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Homework Help: Deviation of wavelengths through barium crown pism

  1. Sep 25, 2010 #1
    1. The problem statement, all variables and given/known data
    The refractive index of an equilateral prism of dense barium crown glass varies with wavelength as given in the table

    wavelength (nm)....n


    making use of Cauchy's dispersion formula determine the minimum angle of deviation for sodium light with wavelength = 589.3 nm

    2. Relevant equations

    A/2 = theta prime

    nsin(theta prime) = sin((A + deviation)/2)

    (weird D)/(deviation) = (nf - nc)/(nd-1)

    3. The attempt at a solution

    2sin^-1((1.635)sin(30)) - 60 = deviation = 49.7 degrees

    2sin^-1((1.646)sin(30)) - 60 = deviation = 50.8 degrees

    weird D = 50.8 - 49.7 = 11.1 degrees

    My problem is that i'm not sure how to find nd to finish the question.

    i know i can use

    n of wavelength = A + B/(wavelength)^2 + C/(wavelength)^4

    but that doesn't leave me with an n between 1.635 and 1.646 and that is my problem
  2. jcsd
  3. Sep 25, 2010 #2


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    You can omit the third term C/(lambda)^4. Find A and B for the two refractive index - wavelength pairs.

  4. Sep 25, 2010 #3

    well i'm not totally sure on this but does this look right, or am i assuming the wrong thing

    1.646 = 1.635 + b/((486.3nm)^2)

    b = 5.9122x10^-15


    n = 1.635 + (5.9122x10^-15)/((589.3x10^-9)^2) = 1.652 , but that is greater than 1.646 which is probably wrong then... I really don't know how to solve for A and B. I looked up their values for barium crown glass and it still doesn't work even with those
  5. Sep 25, 2010 #4


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    No, A is not equal to 1.635, why should it?

    1.635=A + B/656.3^2


    1.646 = A + B/486.1^2.

    Two equations, two unknowns. Solve.

  6. Sep 25, 2010 #5

    :facepalm: why didn't i think of that... homework is 3x harder when you're sick
  7. Sep 25, 2010 #6
    Okay i did that and i got a really high number, it seems unreasonable for an angle of deviation, but my text book examples show ratios of 1/29 1/40... i got for the deviation

    643.8 degrees
  8. Sep 25, 2010 #7


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    Show what did you get for A and B.

  9. Sep 25, 2010 #8
    n1 = A + B/lambda1^2 represent A as a function of B A = n1 - B/lambda1^2
    plug this in for A in the second one

    n2 = A + b/lambda2^2 then n2 = (n1 - B/lambda1^2) + B/lambda2^2

    n2-n1 = B(1/lambda2^2 - 1/lambda1^2) get B = (5.758x10^-15)

    Then plug B in to either equation solve for A to get A = (1.6216)


    n3 = 1.6216 + (5.758x10^-15)/(589.3x10^-9)^2 = 1.638

    then plug the rest of the numbers in and solve for deviation in

    n3 = nd n1 = nc n2 = nf

    (weird D)/(deviation) = (nf-nc)/(nd-1)

    deviation = ((weird D)(nd -1))/(nf-nc) = 643.8
  10. Sep 26, 2010 #9


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    I do not understand what you did. The question was:

    "making use of Cauchy's dispersion formula determine the minimum angle of deviation for sodium light with wavelength = 589.3 nm". You calculated the minimum deviation for the other wavelengths with the equation

    2sin^-1(n sin(30)) - 60 = deviation

    why do not do the same for 589.3 nm? It is about 50 degrees.

  11. Sep 27, 2010 #10
    i'll try that, it seems to make more sense, yet i did the lab on this today and they did it the way i initially was using. But that was a slightly different process.
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