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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 ?
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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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to apply the law of Snell, so we can write: \[\Large {n_{OIL}}\sin {\theta _{OIL}} = {n_{WATER}} \cdot 1\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so: \[\Large \sin {\theta _{OIL}} = \frac{{{n_{WATER}}}}{{{n_{OIL}}}} = \frac{{1.33}}{{1.47}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and what is: arcsin(0.90476)=...?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, it is : 64.791

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh sorry.. it is 64.79 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so our solution is 65º? sorry i was in the wrong mode on my calc :O

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1that's right! make sure to set "DEG" not "RAD" on your calculator
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