1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Natural Light and Linear Polarizer

  1. Dec 2, 2014 #1
    Hi, In the first place I want to apologize for my bad English. In the second, I have a concept doubt, If I make natural light to go on a linear polarizer of angle 0º, the intensity, or irradiance here, It would be I0/√2 and it's Jones vector would be (1, 0) (as column). What would happen If I put infinite polarizators which angles were 0º for the first and Pi/2 for the last, changing infinitesimal angles in between?

    Would I have light which irradiance would be 0 at the end?

    That's what I think that might happen, because the first and last polarizators have an angle of Pi/2 between them, so the irradiance would be 0.
  2. jcsd
  3. Dec 3, 2014 #2


    User Avatar

    Staff: Mentor

    Nope, you'd actually get light through. You can see this by putting a polarizer at a 45 degree angle in between to polarizers set to 0 and 90 degrees. How much gets through I'm not certain.
  4. Dec 3, 2014 #3


    User Avatar
    Science Advisor

    I think that, if the polarizators are ideal, you would still have I0/√2 at the end.
  5. Dec 3, 2014 #4
    Yes, that's right; if you use the fact that Ii+1=Ii*cos^2(b) where b is the differencie between the angles of polarization, adn is constant, you get an infinite sum which result is one.
  6. Dec 3, 2014 #5
    The electric field gets reduced by cos(b). For one filter at 45 deg with respect to the previous one that is 1/sqrt(2), and for two steps 1/2.

    The intensity is proportional to E^2, so if E gets reduced by 1/2, the intensity drops to 1/4. If you then assume that the original light was unpolarized the first filter already cuts out 1/2, and the final intensity is 1/8 of the unpolarized intensity.

    It is not a sum, it is a product. Since all factors are the same the result is trivial.

    If you have N steps, than at each step the E field gets reduced by cos(90 deg/N), i.e. (cos(90deg/N))^N in total for the E-field, and (cos(90deg/N)^(2N) for the intensity, with the additional factor 1/2 for the unpolarized incident light.
  7. Dec 3, 2014 #6

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor

    The transmitted intensity is 25% of the incident intensity.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook