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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.
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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BATMAN31
 one year ago
Best ResponseYou've already chosen the best response.0http://www.education.com/sciencefair/article/refractionfastlighttravelair/

BATMAN31
 one year ago
Best ResponseYou've already chosen the best response.0how does this equation work? walk me through it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you need the numbers that go in the equation ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which is sin and which is V

BATMAN31
 one year ago
Best ResponseYou've already chosen the best response.0neither, from this experiment I do not know any values for 4.

BATMAN31
 one year ago
Best ResponseYou've already chosen the best response.0the WIP equation is \[(\sin 17)/(\sin 27)=v1/v2\]

BATMAN31
 one year ago
Best ResponseYou've already chosen the best response.0I just watched the khan academy vid on snell's law, and it didn't clear much up. @directrix

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i dont understand

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0here is a little explanation: dw:1433353958208:dw the statement of the Snell's law is: \[\Large {n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0since 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: \[\Large \frac{{{n_1}}}{c}\sin {\theta _1} = \frac{{{n_2}}}{c}\sin {\theta _2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0now, according to the previous formula, we have: \[\Large {v_1} = \frac{c}{{{n_1}}},\quad {v_2} = \frac{c}{{{n_2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0then 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: \[\Large \frac{{\sin {\theta _1}}}{{\sin {\theta _2}}} = \frac{{{v_1}}}{{{v_2}}}\]

BATMAN31
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino are you still here? I still don't get it. I was eating

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0please, read my stepbystep explanation
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