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Monochromatic Light and a Prism

  1. Apr 19, 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 = 47.6o with respect to the horizontal and enters the prism at a height h = 11 cm above the base. The beam leaves the prism to the air at a distance d = 27.6 cm along the base as shown.

    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?


    2. Relevant equations
    Snell's law n1*sin(θ1)=n2*sin(θ2)


    3. The attempt at a solution

    Okay, so I started by trying to find the angle (for angle two) for which all the light would be reflected. To find this, I know that the refracted angle will be 90°. Therefore, i think to find this angle it would just be sin^-1(n1/n2). (I used n1 to refer to the air=1, and n2 for the prism which i solved for earlier and got 1.5725). I believe if i take the complement of this angle i would get φ2. I don't really know how to go from here, or even if i did the above correctly.

    https://www.smartphysics.com/Content/Media/Images/EM/25/h25_prism.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 19, 2013 #2

    haruspex

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    That looks right for φ2. Now use some geometry. There's a triangle base d with its apex where the light enters. What's the angle in that apex?
     
  4. Apr 19, 2013 #3
    I don't know what apex your referring to...the other side of the phi 2 line at the top? thats just the same as phi 2

    Well, so i know that the angle that the light enters the prism is with respect to the normal line. So, I tried to figure out the inside (refracted) angle in the prism. If i move H over to wear the light enters i find that top angle is 60 degrees.
     
  5. Apr 19, 2013 #4

    haruspex

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    Say the light enters at point A, leaves at point B, and the 30 degree angle is point C. Then in the triangle ABC (of which A is the apex) you know two of the three angles.
     
  6. Apr 19, 2013 #5
    Okay well the angle at A is just 180-30-phi2
     
  7. Apr 19, 2013 #6

    haruspex

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    Right, and you've calculated φ2 already. How does the angle at A relate to φ1?
     
  8. Apr 19, 2013 #7
    I think it would be equal? Because of some geometry principal that i dont remember the name of..or i just made up

    edit:wait no let me think about more
     
  9. Apr 19, 2013 #8
    Well i thought it would be 180-99.48 (value i got for angle A) - 30 degrees..but that's not right
     
  10. Apr 19, 2013 #9

    haruspex

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    If the angle at A is 99.48 degrees, what angle is the light ray making to the normal (within the prism) at A?
     
  11. Apr 19, 2013 #10
    I believe it would be 99.48-90 = 9.48
     
  12. Apr 19, 2013 #11

    haruspex

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    Right. And what relationship does that have to the angle to the normal at which the light ray originally struck the prism?
     
  13. Apr 19, 2013 #12
    Snell's law n1*sin(θ1)=n2*sin(θ2).
    I do that and get about 15 degrees. Then how do i find the angle phi1 with respect to the horizontal?...I tried adding/subtracting from the original given value...(which yes doesn't make sense). Ahhhhh! I got it. I wasn't really thinking, i should have use the supplementary relationship between the normal 30 and then the deduced 60. So, then i take 60-15 and get the correct phi 1.
     
  14. Apr 19, 2013 #13
    Thank you very much, i really appreciate the help
     
  15. Apr 19, 2013 #14

    haruspex

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    You're welcome.
     
  16. Apr 28, 2013 #15
    Question

    I follow up to how you got 9.48. I have different numbers for my question. I understand snells Law, but I don't know how you got the angle to be about 15 degrees. Can you explain how you got to that number?
     
  17. Apr 29, 2013 #16
    Snell's law, using the correct index of refractions
     
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