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Applying Snell's Law Inside & Outside a Prism

  1. Nov 14, 2016 #1
    1. The problem statement, all variables and given/known data
    I'm having trouble / uncertainties about the third missing angle (from left to right) in the picture. The first angle of reflection was easy as it's just 34 degrees aswell and I used snell's law rearranged to find the first angle of refraction as 13.36 degrees from the rearranged equation.

    I'm wondering whether I should use snell's law again minus the refractive index of 1, as the angle is inside the prism, which would give me 5.48 degrees reflected for the 3rd and 4th angles. I'm not sure whether there is an extension to the snell's law rearranging I should take into account? There must be a relevance for the 17 degrees I've been shown that I'm not including in my workings?

    2. Relevant equations
    n1sinΘ1=n2sinΘ2

    Θ2=sin^-1((1)sin(34) )
    (2.42)

    3. The attempt at a solution
    Sticking with 5.48 degrees and using it for the reflected angle also.
     

    Attached Files:

  2. jcsd
  3. Nov 14, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    Remember that light paths are reversible. So you could shine a ray back along the exit path and it would follow the same path back through the prism and exit where the original ray entered. You can treat the 17° angle like the angle of incidence for that "reverse" path.
     
  4. Nov 14, 2016 #3
    Thanks alot gneill I didn't know that - need to read more carefully in future!
     
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