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Light and refraction in water

  1. Feb 15, 2005 #1


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    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.

  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


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    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
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