23_InstSolManual_PDF_Part5

23_InstSolManual_PDF_Part5 - Electromagnetic Waves 23-5...

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Solve: (a) For the beam, This is spread uniformly over 100 cells, so the energy given to each cell is 80 J. (b) The cross sectional area of each cell is with (c) 23.20. Set Up: The radiation pressure on a totally absorbing surface is Solve: 23.21. Set Up: The radiation pressure is for a totally absorbing surface and for a totally reflecting surface. Solve: (a) (b) Reflect: For the same intensity of light, the radiation pressure is twice as great for a totally reflecting surface versus a totally absorbing surface. For ordinary light intensities, the radiation pressure is very small. 23.22. Set Up: Solve: 23.23. Set Up: and Solve: and Reflect: The power output of the source is proportional to the square of the electric field amplitude in the emitted waves. 23.24. Set Up: For a totally reflecting surface the average pressure is and for a totally reflecting surface it is The radiation is the same so has the same intensity in the two cases. Solve:
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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