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Transcript
EM Waves Friday Honor Lecture This Lecture !More on EM waves !EM spectrum !Polarization Prof. D. Oertel Dept. of Physiology Interpreting Nerve Signals From previous Lecture Displacement currents Maxwell’s equations !EM Waves ! ! 1 Facts on EM waves The 4 Maxwell’s Equations charged particles create an electric field q $ E" dA = # S (Gauss' Law) 0 There are no magnetic monopoles $ B" dA = 0 d& $L E" ds = % dtB (Faraday - Henry) S d& $ LB" ds = µ0I + µ0#0 dtE (Ampere - Maxwell law) An electric field can be created by a changing magnetic field Consequence: induced current ! Lorentz force ! EM waves are solutions of Maxwell’s equations. ! In empty space: sinusoidal wave propagating along x with velocity ! E = E max cos (kx – "t) ! B = B max cos (kx – "t) • E and B are perpendicular oscillating vectors ! Currents create a magnetic field ! A changing electric field can create a magnetic field F = q(E + v " B) 3 Transverse ! waves E!B=0 •The direction of propagation is perpendicular to E and B E x B direction of c ! Quick Quiz on EM waves Quick Quiz pn EM waves An electromagnetic wave propagates in the –y direction. The electric field at a point in space is momentarily oriented in the +x direction. The magnetic field at that point is momentarily oriented in the Which orientation will have the largest induced emf? E y z (a) –x direction x B (b) +y direction (c) +z direction (d) –z direction loop in xz plane loop in xy plane z A B c y x E B C yz in p o lo ane pl Faraday’s law: Loops use B not E! 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. EM Waves from an Antenna EM waves from antennas Sources of EM waves: oscillating charges, accelerated/decellerated charges, electron transitions between energy levels in atoms, nuclei and molecules - Electromagnetic radiation is greatest when charges accelerate at right angles to the line of sight. - Zero radiation is observed when the charges accelerate along the line of sight. - These observations apply to electromagnetic waves of all frequencies. 2 rods connected to AC generator; charges oscillate between the rods (a) As oscillations continue, the rods become less charged, the field near the charges decreases and the field produced at t = 0 moves away from the rod (b) The charges and field reverse (c) The oscillations continue (d) ! ! ! ! 8 Relation between E and B ! E = Emax cos (kx – "t) ! B = Bmax cos (kx – "t) ! First derivatives: From: ! Gamma rays: !~ 10-14- 10-10 m Source: radioactive nuclei, cause serious damage to living tissues "E = #kE max sin(kx # $t) "x "B = $Bmax sin(kx # $t) "t "E "B =# "x "t X-rays: ~10-12 -10-8 m source: deceleration of high-energy electrons striking a metal target Diagnostic tool in medicine Speed of em waves (speed of light in vacuum) UV !~ 6 x 10-10 - 4 x 10-7 m Most UV light from the sun is absorbed in the stratosphere by ozone ! Sources: hot objects and molecules Microwaves: ! ~10-4 -0.3 m This relation comes from Maxwell’s equations! sources: electronic devices radar systems, MW ovens Energy density of E and B field ! ! Rate at which energy flows through a unit area perpendicular to direction of wave propagation In a parallel plate capacitor: C = 1 1" A U 1 U = CV 2 = 0 E 2 d 2 # uE = = "0 E 2 2 2 d Ad 2 This is the power per unit area (J/s.m2 = W/m2) E/B=c ! Radio: ! ~ 10 - 0.1 m Sources: charges accelerating through conducting wires Radio and TV Infrared: ! ~ 7 x 10-7-10-3 m Poynting vector ! Source: atoms and molecules Human eye Visible range from red (700 nm) to violet (400 nm) The EM Spectrum Its direction is the direction of propagation of the EM wave Magnitude: "0 A d True for any geometry ! ! When a battery is connected to a circuit the current does not jump instantaneously from 0 to the final value !/R because there is an induced emf opposing to battery action. By calculating the work done in a solenoid against induced emf arising from increasing B-flux as current goes from 0 to I we can derive the energy density of the B-field associated to the current EB E 2 = µ0 cµ0 This is time dependent ! Its magnitude varies in time ! Its magnitude reaches a maximum at the same instant as E and B do S= ! ! uB = ! B2 2µ0 True for any geometry Energy carried by EM waves Intensity and Poynting vector Total instantaneous energy density of EM waves ! And the intensity (average P/area) ! E/B=c u =uE + uB = 1/2 $oE2 + B2 /(2µo) uE = uB Since B = E/c and ! Iav = 1 B2 E2 uE = "0 E 2 = uB = = 2 2 2µ0 2c µ0 ! ! ! In a given volume, the energy is shared equally by the two fields ! ! Pav = U av uav AL = = uav Ac "t "t ! power per unit area ! ! "p (energy absorbed)/"t Power = = "t c c F Power / A I = = A c c from EM waves transport momentum# pressure on a surface Complete absorption on a surface: total transported energy U ! in time interval %t # total momentum p = U / c and prad=Sav/c (b) very shiny, to reflect as much sunlight as possible Which is the value of the radiation pressure in the above case? (a) P = 2S/c Perfectly reflecting surface: momentum of incoming and reflected light p = U/c # total transferred momentum p = 2U/c and prad = 2Sav/c Direct sunlight pressure ~5 x 10-6 N/m2 (b) P = S/c Circular and elliptical polarization Polarization of Light Waves (34.8) ! ExB (a) very black to absorb as much sunlight as possible Power I= = Save c ! ! EB E 2 = µ0 cµ0 To maximize the radiation pressure on the sails of a spacecraft using solar sailing, the sheets must be momentum p = U / c ! ! S= Quick Quiz on radiation pressure Radiation Pressure = force per unit area prad = I & E2 ! Radiation pressure and momentum F= B 2 EB = µ0 µ0c E max Bmax 2µ0 ! ! u = uE + uB = "0 E 2 = Wave intensity I = time average over one or more cycle <sin2(kx - "t)> = 1/2 then <E2> = Emax2/2 and <B2> = Bmax2/2 ! Iav = uav c = Let’s consider a cylinder with axis along x of area A and length L and the time for the wave to travel L is !t=L/c The average power is: ! Pav = uav c A Linearly polarized waves: E-field oscillates at all times in the plane of polarization Unpolarized light: E-field in random directions. Superposition of waves with E vibrating in many different directions Linearly polarized light: E-field has one spatial orientation ! Circularly polarized light: superposition of 2 waves of equal amplitude with orthogonal linear polarizations, and 90˚ out of phase. The tip of E describes a circle (counterclockwise = RH and clockwise=LH depending on y component ahead or behind) ! The electric field rotates in time with constant magnitude. If amplitudes differ # elliptical polarization ! DEMO with MW generator and metal grid Producing polarized light ! Polarization by selective absorption: material that transmits waves whose E-field vibrates in a plain parallel to a certain direction and absorbs all others pick up antenna connected to Ammeter Metal grid MW generator This polarization absorbed This polarization transmitted transmission axis Long-chain hydrocarbon molecules Polaroid sheet (E. Land 1928) If the wires of the grid are parallel to the plane of polarization the grid absorbs the E-component (electrons oscillate in the wire). The same thing happens to a polaroid: the component parallel to the direction of the chains of hydrocarbons is absorbed. If the grid is horizontal the Ammeter will measure a This not null current since the wave reaches the antenna polarization absorbed pick-up This polarization transmitted transmission axis Detecting polarized light: Malus’ law Polarization by selective absorption If linearly polarized light of intensity I0 passes through a polarizing filter with transmission axis at an angle ' along y y Polaroid sheet ' Einc = E0sin' i + E0 cos' j x transmission axis After the polarizer Etransm = E0cos' j E0cos' So the intensity transmitted is Itransm = E02 cos2' = (0cos2' Long-chain hydrocarbon molecules A polarizer is used to produce ! Ideal polarizer transmits waves with E parallel to transmission axis and absorbs those with E ) axis ! Relative orientation of axis of polarizer and analyzer determines intensity of transmitted light. Transmitted intensity: I = I0cos2' I0 = intensity of polarized beam on analyzer (Malus’ law) ! Allowed component parallel to analyzer axis polarized light of intensity I0 and an analyzer rotated at an angle ': transmission 100% when ' = 0 and zero when ' = 90° Polaroid sheets Quick Quiz on Polarization Law of Malus Example I1= 0.5 I0 I2= I1cos2(45) 1) Light transmitted by first polarizer is vertically polarized. I1 = I0 cos245=I0/2 A B 2) Angle between it and second polarizer is '=45º. I2= I1 cos2 (45º) =0.5I1=0.25 I0 3) Light transmitted through second polarizer is polarized 45º from vertical. Angle between it and third polarizer is '=45º. I3= I2 cos2(45º) = 0.125I0 Relative orientation of polarizers ! ! ! Transmitted amplitude is Eocos' (component of polarization along polarizer axis) Transmitted intensity is Iocos2' ( square of amplitude) Perpendicular polarizers give zero intensity.