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Polarizing filters problem

  • Thread starter bowma166
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  • #1
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Homework Statement


Two sheets of polaroid are aranged as polarized and analyzer. Suppose that the preferential direction of the second sheet is rotated by an angle [tex]\phi[/tex] about the direction of incidence and then rotated by an angle [itex]\alpha[/itex] about the vertical direction. If unpolarized light of intensity [itex]I_0[/itex] is incident from the left, what is the intensity of the light emerging on the right?
http://img22.imageshack.us/img22/8189/polaroiddiagram.jpg [Broken]
Sorry about the crappy diagram quality.

Homework Equations


[tex]I=I_0\cos^2\phi[/tex]
or
[tex]E=E_0\cos\phi[/tex]


The Attempt at a Solution


Well the intensity drops down to [itex]\frac{1}{2}I_0[/itex] after the first polaroid. The second one is hard. The way I'm thinking about it, it basically turns into a geometry problem... Say you're looking at the second polaroid head-on from the first direction. It's turned an angle [itex]\phi[/itex], making the component of the light that can pass through smaller. It's then turned about the vertical axis by an angle [itex]\alpha[/itex], which, when you look at the 2D projection of the filter you see when you look at it head-on, makes the lines seem more vertical and should allow more light to pass through, right?
Here's my even crappier hand drawn diagram of the situation sort of:
http://img254.imageshack.us/img254/9007/polaroid2.jpg [Broken]

So, again, I think the problem is: look perpendicularly at the plane made by the top and bottom two vectors in the above diagram when [itex]\alpha=0[/itex], then rotate it away by [itex]\alpha[/itex] and determine what angle you see the diagonal line makes with the bottom line. Maybe? I just don't know how to do the geometry of that. Sorry for the wordiness and not being very clear, it's hard to express my thoughts on this problem very well.

These points seem clear to me:

at [tex]\alpha=\frac{\pi}{2}[/tex], [tex]I=\frac{1}{2}I_0[/tex]

at [tex]\alpha=0[/tex], [tex]I=\frac{1}{2}I_0\cos^2\phi[/tex]

Thanks.
 
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Answers and Replies

  • #2
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Anyone?
 

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