection For a certain angle, the Brewster angle, θθθp θθθp the reflected light is na completely polarized nb in the horizontal plane. This occurs when the angle between the refl. and o Medium b θθθr refr. rays is 90 . From Maxwell’s eqn. it can be shown that Brewster’s angle is given by nb tan θ p = na Light reflected on dashboard to the windshield will be polarized in the horizontal plane. Using polaroid dark glasses with a vertical polarization axis will remove most of reflected light (glare). Electric field lines from oscillating dipole full computer simulation - a snapshot in time No signal No signal This experiment was done by Hertz with radio waves in the 19 th century Dipole radiation pattern • Oscillating electric dipole generates e-m radiation that is linearly polarized in the direction of the dipole • Radiation pattern is doughnut shaped & outward traveling – zero amplitude above and below dipole (sin 2Θ) – maximum amplitude in-plane x z y Polarization by Scattering • Suppose unpolarized light encounters an atom and scatters (energy absorbed & reradiated). – What happens to the polarization of the scattered light? – The scattered light is preferentially polarized perpendicular to the plane of the scattering. » For example, assume the incident unpolarized light is moving in the z-direction.