Question

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 ?

Answer 1

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

Answer 2

ok!

Answer 3

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

Answer 4

0.90476 ?

Answer 5

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

Answer 6

0.0157916973 ?

Answer 7

no, it is :
64.791

Answer 8

oh sorry.. it is 64.79 ?

Answer 9

so our solution is 65º?

sorry i was in the wrong mode on my calc :O

Answer 10

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

Answer 11

yay! thank you:)

Answer 12

:)