Vertically polarized light from a hene laser passes through a linear

AI Thread Summary
Vertically polarized light from a helium neon laser passing through a linear polarizer at 15° transmits approximately 93.3% of the light, with the transmitted light polarized at 15°. When a second polarizer is added at 45°, the light transmission is calculated to be about 69.98%, with the resultant polarization angle being 45° relative to the vertical axis. The discussion highlights the application of Malus's Law in determining light transmission through polarizers. Overall, the cumulative effect of the two polarizers results in a final polarization angle of 45°.
tizio
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Homework Statement



1. Vertically polarized light from a helium neon laser passes through a linear polarizer with its axis of polarization oriented 15° from the vertical axis. Assuming no absorption or reflection:
(a) What percentage of the light will be transmitted?
(b) What will be the polarization angle of the transmitted light?

2. In question #1, if a second polarizer is placed in the beam after the first polarizer with its axis oriented at 45° from the vertical axis:
(a) How much light is transmitted?
(b) What will be the polarization angle of the transmitted light?


Homework Equations


I believe that Malus's Law applies for Q1: I = Iocos^2 \thetai


The Attempt at a Solution


Q1(a+b): I = Io cos^2 (15°) = Io 0.9330 = 93.3% Io at 15°

Q2(a): Second polarizer uses same formula except that only 93.3% of incident light has gotten through resulting in following change:
I = .933 Io cos^2 (45°) = .933 (0.5) Io = 0.4665 Io at 30° (45-15)

I don't have a lot of confidence in the answer to Q2.
 
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What is the angle between the light and the polarizer for Q2? It is not 45° - this is the angle between polarizer orientation and vertical axis.

at 30° (45-15)
That is not the polarization direction after the polarizer.
 
If the light passes through the first polarizer at 15° and then passes through another polarizer at 45° to vertical, then is the effect cumulative? Would it then be 45° + 15° = 60° overall?
 
OK...I think I figured out some of this:
A2: (a) I think the formula for both polarizers is: I = Io cos^2(15°) Io cos^2 (30°) which ends up to be I = 0.6998 Io or 69.98% (rounded)
A2: (b) Still not quite sure how to figure out the polarization angle mathematically other than I'm pretty sure that the 45° polarizer results in quarter phase shift.
 
A2: (a) I think the formula for both polarizers is: I = Io cos^2(15°) Io cos^2 (30°) which ends up to be I = 0.6998 Io or 69.98% (rounded)
Right.
A2: (b) Still not quite sure how to figure out the polarization angle mathematically other than I'm pretty sure that the 45° polarizer results in quarter phase shift.
You do this way too complicated. It is a polarizer. What does a polarizer do?
 
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Thank you for your help!

A polarizer, as I understand it, blocks/absorbs some light waves and orients the remaining light waves linearly (on a plane, actually, that diverges from the y-axis by θ) as it passes through the "picket fence" openings of the polarizer. If you have 2 polarizers oriented at 90° though, doesn't it block all light (back to Malus's Law-cos 90° = 0)?

Does that mean, in this example, that the resultant light after passing through the two polarizers is oriented at 45°?
 
tizio said:
A polarizer, as I understand it, blocks/absorbs some light waves and orients the remaining light waves linearly (on a plane, actually, that diverges from the y-axis by θ) as it passes through the "picket fence" openings of the polarizer. If you have 2 polarizers oriented at 90° though, doesn't it block all light (back to Malus's Law-cos 90° = 0)?

Does that mean, in this example, that the resultant light after passing through the two polarizers is oriented at 45°?
Yes to both.
 
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