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Deriving Equation Relating Angle of Incidence to the Angle of Refracti

  1. Sep 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Derive an equation relating the angle of incidence and the angle of refraction as light travels from air to glass. Do it in the form of Θ2/Θ1 = expression


    2. Relevant equations
    I think the equation to use here is Snell's law. Right?
    n1sinΘ1 = n2sinΘ2

    n1 = nair = 1.0003 approximately 1
    n2 = nglass
    Θ1 = angle of incidence
    Θ2 = angle of refraction

    3. The attempt at a solution
    n1sinΘ1 = n2sinΘ2
    n1/n2 = (sinΘ2)/(sinΘ1)
    1/n2 = (sinΘ2)/(sinΘ1)

    However, I'm having problems on how to remove the sin on the right side of my equation. I know I can't just apply arcsin to both sides since that wouldn't remove the sin in both the numerator and the denominator of my fraction. Please help!
     
  2. jcsd
  3. Sep 7, 2014 #2

    ehild

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    Solve for Θ2, and divide it by Θ1.

    ehild
     
  4. Sep 7, 2014 #3

    Orodruin

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    For small angles sin x is approximately x (x in radians). For large angles, there really is no way to get a relation of the form you specified.
     
  5. Sep 7, 2014 #4
    I got Θ2/Θ1 = [arcsin((sinΘ1)/n2)]/Θ1

    However, is this really it? Is there no way I can arrive at an expression wherein Θ1 and Θ2 would not appear in the right hand side of the equation?

    I've tried doing it in a calculator. I know I'm working with angles to pi/2. However, as you said, sin x really gets deviated from x as x increases. But it's a good thing you pointed that out. Really simplifies things.

    I'm doing graphical analysis (angle of refraction(y) vs angle of incidence(x)) and I'm trying to figure out what the slope represents. I might use this as an explanation and just say that my slope represents the reciprocal of the refractive index of the glass.
     
  6. Sep 7, 2014 #5

    vela

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    Why don't you plot ##\sin\theta_1## vs. ##\sin\theta_2## instead?
     
  7. Sep 7, 2014 #6
    Yes I did that too, and I've got a linear graph. However, a part of the experiment was to analyze graphs of θ2 vs θ1 too. It's been giving me a hard time. I know that the graph of that one isn't perfectly linear but our instructor advised us to use linear regression on our data. Now it was asking us to know what the slope represents. I've been trying to derive an expression to figure out what it really is.
     
  8. Sep 7, 2014 #7

    ehild

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    ##\sin(θ_2)=\frac{n_1}{n_2}\sin(θ_1)##

    Take the derivative of both sides with respect to θ1. Solve for the slope dθ2/dθ1. If θ1 is small, the slope can be approximated as n1/n2 as the cosines are close to 1.
    I do not see any sense to worry about the slope of the θ2 vs θ1 graph. You can apply linear regression to the graph sinθ2 vs sinθ1.

    ehild
     
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