# Need a little clarification

Consider the following polarization problem.The original light wave is completely un-polarized.The first screen is oriented vertically, while the last (third) screen is oriented horizontally. The middle screen is angled at 30° from the vertical. Assume the original intensity of the unpolarized light is 100%.

I know that after going through the first screen, 50% of the original intensity remains and I also figured out the second screen too. I missed the notes for this so I'm a little stumped with what I would do for the last screen. Would it's % of intensity be half of what came out of the second screen?

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If the intensity of the light before entering the first screen is $I_{0}$, then its intensity after exiting the screen in the middle is $\displaystyle\frac{3I_{0}}{16}$. By passing through the third screen, intensity will be reduced further to [STRIKE]cos^2(70)[/STRIKE] $\displaystyle\frac{3I_{0}}{16}\frac{cos^2(60°)}{2}≈0.04688 I_{0}$. Its 70° because that is the angle between the orientations of screens 2 and 3.

Whatever may be the orientation of the two screens, intensity will certainly be reduced to half of its original value. The angle between the screens will reduce it further.

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TSny
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If the intensity of the light before entering the first screen is $I_{0}$, then its intensity after exiting the screen in the middle is $\displaystyle\frac{3I_{0}}{16}$. By passing through the third screen, intensity will be reduced further to $\displaystyle\frac{3I_{0}}{16}\frac{cos^2(70°)}{2}≈0.01097 I_{0}$. Its 70° because that is the angle between the orientations of screens 2 and 3.
Oops. The factor cos2(70o)/2 is not correct. The angle is not 70o and the division by 2 is incorrect.
[EDIT: I'm afraid the 3/16 factor is also incorrect]
Whatever may be the orientation of the two screens, intensity will certainly be reduced to half of its original value. The angle between the screens will reduce it further.
No. If light is polarized vertically and then passes through a vertically oriented polarizing filter, the intensity will not be reduced at all (assuming ideal filter).

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TSny
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I know that after going through the first screen, 50% of the original intensity remains and I also figured out the second screen too.
Yes, 50% after first filter. Since you didn't show your work for the second filter, I can't know if what you did is correct.

I missed the notes for this so I'm a little stumped with what I would do for the last screen. Would it's % of intensity be half of what came out of the second screen?
No. You will need Malus' Law

Oops. The factor cos2(70o)/2 is not correct. The angle is not 70o and the division by 2 is incorrect.
[EDIT: I'm afraid the 3/16 factor is also incorrect]
Very sorry about that. I was just trying to help.

Silly me, that should be $cos^2(60°)$.

But why is 3I0/16 incorrect? Intensity is reduced by 1/2 after the first screen, another 1/2 because of the second screen, and by $cos^2(30°)$. So that would be 3I0/16. Am I missing anything?

And why is the division by two incorrect? Is the intensity not diminished by 50% after passing through a screen?

Again, very sorry "alittlelost"... TSny
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But why is 3I0/16 incorrect? Intensity is reduced by 1/2 after the first screen, another 1/2 because of the second screen, and by $cos^2(30°)$. So that would be 3I0/16. Am I missing anything?
The intensity of unpolarized light will be reduced by 1/2 when passing through an ideal polarizing filter. But if the incident light is plane polarized, the intensity will be reduced by a factor of cos2θ (not ##\frac{1}{2}\cdot##cos2θ), where θ is the angle between the polarization direction of the incident light and the transmission direction of the filter. For example, if vertically polarized light is incident on an ideal filter with transmission direction also vertical, then 100% of the light gets through.