Light through a polarizer

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Homework Statement



Unpolarized light passes through a polarizer oriented at 33 degrees from the horizontal axis.

1) At what angle (relative to the horizontal axis) would you place a polarizer so that no light passes through it?
2) At what angle (relative to the horizontal axis) would you place a polarizer so that only a third of the light from the

Homework Equations



Malus' law
I/I0 = cos(θ)[itex]^{2}[/itex]

The Attempt at a Solution



1) Light into the second polarizer is polarized at an angle of 33 degrees. No light will pass through the second polarizer if the difference in angles is 90°, so the second polarizer should be at an angle of 90° + 33° = 123°.

2)
[itex]\frac{1}{3}= cos(θ)^{2}[/itex] where θ is the difference in angles between the two polarizers.
[itex]θ difference = arccos(\sqrt{\frac{1}{3}})[/itex]
[itex]θ_{final} - θ{_initial} = arccos(\sqrt{\frac{1}{3}})[/itex]
[itex]θ_{final} - 33° = arccos(\sqrt{\frac{1}{3}})[/itex]
θ = 54.7° + 33° = 87.7° from the horizontal axis

But my instructor says that the left-hand side of the equation is [itex]θ_{final} - θ{_initial}[/itex], which results in θ = 54.7° - 33°, so I'm confused now... have I done these two correctly?
 

Answers and Replies

  • #2
HallsofIvy
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This is a physics problem, not math so I am moving it.
 
  • #3
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I'm sorry, I must have been looking at the wrong section when I posted it.
 

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