Question

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.

Answer 1

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

Answer 2

LINK

Answer 3

number 9.

Answer 4

how does this equation work? walk me through it

Answer 5

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

Answer 6

27 and 17

Answer 7

which is sin and which is V

Answer 8

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

Answer 9

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

Answer 10

@Nnesha
@iGreen

Answer 11

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

Answer 12

@amistre64

Answer 13

yeah i dont understand

Answer 14

@Michele_Laino

Answer 15

here is a little explanation:

the statement of the Snell's law is:
\[\Large {n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}\]

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

@paki @texaschic101

Answer 21

why?

Answer 22

please, read my step-by-step explanation