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- BATMAN31

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

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- BATMAN31

http://www.education.com/science-fair/article/refraction-fast-light-travel-air/

- anonymous

LINK

- BATMAN31

number 9.

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- BATMAN31

how does this equation work? walk me through it

- anonymous

you need the numbers that go in the equation ...

- BATMAN31

27 and 17

- anonymous

which is sin and which is V

- BATMAN31

neither, from this experiment I do not know any values for 4.

- BATMAN31

the WIP equation is \[(\sin 17)/(\sin 27)=v1/v2\]

- BATMAN31

I just watched the khan academy vid on snell's law, and it didn't clear much up. @directrix

- anonymous

- anonymous

yeah i dont understand

- BATMAN31

- Michele_Laino

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

- Michele_Laino

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:
\[\Large \frac{{{n_1}}}{c}\sin {\theta _1} = \frac{{{n_2}}}{c}\sin {\theta _2}\]

- Michele_Laino

now, according to the previous formula, we have:
\[\Large {v_1} = \frac{c}{{{n_1}}},\quad {v_2} = \frac{c}{{{n_2}}}\]

- Michele_Laino

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:
\[\Large \frac{{\sin {\theta _1}}}{{\sin {\theta _2}}} = \frac{{{v_1}}}{{{v_2}}}\]

- BATMAN31

@Michele_Laino are you still here? I still don't get it. I was eating

- BATMAN31

- Michele_Laino

why?

- Michele_Laino

please, read my step-by-step explanation

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