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Polarization of Light: Polarizer

  • Thread starter Beth N
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  • #1
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Homework Statement



When unpolarized light passes through a polarizer, what happens?
A.The light emerges polarized in the polarizer axis direction with about 12the intensity of the incident beam.
B.The light emerges polarized perpendicular to the polarizer axis direction with about 12the intensity of the incident beam.
C.The light emerges polarized in the polarizer axis direction with about 14the intensity of the incident beam.
D.The light emerges polarized perpendicular to the polarizer axis direction with about 14the intensity of the incident beam.E.The light emerges unpolarized.
Screen Shot 2018-10-27 at 8.23.46 PM.png

Homework Equations


##Itransmitted=Iincident(cosθ)2##

The Attempt at a Solution


[/B]
The answer is A.
Aligned conducting chains of molecules absorbs and blocks that component of the electric field.
The light emerges polarized in the polarizer axis direction with about 1/ 2
the intensity of the incident beam.
The intensity of light is proportional to the square of the electric field.
Screen Shot 2018-10-27 at 8.23.41 PM.png


I don't understand why it is half as much though. Where do you get that number 1/2 from? Wouldn't the answer depends on the angle (yet we are not given the angle here). Imagine the unpolarized light go in all direction... now there is only 1 direction passing through the polarizer. Why would it not be less than 1/2 passing through?
 

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Answers and Replies

  • #2
TSny
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##I_{\rm transmitted} = I_{\rm incident} \cos^2(\theta)##.
Note that in this formula, ##\theta## does not represent the angle labeled as ##\theta## in the picture.
What is the meaning of ##\theta## in the formula ##I_{\rm transmitted} = I_{\rm incident} \cos^2(\theta)##?
 
  • #3
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4
##I_{\rm transmitted} = I_{\rm incident} \cos^2(\theta)##.
Note that in this formula, ##\theta## does not represent the angle labeled as ##\theta## in the picture. What is the meaning of ##\theta## in the formula ##I_{\rm transmitted} = I_{\rm incident} \cos^2(\theta)##?
In the formula, ##\theta## is the angle between the polarizer axis and the electric field? I think this is where I'm confused too. How do we know which direction the electric field wave goes?

That is a good point though, I totally thought that the angle is that one in the picture.
 
  • #4
TSny
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In the formula, ##\theta## is the angle between the polarizer axis and the electric field?
Yes.
I think this is where I'm confused too. How do we know which direction the electric field wave goes?
In unpolarized light, the electric field fluctuates rapidly over all directions in the plane perpendicular to the direction of travel. So, the transmitted intensity is found by averaging ##\cos ^2 \theta## over all angles between 0 and ##2 \pi## radians. So, you have the math problem of finding the average value of ##\cos ^2 \theta##.
 
  • #5
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Yes.
In unpolarized light, the electric field fluctuates rapidly over all directions in the plane perpendicular to the direction of travel. So, the transmitted intensity is found by averaging ##\cos ^2 \theta## over all angles between 0 and ##2 \pi## radians. So, you have the math problem of finding the average value of ##\cos ^2 \theta##.
Oh, so ##\theta## would always be equal to 45 degree (and therefore ##cos^2 {\theta} =1/2 ## regardless of where the polarization axis is oriented?
So that means if we have a second polarizer, that angle is gonna be important because we no longer taking the average and the intensity that passes through could go to 0 if the angle is 90 degree?
 
  • #6
TSny
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Oh, so ##\theta## would always be equal to 45 degree (and therefore ##cos^2 {\theta} =1/2 ## regardless of where the polarization axis is oriented?
For unpolarized light, ##\theta## would not always equal 45 degrees. ##\theta## changes rapidly and randomly over all possible values. At one instant of time, ##\theta## might be 12o while at another instant of time it might be 57o. These changes in ##\theta## for unpolarized light are taking place extremely rapidly. So, we take the intensity of the transmitted light to correspond to the time average of these fluctuations. This amounts to taking the average value of ##\cos^2\theta## over all possible values of ##\theta##. There are ways to show that the average value of ##\cos^2 \theta## is 1/2.
So that means if we have a second polarizer, that angle is gonna be important because we no longer taking the average and the intensity that passes through could go to 0 if the angle is 90 degree?
Yes.
 
  • #7
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Thank you so much that resolves a big misconception I have!
 
  • #8
TSny
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Glad I could help. Hope you enjoy your study of physics.
 

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