Natural Light and Linear Polarizer

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Discussion Overview

The discussion revolves around the behavior of natural light passing through a series of linear polarizers set at varying angles, specifically examining the resulting intensity and the implications of using multiple polarizers. Participants explore the theoretical outcomes of this setup, including calculations related to intensity reduction and the effects of intermediate polarizers.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that with infinite polarizers from 0º to 90º, the final irradiance would be 0 due to the angle difference between the first and last polarizers.
  • Another participant counters that light would still pass through, referencing the effect of inserting a polarizer at 45 degrees between the two extremes.
  • A different viewpoint posits that if the polarizers are ideal, the final intensity could still be I0/√2, indicating a potential misunderstanding of the cumulative effects of multiple polarizers.
  • One participant discusses the mathematical relationship governing intensity reduction, stating that the intensity after each polarizer can be calculated using Ii+1=Ii*cos²(b), where b is the angle difference.
  • Another participant elaborates on the relationship between electric field reduction and intensity, noting that the intensity is proportional to the square of the electric field and providing a calculation for the final intensity based on the number of polarizers.
  • There is a correction regarding the nature of the calculation, with one participant asserting that it is a product rather than a sum, emphasizing the cumulative effect of multiple polarizers on the electric field and intensity.

Areas of Agreement / Disagreement

Participants express differing views on the final intensity of light after passing through multiple polarizers, with no consensus reached on whether the final irradiance would be zero or a non-zero value. The discussion remains unresolved regarding the exact outcome of the proposed setup.

Contextual Notes

Participants reference ideal polarizers and make assumptions about the nature of the incident light, but the discussion does not clarify the conditions under which these assumptions hold true. There are also unresolved mathematical steps regarding the cumulative effects of multiple polarizers.

Frank Einstein
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Hi, In the first place I want to apologize for my bad English. In the second, I have a concept doubt, If I make natural light to go on a linear polarizer of angle 0º, the intensity, or irradiance here, It would be I0/√2 and it's Jones vector would be (1, 0) (as column). What would happen If I put infinite polarizators which angles were 0º for the first and Pi/2 for the last, changing infinitesimal angles in between?

Would I have light which irradiance would be 0 at the end?

That's what I think that might happen, because the first and last polarizators have an angle of Pi/2 between them, so the irradiance would be 0.
 
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Nope, you'd actually get light through. You can see this by putting a polarizer at a 45 degree angle in between to polarizers set to 0 and 90 degrees. How much gets through I'm not certain.
 
Frank Einstein said:
Would I have light which irradiance would be 0 at the end?
I think that, if the polarizators are ideal, you would still have I0/√2 at the end.
 
Yes, that's right; if you use the fact that Ii+1=Ii*cos^2(b) where b is the differencie between the angles of polarization, adn is constant, you get an infinite sum which result is one.
 
The electric field gets reduced by cos(b). For one filter at 45 deg with respect to the previous one that is 1/sqrt(2), and for two steps 1/2.

The intensity is proportional to E^2, so if E gets reduced by 1/2, the intensity drops to 1/4. If you then assume that the original light was unpolarized the first filter already cuts out 1/2, and the final intensity is 1/8 of the unpolarized intensity.

Frank Einstein said:
Yes, that's right; if you use the fact that Ii+1=Ii*cos^2(b) where b is the differencie between the angles of polarization, adn is constant, you get an infinite sum which result is one.
It is not a sum, it is a product. Since all factors are the same the result is trivial.

If you have N steps, than at each step the E field gets reduced by cos(90 deg/N), i.e. (cos(90deg/N))^N in total for the E-field, and (cos(90deg/N)^(2N) for the intensity, with the additional factor 1/2 for the unpolarized incident light.
 
Drakkith said:
Nope, you'd actually get light through. You can see this by putting a polarizer at a 45 degree angle in between to polarizers set to 0 and 90 degrees. How much gets through I'm not certain.

The transmitted intensity is 25% of the incident intensity.
 

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