Question

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°

? :/

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

0.850787638?

Answer 5

no, it is:
48.753

Answer 6

ohh :/ oops :P

okay! so that means our solution is 49º?

Answer 7

yes! that's right!

Answer 8

yay!! thank you!

Answer 9

:)