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

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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 μ 0 ! 0 u = 1 2 ! 0 E 2 + 1 2 μ 0 E 2 c 2 = 1 2 ! 0 E 2 + 1 2 μ 0 μ 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 μ 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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