Physics Help Forum Vertically polarized light from a HeNe laser passes through a linear polarizer

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 Apr 13th 2014, 07:19 AM #1 Junior Member   Join Date: Apr 2014 Posts: 3 Vertically polarized light from a HeNe laser passes through a linear polarizer Q1. 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? A1: (a) I am reasonably certain that Malus's Law applies here: I = I0 cos^2 θi I = I0 cos^2 (15) = Io 0.9330 = 93.3% Io A1: (b) 15° Q2. 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? A2: (a) Here is where I get messed up....Is the effect cumulative on the resulting incident light? In other words, the light passes through the first polarizer at 15° and then passes through a second polarizer oriented at 45° to the original vertical, do I add the two together (15 + 45) and use Malus's Law again for cos^2 (60)? What am I missing?
 Apr 13th 2014, 09:35 AM #2 Junior Member   Join Date: Apr 2014 Posts: 3 I think I figured some of this out. 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.
 Apr 15th 2014, 09:47 AM #3 Senior Member     Join Date: Apr 2008 Location: Bedford, England Posts: 668 I am not familiar with Malus's Law, but I think you have a flaw in the maths of A2: The answer of A1 surely becomes the starting point for A2, thus: I = {Io cos^2(15°)} cos^2 (30°)
 Apr 16th 2014, 02:46 PM #4 Junior Member   Join Date: Apr 2014 Posts: 3 Yes....thank you! One Io is certainly sufficient; better than Io^2 anyway. I = Io {cos^2(15°) cos^2 (30°)}

 Tags hene, laser, light, linear, passes, polarized, polarizer, vertically

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