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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) θ = while 90 θ = ° for polarizer a . Polarizers c and e are at the same angle θ 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 θ is 1 2 , the intensity transmitted by a polarizing filter is 1 transmitted 0 2 I I = . On the other hand, for polarized light I transmitted = I 0 cos 2 θ . Therefore, ( ) 2 2 2 2 2 1 1 transmitted 2 transmitted 1 0 2 2 cos cos 350 W/m cos 30 131 W/m I I I θ θ = = = ° = 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.
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 I C ω ε = for all parts of this question.
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