week6 - Physics 1C W. Gekelman Class Notes Sixth Week...

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1 Physics 1C Class Notes© Winter 2009 W. Gekelman Sixth Week chapter 33 Consider the energy density of an electromagnetic wave. To get it let us add the energy density (energy/volume) of electric fields and of magnetic fields. u = 1 2 ! 0 E 2 + 1 2 μ 0 B 2 but E B = c = 1 0 0 u = 1 2 0 E 2 + 1 2 0 E 2 c 2 = 1 2 0 E 2 + 1 2 0 0 0 E 2 = 0 E 2 What is the energy of the electromagnetic wave that crosses an area A, per unit time. Let U be the energy (not energy density) of the wave. That is U = udV ! . Suppose the light is moving in the y direction and crossing a fixed area A. Therefore: dU = udV ; dU dt = u dV dt = uA dy dt . Here dy/dt is from how far the light has traveled in time dt. Call the quantity S= (1/A)*dU/dt. The energy that crosses an area A per unit time. We see from this that S = 1 A uA dy dt . But dy/dt is how far the light has cone in time dt. This is the sped of light c. Thus S=uc. Substituting from the expression we derived for u S = c 0 E 2 . We can make S into a vector recognizing that the k direction , the direction that light travels is perpendicular to both E and B. ! S = 1 0 ! E ! ! B ( ) . This is the Poynting flux vector.
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2 Physics 1C Lecture Notes Week 6 Using Euler’s theorem we can write the solution to the wave equation as: E 0 e i ! k i ! r ! " t ( ) . The energy of a light wave is proportional to the frequency. In quantum mechanics we learn that the expression connection the energy of a “packet” of light (a photon) to the frequency is : E = ! ! , where ! is a constant. Ultraviolet light has higher frequency (and more energy) than green light . You cant get sunburned from green light reflected off of glass. You can get sunburned from the ultraviolet component of sunlight. When light reflects off a surface or moves through glass, its energy cant spontaneously change, thus on any reflection is constant. If there are two media and the light goes from one to another, or gets reflected off of one, the incoming and outgoing electric fields must match to conserve energy. If r stands for reflected and t for transmitted and i for incident then: ! E i + ! E r = ! E t E i e i ! k i i ! r ! t ( ) = E r e i ! k r i ! r ! t ( ) + E t e i ! k r i ! r ! t ( ) The E’s are the magnitude of the electric field inside and outside the material. This expression can be rewritten as: ( incident coefficient)e i ! k i ! r ! t ( ) + ( reflected coefficient)e i ! k r ! r ! t ( ) = ( transmitted coefficient)e i ! k t ! r ! t ( ) The only way to satisfy this for every t and r is that the phases (the exponents) must all be equal. Since e i ! k i ! r ! t ( ) = e i ! k i ! r ( ) e ! i t and all the frequencies are the same: ! k i ! r i = ! k r ! r r = ! k t ! r t ! k i ! r i = k ix x + k iy y ! k r ! r r = k rx x + k ry y and ! k t ! r t = k tx x + k ty y The geometry is shown in the figure below:
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3 Figure 1 Consider a wave traveling at an angle and hitting a surface (figure 1) . The velocity of light in a media under than vacuum is less than the speed of light in vacuum. The decrease in speed is given by the index of refraction n. The index of refraction is n = kc !
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This note was uploaded on 01/21/2010 for the course PHYS 1C taught by Professor Whitten during the Winter '07 term at UCLA.

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week6 - Physics 1C W. Gekelman Class Notes Sixth Week...

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