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Find angle of polarizing sheet given original & transmitted intensity

  • Thread starter MrMoose
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  • #1
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Homework Statement



A beam of polarized light is sent through a system of two polarizing sheets. Relative to the polarization direction of that incident light, the polarizing directions of the sheets are at angles θ for the first sheet and 90° for the second sheet. If 0.10 of the incident intensity is transmitted by the two sheets, what is θ?

Homework Equations



I1 = Io /2

I2 = I1 * cos^2(θ)

The Attempt at a Solution



The intensity of the beam after the first polarizing sheet is as follows:

I1 = Io /2

The intensity of the beam after the second polarizing sheet is as follows:

I2 = I1 * cos^2(θ)

Where θ is the angle difference between θ1 and θ2.

The problem tells you that the transmitted beam is 0.10 of the original beam:

I2 = 0.10 * Io

Substituting with the equations above:

I2 = 0.10 * Io = (Io/2) * cos^2(θ) and therefore:

0.10 = cos^2(θ)/2

Solving for θ:

θ =arccos([itex]\sqrt{0.2}[/itex])= 63°

Which means the angle of the first polarizing sheet is 63° out from 90° so the answer I get is:

27° or 153°

But this isn't correct according to the back of the book. Please help. Thanks in advance, MrMoose

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
haruspex
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I1 = Io /2
That would be true if the original beam were unpolarized, but it isn't.
 
  • #3
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Thank you! I read that question over so many times and missed that part every time. I think I got it now:

I1 = Io*cos^2(θ1)

I2 = I1*cos^2(θ2)

Where θ2 = 90 - θ1

We also know that I2 = 0.10 * Io, therefore:

I2 = 0.1 * Io = Io * cos^2(θ1) * cos^2(90-θ1)

Io cancels out and you have:

0.1 = cos^2(θ1) * cos^2(90-θ1) = (cos(θ1) * cos(90-θ1))^2

Had to look up how to combine cosine functions for this next part:

2*Sqrt(0.1) = cos(2*θ1- 90) + cos(90)

Solving for θ1

θ1 = (arccos(2*sqrt(0.1)) + 90) / 2

θ1 = 70.38° which is one of the correct answers.

My final question is this: The book also says that 20° is correct. How do they get this solution?
 
  • #4
haruspex
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My final question is this: The book also says that 20° is correct. How do they get this solution?
Whenever you execute an inverse trig function you need to be aware that there may be multiple valid answers. In this case, consider that cos x = cos -x.
 
  • #5
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Great, Thanks Haruspex.

So when I consider the following:

θ1 = (arccos(-2*sqrt(0.1)) + 90) / 2

I get θ1 = 109.62° , which is also 19.62° off the normal. That wraps it up!
 

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