Find % Light Through 3 Polarizers

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AI Thread Summary
The discussion revolves around calculating the percentage of light intensity that passes through three polarizers, where the first and third are perpendicular, and the second is at a 42° angle to the first. Initially, no light passes through the first and third polarizers alone, but introducing the second allows some light to pass. The relevant equation used is I = I(nought) * cos²(theta), which is applied to find the intensity after each polarizer. There is confusion about the angle calculations, particularly regarding the use of "90 - angle given." Ultimately, the participants clarify their understanding of the polarizer interactions and the correct application of the formulas. The discussion emphasizes careful application of previously learned concepts to new problems.
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Homework Statement



Polarizers 1 and 3 have their axes of polarization, indicated by the black solid lines, perpendicular to each other. If you try to shine light through only the combination of 1 and 3, you will find that none passes through. However, now we put in another polarizer (number 2 in the figure) between number 1 and number 3. This polarizer has an axes of polarization that has an angle of θ = 42° with respect to the polarization axes of polarizer 1. Surprisingly, now some light passes through the combination 1+2+3. What percentage of the initial (unpolarized) light intensity passes through?

Homework Equations


I=I(nought) * cos^2 (theta)

The Attempt at a Solution


1. 100/2 = 50
2. 50*cos^2(90-Angle given)= x
3. x*cos^2 ((90-angle given)-angle given)= Answer
 
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ScrubTier said:
2. 50*cos^2(90-Angle given)= x
Can you explain your thinking here?
 
We had had a similar problem but using only two polarizers so I was thinking after the first one the same ideas could be applied for the last two.
 
I don't understand why you take "90 - Angle given".
 
Would it just be angel given then?
 
Okay you pointing out how stupid I was being really helped, I got it! Thank you
 
ScrubTier said:
Okay you pointing out how stupid I was being really helped, I got it! Thank you
I would have said "distracted" :smile:

Be careful when using solutions from other problems that you are applying them correctly for the problem at hand.
 
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