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Index of refraction of the prism

  1. Aug 19, 2009 #1
    1. The problem statement, all variables and given/known data
    5) A beam of monochromatic light enters an equilateral triangular prism at an angle of incidence of 30.0º. If the index of refraction of the prism is 1.53, at what angle with respect to the surface of the prism does the beam emerge from the other side of the prism?


    2. Relevant equations



    3. The attempt at a solution
    i tried drawing it out, but i have absolutely no idea.
     
  2. jcsd
  3. Aug 19, 2009 #2

    berkeman

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    Re: angles

    Show us a drawing and your work. A "relevant equation" might be _____'s Law, eh?
     
  4. Aug 19, 2009 #3

    jgens

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    Re: angles

    Use Snell's Law: [itex]n_1\sin{\theta_1} = n_2\sin{\theta_2}[/itex]
     
  5. Aug 19, 2009 #4
    Re: angles

    i dont know how to draw on here. :( and Snells law right? i cannot figure out how to do that. if Sin(feta)1/Sin(feta)2 = N2/N1, i only have two of the variables! dont i at least need three?
     
  6. Aug 19, 2009 #5

    berkeman

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    Re: angles

    What are your unknowns?
     
  7. Aug 19, 2009 #6

    jgens

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    Re: angles

    Assume that the ray of light is initially in air. The index of refraction in air is approximately one. Also, the symbol [itex]\theta[/itex] is called theta, not feta.
     
  8. Aug 19, 2009 #7
    Re: angles

    we dont know Sin(feta) 1 or N1. we only know the angle of refraction and the index of refraction.
     
  9. Aug 19, 2009 #8
    Re: angles

    so sinθ/Sin30=1.53/1
    =1.53sin(30)??
     
  10. Aug 20, 2009 #9
    Re: angles

    ok let me try this again,

    2. Relevant equations
    N1sinθ=N2sinθ


    3. The attempt at a solution
    The light will enter the prism and refract, cross the prism, hit the other side, exit and refract. so there will be lots of angles for me to find, i jsut dont know how to solve for them except for this one.

    N1sinθ=N2sinθ
    Θ2=sin-1((1.0003)(sin30)/1.53)
    Θ= 19 degrees
     
  11. Aug 20, 2009 #10

    drizzle

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    Re: angles

    hey {smile}

    you need to sketch so you can understand it well, you need to do it twice [Snell's Law] so you can get the angle of the emerged beam with respect to the surface of the prism [it’s theta 3], try to work out the other theta illustrated in the figure.


    http://img197.imageshack.us/img197/1777/13527457.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  12. Aug 20, 2009 #11

    kuruman

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    Re: angles

    Note that, in the above drawing, θ1 and θ3 are defined in such a way that Snell's Law should be written in terms of their cosines on the "air side" of the ray.
     
  13. Aug 20, 2009 #12

    drizzle

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    Re: angles


    you’re right [we’ll get the same answers], but according to Snell's law, these angles are measured with respect to the normal lines which is perpendicular to the boundaries [that’s why I drew them, but then forgot about them:tongue:], anyhow, here’s the exact one;

    http://img30.imageshack.us/img30/1359/10703724.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  14. Aug 20, 2009 #13

    kuruman

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    Re: angles

    Great. Now see if you can find what the sum of the two internal angles is

    θ2 + θ = ?

    This will help you connect the first angle of refraction θ2 to the internal angle of incidence θ.
     
  15. Aug 20, 2009 #14

    kuruman

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    Re: angles

    Theta, feta what's the difference - it's all Greek anyway. :rofl:
     
  16. Aug 20, 2009 #15

    drizzle

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    Re: angles

    was that directed to me? :biggrin:

    I'm only helping the OPer
     
  17. Aug 20, 2009 #16

    kuruman

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    Re: angles

    It was directed to {smile} who I believe is trying to solve this.
     
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