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Total Internal Reflection

  1. Nov 20, 2013 #1
    1. The problem statement, all variables and given/known data


    Red light is incident in air on a 30o - 60o - 90 o prism as shown. The incident beam is directed at an angle of φ1 = 37.6o with respect to the horizontal and enters the prism at a height h = 26 cm above the base. The beam leaves the prism to the air at a distance d = 73.1 cm.

    What is φ1,max, the maximum value of φ1 for which the incident beam experiences total internal reflection at the horizontal face of the prism?

    http://imgur.com/hJ5oid3
    2. Relevant equations
    Snell's Law : n1sin(φ1) = n2sin(φ2)
    geometric relations of a 30-60-90 triangle


    3. The attempt at a solution

    I used what I got for my n of the triangle, which is correct, at 1.289, and got a critical angle using the fact that for total internal reflection, the angle must be 90 so i got φ = sin^-1(1/1.289) which equals 50.877 degrees. My problem lies within the geometry of the situation, I've tried adding/subtracting all combinations of 90, 60, and 30. I can't seem to get what the φ1, max would be though.
     
  2. jcsd
  3. Nov 20, 2013 #2
    Does that help?
     

    Attached Files:

  4. Nov 21, 2013 #3
    Sort of... I assume the 2 and three have nothing to do with the φ2 and φ3? I don't get exactly what to get with them. Like i understand 3 would be 50.877 degrees right? And that 2 is the actual φ1 I need, but I dont get how to find the angle underneath 2 because if i get that one can't i just solve for 2?
     
  5. Nov 21, 2013 #4
    3) is what is termed the incident angle, the angle of the incident ray with the normal at the interface. 1) is the angle of the slope of the interface, a given. 2) is the angle of the incident angle with the horizontal. The three angles are related, so if you know any two of them, you can easily find out the third.

    Is this where you are stuck?
     
  6. Nov 21, 2013 #5
    Yeah thats exactly where I'm not sure how to work them. I only know one of the three, the incident ray. RIght? Because when i use the critical angle calculation thats the angle im getting.
     
  7. Nov 21, 2013 #6
    You always know 1). You know 2) or 3) depending on which part of the problem you are on. Now, 1) + 2) + 3) = ? Look at the diagram.
     
  8. Nov 21, 2013 #7
    OHHHH, so the first angle is 30, and if I know the angle of incident, then i can just do 90 - 3) - 30 = 2) right? If I do that with the angle of incident i got (50.877 degrees), then I still don't get the right answer.
     
  9. Nov 21, 2013 #8
    But that is something else. You have not shown how you got 50.877 degrees, so I cannot comment on that.
     
  10. Nov 21, 2013 #9
    Ok I'm using Snell's Law, n1 = 1 (its in air) and n2 = 1.289, if total internal reflection is required, then you can solve for the incident angle by φ = sin^-1(n1/n2). When you plug in n1 and n2 you get 50.877 degrees.
     
  11. Nov 21, 2013 #10
    The problem wants total internal reflection at the horizontal face of the prism. So ##n_1## must be that of glass. And the angle you will obtain will be that with the normal to the horizontal face, i.e., with the vertical. So you need quite a bit more effort to get the answer.
     
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