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Dispersion through an equilateral prism

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data
    white light propagating in air is incident at 45 degrees on an equilateral prism Find the angular dispersion "gamma" of the outgoing beam if the prism has refractive indices n(red) = 1.582 and n(violet) = 1.633.
    NOTE: "gamma" is just a variable for describing the angular dispersion.

    2. Relevant equations
    Snell's law n1sin(theta)=n2sin(theta)
    adjusted for our purposes I came up with:

    For incoming wight light:

    n(air)sin(45) = n(red)*sin(theta(r1)) and n(air)sin(45) = n(violet)sin(theta(v))

    solving these equations we can find our angles and use those to determine our outgoing theta for each wave:

    n(red)sin(theta(r1)) = n(air)sin(theta(r2)) and


    3. The attempt at a solution

    Well this seems like a simple plug and chuck, but the approach seems like it may be oversimplified, I think there may be some geometry somewhere that needs to be added, (i.e. should I add 45 degrees to the theta(r1) and use that degree value for the "theta(r1)" for the outgoing red beam angle?

    Assuming I find the output angles then all I would need to do is take the difference of those angles to determine the angular dispersion "gamma"?

    The given text available provides about half a paragraph talking about angular dispersion for the entire chapter on reflection and refraction, so if you guy's happen to know of any useful links I could refer to for FYI purposes, that would be appreciated as well, but in the mean time am I doing this correctly?

    When I solved the numbers I basically got output angles of red and violate that were suspiciously close making the difference near zero... the only thing that seems correct is that my angle for red is above the angle for violate which is expected...
  2. jcsd
  3. Nov 16, 2008 #2

    Doc Al

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    Staff: Mentor


    Instead of theta(r1) & theta(v1), which are angles with respect to the first surface normal, you need angles with respect to the second surface normal.

    Yes, you'll need a bit of geometry to find the incident angles at the second surface. But why 45 degrees?
  4. Nov 16, 2008 #3
    Right, that's the ambiguity that needed addressing, we'll in that case because this is a equilateral triangle the incident angle of the light on the interior of the prism should be equal the incoming angle being subtracted from the axis normal to the prism surface.

    The reference of 45 degrees being added to the initial angle was purely arbitrary for example purposes.
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