# Modern Optics: Polarization

• Shackleford

#### Shackleford

Whenever there is natural light, you always assume the amplitude is 1/2 of the incident light amplitude after the first polarizer, right?

http://i111.photobucket.com/albums/n149/camarolt4z28/a1.png [Broken]

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20111030_161325.jpg [Broken]

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Whenever there is natural light, you always assume the amplitude is 1/2 of the incident light amplitude after the first polarizer, right?
No. Don't worry about amplitude, just work with the irradiance -- which is different than amplitude.

http://i111.photobucket.com/albums/n149/camarolt4z28/a1.png [Broken]

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20111030_161325.jpg [Broken]
If I may requote the stated problem, it will be easier for us to follow what is going on:
Natural light having a flux density I0 is incident on a sheet of HN-38. After the light passes through the sheet of HN-38, it transmits through a sheet of HN-32 whose axis of transmission makes an angle of 30 degrees with respect to the transmission axis of the first sheet. Find an expression for the irradiance of the beam after it emerges from the second polarizer.
In an ideal polarizer, the irradiance of natural light is 1/2 after transmission through it. However, for real polarizers HN-38 and HN-32, the irradiance will be less than 1/2.

If your professor wants you to use the actual transmissions of HN-38 and HN-32, then that information should have been provided to you somehow -- either in your lecture notes, class handouts, course website, textbook, or somewhere in the problem set you have been given.

(And if your professor wants you to use the transmission of an ideal polarizer, then I don't think he would have specified two different polarizing materials.)

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