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Two Prisms connected by a material of refractive index n

  1. Sep 28, 2015 #1
    1. The problem statement, all variables and given/known data
    The prisms have a refractive index ##n_1=1.3## at wavelength ##\lambda=300nm## and are connected by a material with refractive index ##n_2##. What is the refractive index ##n_2## so that light of wavelength ##>\lambda## follow path 2, and light at ##<\lambda## follow path 3?
    Untitled.png

    2. Relevant equations
    ##\sin\theta_c=\frac{n_2}{n_1}##
    ##n=\frac{c}{v}=\frac{c}{f\lambda}##

    3. The attempt at a solution

    I am very confused as to how the wavelength of light effects total internal reflection. I can solve for the frequency of the light by using ##n=\frac{c}{f\lambda}\Rightarrow f=\frac{c}{n\lambda}## but I'm not entirely sure this is useful. I also have that ##\sin\theta_c=\frac{n_2}{n_1}## which can be rewritten as ##\sin\theta_c=\frac{\lambda_1}{\lambda_2}## The critical angle is a constant in this case I think? it's 45 degrees since the light is always coming in parallel to the bottom of the prisms but I don't understand how varying ##\lambda## is going to change whether or not it exits the prisms.
     
  2. jcsd
  3. Sep 28, 2015 #2

    Daz

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    Gold Member

    The question only says that n1 is 1.3 at a particular wavelength: 300nm. Do you suppose it could be different at other wavelengths? How would it vary?
     
  4. Sep 28, 2015 #3

    mfb

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    2016 Award

    Staff: Mentor

    The refractive index depends on the wavelength, so the critical angle depends on it as well. Here, the transition between total reflection and transmission is supposed to be at a wavelength of 300 nm, so you can calculate the necessary refractive index at 300 nm to be directly at the transition.
     
  5. Sep 28, 2015 #4
    So the correct refractive index then is ##n_1=\frac{n_2}{\sin\theta_1}##, edit: I mean ##n_2=n_1\sin\theta_1## sorry for all the mistakes I'm on my phone
     
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