23_InstSolManual_PDF_Part23 - Electromagnetic Waves 23-23...

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23.86. Set Up: For linearly polarized light the intensity passed by a filter is where is the incident intensity and I is the transmitted intensity. After passing through the filter the light is linearly polarized along the filter axis. Solve: (a) Use two filters, with the axis of the second filter at from the original direction of polarization of the light. (b) Let be the angle between the original direction of polarization of the light and the axis of the first filter. The angle between the axes of the two filters therefore is The intensity passed by the two filters is so so The transmitted intensity has a maximum value of when 23.87. Set Up: Let the light initially be in the material with refractive index and let the final slab have refractive index In part (a) let the middle slab have refractive index Apply Snell’s law to each refraction. Solve: (a) 1 st interface: 2 nd interface: Combining the two equations gives This is the equation that would apply if the middle slab were absent. (b)
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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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