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Refraction and Snell's Law, A complex question

  1. Oct 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculating the refraction of index (n) with given information.

    1.png
    2. Relevant equations

    3. The attempt at a solution
    I tried to use Snell's Law but I have no idea about how i'm supposed to use it without angles. Instead, question gives distances. Need help very badly.
     
  2. jcsd
  3. Oct 6, 2015 #2
    You can get the necessary angles from the given distances. To find the angle θ in the picture, it is arctan(x2/(D/h)). Do you see why?
     
  4. Oct 6, 2015 #3
    Well, I'm not good at trigonometry so I didn't get where to find that Q angle. I'll be thankful if you show it with a drawing. Let's say I get Q with that formula. How is it going to help me about finding that refractive index?
     
  5. Oct 6, 2015 #4

    berkeman

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    Staff: Mentor

    What is the definition of the tangent of an angle? What sides of a triangle does it involve? You need to shoe more effort on your schoolwork questions here -- please do your best to figure this out on your own. :smile:
     
  6. Oct 6, 2015 #5
    Of course I know what does tangent of an angle means, it is a ratio between the opposite and adjacent side of a triangle containing that angle. That's not the part that i'm worried about. I need help about finding angle of refraction so i can apple Snell's Law. Be sure that i'm really paying attention to my homework.
     
  7. Oct 6, 2015 #6

    berkeman

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    Staff: Mentor

    Good. You should be able to figure out the internal angle given the distances that you are shown outside and the thickness of the glass. Can you make a sketch that marks up the figure above, showing how you would work your way back to that angle?

    It doesn't help you if we just tell you how to do it. You need to get used to figuring this kind of problem out on your own. Please give it a try and post your sketch. Thanks.
     
  8. Oct 6, 2015 #7
    Here is what I've got so far. DSC_4817.JPG
     
  9. Oct 6, 2015 #8

    berkeman

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    Good. And X2-X1 gives you how far the beam is translated from where it would have exited with no deflection, right? How can you use that to help you find θ2?
     
  10. Oct 7, 2015 #9
    I really don't know how that 1cm difference help me to find Q2. If i knew that, I would solve it, right? I told before I'm not good at Geometry and Trigonometry and i guess that's why i can't solve this question.
     
  11. Oct 7, 2015 #10

    berkeman

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    Do you see how tan(θ2) is formed? It is related to the thickness of the glass, and the path of the beam in the glass. So you need to figure out that opposite side of the triangle that forms tan(θ2). To figure that out, you have to use the displacement of the beam, and something else...
     
  12. Oct 7, 2015 #11
    Yes, i saw how tan(Q2) formed and i know that adjacent side which is thickness of the glass. If i knew the opposite side it would be easy to find with Arctan but it's not easy for me find it. By the way you're always giving clues and i'm appreciated with your interest but it's not helping. Because it's about something i was never successful about and i will never will be.
     
  13. Oct 7, 2015 #12

    berkeman

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    Sure you can and will be successful. You just need practice, which this problem is giving you. Just keep looking at the triangles formed in the lower right of the problem, and see how you can work out that opposite side of the triangle. What do you need to get it? How can you find that length?

    A very important theme here at the PF is helping students learn how to learn. Telling you the answer is not helping you learn how to learn. :smile:
     
  14. Oct 7, 2015 #13
    I may be found a way but i'm not sure if it's correct or not.

    -We said tan(Q1)=13/11 in biggest triangle right?
    -So in the biggest triangle in the glass, Q2+alfa=Q1, so can I use tan(49) is also equals to (a+1)/1 ?
    DSC_4821.JPG
     

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