Hwksolns4_Ungraded_Publishers

Hwksolns4_Ungraded_Publishers - 35.10. Id > Ic = Ie > Ib >...

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35.10. I d > I c = I e > I b > I a . The intensity passing through a polarizer is 2 0 cos . I θ Polarizer d is aligned with the incident wave (0 ) = while 90 for polarizer a . Polarizers c and e are at the same angle from the vertical.
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35.27. Model: Use Malus’s law for polarized light. Visualize: Solve: For unpolarized light, the electric field vector varies randomly through all possible values of θ . Because the average value of cos 2 is 1 2 , the intensity transmitted by a polarizing filter is 1 transmitted 0 2 II = . On the other hand, for polarized light I transmitted = I 0 cos 2 . Therefore, ( ) 22 2 2 2 11 transmitted 2 transmitted 1 0 cos cos 350 W/m cos 30 131 W/m I θθ == = ° = Assess: Note that any particular wave has a clear polarization. It is only in a “sea” of waves that the resultant wave has no polarization.
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36.2. (a) 1.0 A. Use R R V I R = . (b) 4.0 A. Use 0 R R V I R R ε == . (c) 2.0 A. I R is not dependent on the frequency.
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36.3. (a) 4.0 A. Use 0 c IC ωε = for all parts of this question.
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This note was uploaded on 12/26/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.

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Hwksolns4_Ungraded_Publishers - 35.10. Id > Ic = Ie > Ib >...

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