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Snell's Law Application Homework Problem

  1. Apr 28, 2014 #1
    1. The problem statement, all variables and given/known data

    snellslaw_zps1120239d.png

    I apologize for not utilizing built-in Latex, I'll have to work with that a little more.


    Question statement: Explain the paths, both transmitted and reflected, that a light ray of 400nm would take if it was incident perpendicular to the bottom face ofo the crystal. If the light was in the shape below (just an arrow shown in the positive Y-direction of an X-Y coordinate plane), what would the relative orientation be when it leaves the crystal.

    2. Relevant equations

    (sin(θ1))*(n1) = (sin(θ2))*(n2) = λ1sin(θ2) = λ2sin(θ1)


    3. The attempt at a solution

    To be honest, I'm trying to figure out where to start with this one. I've used the provided equation and calculated n to be 1.682725, however I don't know if that corresponds to the incident or refracted side of the light, and if the light is coming in through the bottom face at the center, I'm not sure if I should consider the transmission through the assumed air medium before it hits the crystal. I could very well be overthinking it, but I'd love some direction.

    Additionally, I understand that Snell's Law has the ability to predict how these light rays will trace through mediums based on wanting to take the path that's quickest through the medium. I don't know if I have to consider light that's lost at the top surface though--will some light exit the crystal?
     
  2. jcsd
  3. Apr 28, 2014 #2

    SammyS

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    You should not have that middle equal sign.

    (sin(θ1))*(n1) = (sin(θ2))*(n2)

    λ1sin(θ2) = λ2sin(θ1)

    Start with the incident ray approaching the crystal from below.

    A fraction of the intensity will be reflected at the surface. This is a reflected ray.

    A fraction will pass into the crystal, its direction determined by Snell's Law. This is a refracted ray.
    Whenever the ray strikes a surface, it will be partially reflected and partially refracted -- unless Snell's Law says the refracted beam is impossible -- this may well be a part of this exercise.

    I would assume that the crystal is surrounded by air, the index of refraction of which is very close to 1.

    In applying Snell's Law: Any ray approaching a surface is the incident ray. Any ray passing into the neighboring medium is the refracted ray.

    Use ncrystal = 1.682725 and nair = 1 appropriately .
    I doubt that you are expected to use Fermat's Principle, the principle of least time.
     

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