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anonymous

  • one year ago

The refractive index of cooking oil is 1.47, and the refractive index of water is 1.33. A thick layer of cooking oil is floating on top of water in a beaker. What is the critical angle for light passing from the oil into the water? 25º, 43º , 49º , or 65º ?

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  1. Michele_Laino
    • one year ago
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    we have to apply the law of Snell, so we can write: \[\Large {n_{OIL}}\sin {\theta _{OIL}} = {n_{WATER}} \cdot 1\]

  2. anonymous
    • one year ago
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    ok!

  3. Michele_Laino
    • one year ago
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    so: \[\Large \sin {\theta _{OIL}} = \frac{{{n_{WATER}}}}{{{n_{OIL}}}} = \frac{{1.33}}{{1.47}}\]

  4. anonymous
    • one year ago
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    0.90476 ?

  5. Michele_Laino
    • one year ago
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    and what is: arcsin(0.90476)=...?

  6. anonymous
    • one year ago
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    0.0157916973 ?

  7. Michele_Laino
    • one year ago
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    no, it is : 64.791

  8. anonymous
    • one year ago
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    oh sorry.. it is 64.79 ?

  9. anonymous
    • one year ago
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    so our solution is 65º? sorry i was in the wrong mode on my calc :O

  10. Michele_Laino
    • one year ago
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    that's right! make sure to set "DEG" not "RAD" on your calculator

  11. anonymous
    • one year ago
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    yay! thank you:)

  12. Michele_Laino
    • one year ago
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    :)

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