Unpolarized Light Intensity problem

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SUMMARY

The discussion centers on the calculation of light intensity after passing through polarizers. An unpolarized light with an initial intensity of 15 W/m² is transmitted through a first polarizer at a 25-degree angle, resulting in an intensity of 7.5 W/m². The second polarizer, aligned horizontally, further reduces the intensity to 1.3 W/m², calculated using the formula I = I₀(cos θ)², where θ is the angle between the transmission axes of the polarizers.

PREREQUISITES
  • Understanding of unpolarized light and its intensity
  • Knowledge of polarizing materials and their transmission axes
  • Familiarity with the equations I = (1/2)I₀ and I = I₀(cos θ)²
  • Basic trigonometry for angle calculations
NEXT STEPS
  • Study the principles of light polarization and its applications
  • Learn about the effects of multiple polarizers on light intensity
  • Explore the derivation and applications of Malus's Law
  • Investigate the behavior of light in different mediums and angles
USEFUL FOR

Students studying optics, physics educators, and anyone interested in the principles of light behavior and polarization effects.

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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