Vertically polarized light from a hene laser passes through a linear

[tizio]

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 = I_ocos^2 [itex]\theta[/itex]_i

The Attempt at a Solution

Q1(a+b): I = I_o cos^2 (15°) = I_o 0.9330 = 93.3% I_o 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 I_o cos^2 (45°) = .933 (0.5) I_o = 0.4665 I_o at 30° (45-15)

I don't have a lot of confidence in the answer to Q2.

 

[mfb]

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.

 

[tizio]

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?

 

[tizio]

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.

 

[mfb]

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?

 

[tizio]

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°?

 

[mfb]

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.