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Index of refraction and electromagnetic radiation

  1. Aug 1, 2006 #1
    "An electromagnetic radiation has a frequency of 5 x 10^14Hz

    a) (wave in vacuum = 600nm)
    b) (wave in water = 440 nm)
    c) what is the index of refraction of a medium in which the speed of this radiation is 2.54 x 10^8 m/s?" :yuck:

    so given: c = 2.54 x 10^8 m/s
    f = 5 x 10^14Hz

    required: ni

    im not sure where to start on this one. any help will be appreciated.

  2. jcsd
  3. Aug 1, 2006 #2
    nevermind, i figured it out! :biggrin:

    n = C/v

    i dunno how to delete my original post.

  4. Aug 1, 2006 #3
    I can help with part c, the index of refraction is a ratio between the speed of light in a vacuum to the speed of light in the medium, so if we call the index of refraction n, then

    n = (speed of light in vacuum)/(speed of light in medium)

    For parts a and b, I think you can do a similar thing with a ratio of wavelengths, but I'm not sure if that's right or not so you may want to wait for someone to give you a more concrete answer about that.
  5. Aug 2, 2006 #4
    thanks dLeet!

    im working on another problem if anyone can share some insight..

    Someone is on their boat and their eye level is 1 m above the water. and they guess the apparent depth of an object is 2 m below the water surface. the angle of which the person is looking at the object is 45 degrees.

    so ni = 1.00029 (air)
    nR = 1.33 (water)

    so is this just a matter of (2m x 1.33) + 1m = 3.66m actual depth? :confused:

  6. Aug 2, 2006 #5
    ok.. did some reading and looks like i was wrong (surprise, surprise).

    so i go:
    sin 0i = (ni/nR) x sin 45 degrees
    sin 0i = (1.00029/1.33) x sin 45 degrees
    sin 0i = 0.5318
    0i = 32.1 degrees

    Dactual = 2.0/tan 32.1 degrees
    = 3.18 meters


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