1. Not finding help here? Sign up for a free 30min 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!

Light and refraction in water

  1. Feb 15, 2005 #1

    Bri

    User Avatar

    Could someone tell me if I did this right?
    A narrow beam of sodium yellow light, with wavelength 589 nm in vacuum, is incident from air onto a smooth water surface at an angle of incidence of 35.0 degrees. Determine the angle of refraction and the wavelength of the light in water.
    I did...
    sin(35 degrees)=1.333*sin(theta)
    theta = 25.486 degrees

    589 nm = 1.333*lambda
    lambda = 441.86

    I'm just not sure I used the right equation/setup.

    And I can't figure out this problem...
    Unpolarized light in vacuum is incident onto a sheet of glass with index of refraction n. The reflected and refracted rays are perpendicular to each other. Find the angle of incidence.
    I figured Theta1 + Theta2 = 90 degrees
    and I tried to find a way to solve it using sin(Theta1) = n*sin(Theta2) but it's not working out.

    Thanks.
     
  2. jcsd
  3. Feb 15, 2005 #2
    The first part looks right to me I guess. Think about [tex]\Theta_2 = 90 - \Theta_1[/tex] and plug it into your formula. Remember that sin(90-x) = cos(x)
     
  4. Feb 15, 2005 #3

    Bri

    User Avatar

    I've tried it that way, I wasn't able to get anywhere with it.
     
  5. Feb 16, 2005 #4
    I don't really see a definitive answer for part 2, but since [tex]sin(\theta_1) = n\times cos(\theta_1)[/tex], [tex]\theta_1 = tan^{-1}(n)[/tex]
     
    Last edited by a moderator: Feb 16, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Light and refraction in water
Loading...