How Does Changing the Angle of a Third Polarizer Affect Light Intensity?

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SUMMARY

The discussion centers on calculating the intensity of light passing through three Polaroid filters, specifically when the third filter is set at an angle of 50 degrees. The initial intensity of the helium-neon laser beam is denoted as I₀. The intensity at point B was calculated to be 0.375 I₀ using the formula I = I₀/2 cos²(30°). The final intensity at point C was incorrectly estimated using 0.375 I₀ cos²(50°), leading to confusion regarding the correct application of the polarization angle. The correct answer provided by the teacher is 0.331 I₀, indicating the necessity of considering the angular difference between the polarization angle at point B and the angle of the third polarizer.

PREREQUISITES
  • Understanding of light polarization and the behavior of unpolarized light through polarizers.
  • Familiarity with the equations I = I₀ cos²(x) and I = I₀/2 for unpolarized light.
  • Knowledge of angular measurements in degrees.
  • Basic skills in trigonometry to apply cosine functions.
NEXT STEPS
  • Study the principles of light polarization and how multiple polarizers interact.
  • Learn about the mathematical derivation of intensity calculations through polarizers.
  • Explore the implications of angular differences in polarization effects.
  • Investigate practical applications of polarizers in optical devices.
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Students studying optics, physics educators, and anyone interested in the practical applications of light polarization and intensity calculations in optical systems.

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Homework Statement


A helium-neon laser emits a beam of unpolarized light thatpasses through three Polaroid filters, as shown in the figure . The intensity of the laser beam is I_{o}.
Walker.25.72.jpg


Suppose the third filter were at an angle of 50˚, what would be the intensity at point C?

Homework Equations


I = Io cos^2(x)

I = Io/2 (Unpolarized light through transmission axis)

The Attempt at a Solution


I've tried a bunch of things...

First I calculated the intensity at point B, which I found to be .375 Io via (1/2)Io*cos^2(30˚)

Then I tried .375Io*cos^2(50˚)

and also tried replacing .375 with .375/2 and just 1/2

But I can't get my teacher's answer, which is .331 Io
 
Last edited:
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ObviousManiac said:
First I calculated the intensity at point B, which I found to be .375 Io via (1/2)Io*cos^2(30˚)
So far so good! :approve:
Then I tried .375Io*cos^2(50˚)
But the light just passed through polarizer 2 which was configured at 30o. That means not only the light is already polarized, but the light already has a polarization angle of 30o before it even gets to polarizer 3 (the one configured at 50o).

So what's the angular difference between polarizer 3's angle and the polarization angle of the light at B?
and also tried replacing .375 with .375/2 and just 1/2
Now you're just randomly guessing.
 
Last edited:

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