Physics 2102 Gabriela Gonz lez Electromagnetic waves Electromagnetic waves A solution to Maxwell s equations in free space E E m sin k x t c k speed of propagation B Bm sin k x t c Em 1 Bm 0 0 299 462 954 http phys23p sl psu edu CWIS m 187 163mph s Visible light infrared ultraviolet radio waves X rays Gamma rays are all electromagnetic waves The Poynting vector Electromagnetic waves are able to transport energy from transmitter to receiver example from the Sun to our skin The power transported by the wave and its direction is quantified by the Poynting vector 1 S E B John Henry Poynting 1852 1914 1 1 2 For a wave since S EB E E is perpendicular to B c Units Watt m2 E B S In a wave the fields change with time Therefore the Poynting vector changes too The direction is constant but the magnitude changes from 0 to a maximum value EM wave intensity energy density A better measure of the amount of energy in an EM wave is obtained by averaging the Poynting vector over one wave cycle The resulting quantity is called intensity 2 2 2 1 1 I S Em sin kx t E c c 1 I Em 2 2c Both fields have the same energy density or The average of sin2 over one cycle is 1 I Erms 2 c 2 1 1 1 B u E E 2 cB 2 0 u B 2 2 2 The total EM energy density is then 2 2 u 0 E B 0 Solar Energy The light from the sun has an intensity of about 1kW m2 What would be the total power incident on a roof of dimensions 8x20m I 1kW m2 is power per unit area P IA 103 W m2 x 8m x 20m 0 16 MW The solar panel shown Sunpower E19 is 61in x 41in The actual solar panel delivers 6A at 50V What is its efficiency http us sunpowercorp com homes products services solar panels EM spherical waves The intensity of a wave is power per unit area If one has a source that emits isotropically equally in all directions the power emitted by the source pierces a larger and larger sphere as the wave travels outwards P I s 4 r 2 So the power per unit area decreases as the inverse of distance squared Example A radio station transmits a 10 kW signal at a frequency of 100 MHz We will assume it radiates as a point source At a distance of 1km from the antenna find a the amplitude of the electric and magnetic field strengths and b the energy incident normally on a square plate of side 10cm in 5min Ps 10kW 2 I 0 8 mW m 4 r 2 4 1km 2 1 2 I Em Em 2c I 0 775V m 2c Bm Em c 2 58 nT Received energy P U t S U SA 2 4 m A A Radiation Pressure Waves not only carry energy but also momentum The effect is very small we don t ordinarily feel pressure from light If light is completely absorbed during an interval Dt the momentum transferred is given by p u and twice as much if reflected c Newton s law F p t Now supposing one has a wave that hits a surface of area A perpendicularly the amount of energy transferred to that surface in time Dt will be IA t p U IA t therefore c Radiation pressure A I IA F c I 2I pr total absorption pr total reflection c c Radiation pressure examples Solar mills Not radiation pressure Solar sails From the Planetary Society Comet tails Sun radiation I 1 KW m2 Area 30m2 F IA c 0 1 mN Mass m 5 kg a F m 2 10 5 m s2 When does it reach 10mph 4 4 m s V at t V a 2 105 s 2 3 days EM waves polarization Radio transmitter If the dipole antenna is vertical so will be the electric fields The magnetic field will be horizontal The radio wave generated is said to be polarized In general light sources produce unpolarized waves emitted by atomic motions in random directions Completely unpolarized light will have equal components in horizontal and vertical directions Therefore running the light through a polarizer will cut the intensity in half I I0 2 When polarized light hits a polarizing sheet only the component of the field aligned with the sheet will get through E y E cos And therefore I I 0 cos 2 Light reflected from surfaces is usually polarized horizontally Polarized sunglasses take advantage of this they are vertical polarizing sheets so that they cut the horizontally polarized light from glare reflections on roads cars etc Example Initially unpolarized light of intensity I0 is sent into a system of three polarizers as shown What fraction of the initial intensity emerges from the system What is the polarization of the exiting light Through the first polarizer unpolarized to polarized so I 1 I0 Into the second polarizer the light is now vertically polarized Then I 2 I1cos26 o 1 4 I1 1 8 I0 Now the light is again polarized but at 60o The last polarizer is horizontal so I3 I2cos23 o 3 4 I2 3 32 I0 0 094 I0 The exiting light is horizontally polarized and has 9 of the original amplitude Reflection and refraction When light finds a surface separating two media air and water for example a beam gets reflected and another gets refracted transmitted Law of reflection the angle of incidence q1 equals the angle of reflection q 1 Law of refraction n2 sin 2 n1 sin 1 Snell s law n is the index of refraction of the medium In vacuum n 1 In air n 1 In all other media n 1 Example Water has n 1 33 How much does a beam incident at 45o refracts n2 sin q2 n1 sin q1 sin q2 n1 n2 sin q1 1 1 33 sin 45o 0 0098 q2 32o Image of the object Actual object Actual light ray Light ray the brain imagines as if in air Example an optical illusion The index of refraction decreases with temperature the light gets refracted and ends up bending upwards We seem to see water on the road but in fact we are looking at the sky Chromatic dispersion The index of refraction depends on the wavelength color of the light n2 sin 2 n1 sin 1 Examples Prisms Rainbows water drops act as reflecting prisms Total internal reflection From glass to air the law of refraction uses n2 n1 so q2 q1 it may reach 90o or more the ray is reflected instead of refracted q2 n2 sin q2 n1 sin q1 n2 1 q1 n1 For glass fused quartz n 1 46 and the critical angle is 43o optical fibers Polarization by reflection Different polarization of light get reflected and refracted with different amplitudes birefringence At one particular angle the parallel polarization is NOT reflected at …
View Full Document
Unlocking...