Sol4 - the magnetic field in the wave and also...

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UAHuntsville Spring, 2010 PH 113: Homework Solutions No. 4 Solution 21: Equation (E16.13) is for the average intensity of an EM wave, and is W=c <E^2>/(8 \pi k). Solution 22: Equation (E16.13) is for the average intensity of an EM wave, and is W=c <E^2>/(8 \pi k).

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Solution 23:

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Solution 24: Solution 25: From the figure, we know that when the middle polarizer is at θ 2 = 0 o ,90 o there is no light going through the system, which means that the middle polarizer is parallel to either the first or last one, so that effectively the system is crossed. With no loss of generality, we assume θ 1 = 0 o and θ 3 = 90 o . Therefore, the total intensity is given by I = 1 2 cos 2 0 o θ 2 180 o π Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ cos 2 90 o θ 2 180 o π Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ and for θ 2 = 30 o we find I = 0.09375 . Solution 26:
Solution 27: (a) The polarization direction is defined by the electric field (which is perpendicular to
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Unformatted text preview: the magnetic field in the wave, and also perpendicular to the direction of wave travel). The given function indicates the magnetic field is along the x axis (by the subscript on B ) and the wave motion is along – y axis (see the argument of the sine function). Thus, the electric field direction must be parallel to the z axis. (b) Since k is given as 1.57 × 10 7 /m, then λ = 2 π /k = 4.0 × 10 − 7 m, which means f = c / λ = 7.5 × 10 14 Hz. (c) The magnetic field amplitude is given as B m = 4.0 × 10 − 6 T. The electric field amplitude E m is equal to B m divided by the speed of light c . The rms value of the electric field is then E m divided by 2 . Eq. 33-26 then gives I = 1.9 kW/m 2 ....
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