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Snells Law and the Refraction of Light

  1. May 30, 2006 #1
    The drawing shows a rectangular block of glass (n = 1.52) surrounded by liquid carbon disulfide (n = 1.63). A ray of light is incident on the glass at point A with a theta = 23.0° angle of incidence. At what angle of refraction does the ray leave the glass at point B?


    I know I have to use snells law n1theta1=n2theta2. I also know I have to use it twice, once for entering the glass and then again for when the light ray exits the glass. However, as the drawing shows, the ray that exits will not be parellel to the original ray, so I know I need to do some geometry to figure out the angles, but I have no idea where to start with the geometry. Please Help. Thanks.
  2. jcsd
  3. May 30, 2006 #2

    Doc Al

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    Staff: Mentor

    Draw the refracted ray going from side A to side B. Consider that refracted ray to be the hypotenuse of a right triangle whose sides are the normals to sides A and B. Use that triangle to relate the angle of refraction at side A to the angle of incidence at side B.
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