## 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 ?

1. Michele_Laino

we have to apply the law of Snell, so we can write: $\Large {n_{OIL}}\sin {\theta _{OIL}} = {n_{WATER}} \cdot 1$

2. anonymous

ok!

3. Michele_Laino

so: $\Large \sin {\theta _{OIL}} = \frac{{{n_{WATER}}}}{{{n_{OIL}}}} = \frac{{1.33}}{{1.47}}$

4. anonymous

0.90476 ?

5. Michele_Laino

and what is: arcsin(0.90476)=...?

6. anonymous

0.0157916973 ?

7. Michele_Laino

no, it is : 64.791

8. anonymous

oh sorry.. it is 64.79 ?

9. anonymous

so our solution is 65º? sorry i was in the wrong mode on my calc :O

10. Michele_Laino

that's right! make sure to set "DEG" not "RAD" on your calculator

11. anonymous

yay! thank you:)

12. Michele_Laino

:)