Need help with question regarding polarization filters

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Homework Help Overview

The discussion revolves around a problem involving un-polarized light passing through two polarization filters with misaligned transmission axes. The original poster is attempting to determine the angle between the filters based on the percentage of light transmitted.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster uses a formula related to intensity and polarization to calculate the angle but arrives at a different result than expected. Some participants inquire about the effects of the first filter on the light and discuss the implications of the intensity reduction.

Discussion Status

Participants are engaging in clarifying the effects of the first filter on the light and how it relates to the calculations. There is acknowledgment of the original poster's confusion, and some guidance is provided regarding the nature of polarized light and intensity changes.

Contextual Notes

The original poster's calculations and assumptions about the filters and their alignment are under examination, with a focus on understanding the principles of polarization rather than resolving the numerical discrepancy.

ZHIHUI
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Moved from a technical forum, so homework template missing
Un-polarized light is incident upon two polarization filters and do not have their transmission axes aligned. If 18% of the light passes through this combination of filters, what is the angle between the transmission axes of the filters.

I got 64 degrees using I = Imax(cosΦ)^2 but the correct answer is 53 degrees.

Thank you very much!
 
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What happens when the light hits the first filter?
 
DrClaude said:
What happens when the light hits the first filter?
The light will be linearly polarized. If the first filter is ideal, intensity of the transmitted light will be exactly half that of the incident un-polarized light.
 
ZHIHUI said:
The light will be linearly polarized. If the first filter is ideal, intensity of the transmitted light will be exactly half that of the incident un-polarized light.
Correct. So how does that affect the terms in your equation?
 
DrClaude said:
Correct. So how does that affect the terms in your equation?
I got it!
Thank you very much!
Sorry for the trouble!
 
ZHIHUI said:
I got it!
Thank you very much!
Sorry for the trouble!
No trouble, that what PF is for!
 

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