## anonymous one year ago Water has an index of refraction of 1.33. What is the critical angle for light leaving a pool of water into air? 0°, 90°, 37° , or 49° ? :/

1. Michele_Laino

I remember that the critical angle is given by the subsequent formula: $\Large \sin \left( {{\theta _c}} \right) = \frac{1}{n} = \frac{1}{{1.33}}$

2. anonymous

okay! so that gets 0.751879 right? what happens next?

3. Michele_Laino

we have to compute this value: $\Large \arcsin \left( {0.7518} \right) = ...$

4. anonymous

0.850787638?

5. Michele_Laino

no, it is: 48.753

6. anonymous

ohh :/ oops :P okay! so that means our solution is 49º?

7. Michele_Laino

yes! that's right!

8. anonymous

yay!! thank you!

9. Michele_Laino

:)