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Find index of refraction of the prism for 2 wavelengths

  1. Jul 25, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    2. Relevant equations
    Snell's Law: na*sin(theta of a) = nb*sin(theta of b)
    wavelength = index of refraction / n
    n = c / v

    3. The attempt at a solution
    The only information given in the problem is the 2 angles of refraction and of course n of air is 1, I do not understand how to get the index of refraction for both wavelengths with just the angles given and nothing more. They do not state the wavelengths of the rays. I assume it is solvable because it is in the text but if someone could maybe lead me in any direction I would appreciate it.
    Thanks
     
  2. jcsd
  3. Jul 25, 2008 #2

    mgb_phys

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    It's not asking for their wavelengths, you just have to assume the wavelengths are different so that the refractive index can be different.
    It's simply a geometry question, find the angle from the normal for each of the exit rays and then just use Snell's law to calculate 'n' for each of them.
     
  4. Jul 25, 2008 #3
    Thats what I was thinking so you will have na*sin(theta of a) equals to both nb*sin(0b) and nc*sin(0c) (b and c the 2 refractions) so if u set the 2 equal to each other:
    nb*sin(20.5) = nc*sin(12)
    dont you need more info to get the exact values instead of just a ratio?
    thanks for the quick response
     
  5. Jul 25, 2008 #4

    mgb_phys

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    Assuming the prism is in air, anf consideringthe refraction at the exit face - isn't it just
    na sin(internal angle ) = 1 * sin(12+external angle)
    nb sin(internal angle ) = 1 * sin(12+8.5+external angle)

    And the angle from normal inside the prism is just the angle from normal to the horizontal dashed line outside ( and is equal to the prism wedge angle)
    na sin(x) = sin (12 + x)
    nb sin(x) = sin(20.5 + x)
    It should be possible to rearrange this to get exact values of na and nb without knowingx
     
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