What is the angle for perfect light blockage through a polarizer?

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SUMMARY

The discussion focuses on the application of Malus' Law to determine the angles for light blockage through polarizers. For complete light blockage, the second polarizer must be oriented at 123 degrees relative to the horizontal axis, calculated by adding 90 degrees to the initial 33-degree angle. To achieve one-third light transmission, the final angle should be 87.7 degrees, derived from the equation I/I0 = cos(θ)², where θ is the difference in angles between the two polarizers. Clarification is sought regarding the correct interpretation of angle differences in the context of the problem.

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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)^{2}

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

But my instructor says that the left-hand side of the equation is θ_{final} - θ{_initial}, which results in θ = 54.7° - 33°, so I'm confused now... have I done these two correctly?
 
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This is a physics problem, not math so I am moving it.
 
I'm sorry, I must have been looking at the wrong section when I posted it.
 

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