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!

Calculate the desired incident polarization of a light beam

  1. Oct 29, 2015 #1
    Hi I want to calculate the necessary incident polarization of a light beam at a given angle of incidence (theta_i) that reflects off BK7 glass (n = 1.5168) and is linearly polarized (i.e., 45 degrees). I know how to do similar calculations for incident natural unpolarized light, but not in the case of incoming polarized light.

    Cheers
     
  2. jcsd
  3. Oct 29, 2015 #2

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    What type of polarization is it, linear?
     
  4. Oct 29, 2015 #3
    The reflected beam comes out linearly polarized (50/50 s/p). I'm trying to find the incoming beam's polarization (assume it is not circularly or ellipitically polarized).
     
  5. Oct 29, 2015 #4

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Forget my previous comment, if the reflected light is linearly polarized then so is the incoming one for the case of external reflection (##n_2 > n_1##).
    Anyway, you have Fresnel equations for TE and TM components:
    $$
    \frac{E_r^{TE}}{E_i^{TE}} = \frac{n_1\cos \theta_1 - n_2\cos \theta_2}{n_1\cos \theta_1 + n_2\cos \theta_2}
    $$
    $$
    \frac{E_r^{TM}}{E_i^{TM}} = \frac{n_2\cos \theta_1 - n_1\cos \theta_2}{n_2\cos \theta_1 + n_1\cos \theta_2}
    $$
    Now it's required that ##E_r^{TE}=E_r^{TM}## and that ##\theta_1## is given, from which ##\theta_2## will follow from Snell's law. So, isn't it straightforward to get the ratio of the components of the incoming light?
     
  6. Oct 29, 2015 #5
    This is completely correct. Those can be called the reflection coefficients of the s and p polarized components. I believe the correct way of determining the polarization of the input is to find the degree of polarization:

    V = Ip/(Ip+In)
    And I believe, although am not sure, that Ip = Rs + Rp and In = 1/2(Rp + Rn).

    Hence given the specs n = 1.5154 @ 650 nm, and the incidence/reflectance angle = 50 degrees, I compute the degree of polarization to be ~66.67%.

    I don't know if this is right though.
     
  7. Oct 29, 2015 #6
    Correction, n =1.5145
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Calculate the desired incident polarization of a light beam
  1. Beams of light (Replies: 4)

  2. Polarization of light (Replies: 17)

  3. Polarization of light (Replies: 4)

Loading...