## 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.

1. BATMAN31
2. anonymous

3. BATMAN31

number 9.

4. BATMAN31

how does this equation work? walk me through it

5. anonymous

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

6. BATMAN31

27 and 17

7. anonymous

which is sin and which is V

8. BATMAN31

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

9. BATMAN31

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

10. anonymous

@Nnesha @iGreen

11. BATMAN31

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

12. anonymous

@amistre64

13. anonymous

yeah i dont understand

14. BATMAN31

@Michele_Laino

15. Michele_Laino

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

16. 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: $\Large \frac{{{n_1}}}{c}\sin {\theta _1} = \frac{{{n_2}}}{c}\sin {\theta _2}$

17. Michele_Laino

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

18. 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: $\Large \frac{{\sin {\theta _1}}}{{\sin {\theta _2}}} = \frac{{{v_1}}}{{{v_2}}}$

19. BATMAN31

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

20. BATMAN31

@paki @texaschic101

21. Michele_Laino

why?

22. Michele_Laino