Unpolarized Light Intensity problem

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Homework Help Overview

The discussion revolves around a problem involving unpolarized light and its interaction with polarizing materials. The original poster presents a scenario where unpolarized light with a given intensity passes through two polarizers, with specific angles affecting the transmitted intensity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of intensity after passing through the first polarizer and question the setup for the second polarizer. There is an exploration of the angle adjustments needed for the calculations.

Discussion Status

Some participants have provided calculations for the intensity after the first polarizer and have confirmed the arithmetic. There is ongoing dialogue about the setup for the second polarizer, with some guidance offered regarding angle adjustments.

Contextual Notes

Participants are working within the constraints of homework rules, which may limit the amount of direct assistance they can provide. There is a focus on ensuring the correct application of formulas related to intensity and angles.

hardwork
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Homework Statement


Unpolarized light has intensity 15W/m^2 and is incident on a sheet of polarizing material whose transmission axis makes an angle of 25 degrees with the vertical. A second polarizer, whose transmission axis is horizontal, is located just past the first.

(a) What is the intensity of the light after it is transmitted through the first polarizer?
(b) What is the intensity of the light after it is transmitted through the second polarizer?

Homework Equations


I = (1/2)Io
I = Io(cos theta)^2

The Attempt at a Solution


a) I = (1/2)(15W/m^2) = 7.5W/m^2 -- haha sorry. I had it set up differently before, and put in the wrong answer.

b) How would I set this up? Would the angle be 90-25?
 
Last edited:
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hardwork said:
a) I = (1/2)(15W/m^2) = 30W/m^2
Check that arithmetic. :rolleyes:

b) How would I set this up? Would the angle be 90-25?
Yes. So by what additional factor would the intensity be reduced?
 
b) I = (7.5W/m^2)(cos 65)^2 = 1.3 W/m^2

Thank you for your help today!
 
hardwork said:
b) I = (7.5W/m^2)(cos 65)^2 = 1.3 W/m^2
Looks good. (And you're welcome.)
 

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