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BATMAN31

  • one year ago

I am doing this experiment, and have reached the point where you apply snell's law, and I don't get it. Link is in the comments.

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  1. BATMAN31
    • one year ago
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    http://www.education.com/science-fair/article/refraction-fast-light-travel-air/

  2. anonymous
    • one year ago
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    LINK

  3. BATMAN31
    • one year ago
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    number 9.

  4. BATMAN31
    • one year ago
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    how does this equation work? walk me through it

  5. anonymous
    • one year ago
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    you need the numbers that go in the equation ...

  6. BATMAN31
    • one year ago
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    27 and 17

  7. anonymous
    • one year ago
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    which is sin and which is V

  8. BATMAN31
    • one year ago
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    neither, from this experiment I do not know any values for 4.

  9. BATMAN31
    • one year ago
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    the WIP equation is \[(\sin 17)/(\sin 27)=v1/v2\]

  10. anonymous
    • one year ago
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    @Nnesha @iGreen

  11. BATMAN31
    • one year ago
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    I just watched the khan academy vid on snell's law, and it didn't clear much up. @directrix

  12. anonymous
    • one year ago
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    @amistre64

  13. anonymous
    • one year ago
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    yeah i dont understand

  14. BATMAN31
    • one year ago
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    @Michele_Laino

  15. Michele_Laino
    • one year ago
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    here is a little explanation: |dw:1433353958208:dw| the statement of the Snell's law is: \[\Large {n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}\]

  16. Michele_Laino
    • one year ago
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    since the light speed v in amedium with refractive index n, is: \[\Large v = \frac{c}{n}\] then we can divide both sides of the Snell's law by c, so we get: \[\Large \frac{{{n_1}}}{c}\sin {\theta _1} = \frac{{{n_2}}}{c}\sin {\theta _2}\]

  17. Michele_Laino
    • one year ago
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    now, according to the previous formula, we have: \[\Large {v_1} = \frac{c}{{{n_1}}},\quad {v_2} = \frac{c}{{{n_2}}}\]

  18. Michele_Laino
    • one year ago
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    then the statement of the Snell's law can be rewritten as follows: \[\Large \frac{{\sin {\theta _1}}}{{{v_1}}} = \frac{{\sin {\theta _2}}}{{{v_2}}}\] and finally: \[\Large \frac{{\sin {\theta _1}}}{{\sin {\theta _2}}} = \frac{{{v_1}}}{{{v_2}}}\]

  19. BATMAN31
    • one year ago
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    @Michele_Laino are you still here? I still don't get it. I was eating

  20. BATMAN31
    • one year ago
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    @paki @texaschic101

  21. Michele_Laino
    • one year ago
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    why?

  22. Michele_Laino
    • one year ago
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    please, read my step-by-step explanation

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