anonymous
  • anonymous
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° ? :/
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Michele_Laino
  • Michele_Laino
I remember that the critical angle is given by the subsequent formula: \[\Large \sin \left( {{\theta _c}} \right) = \frac{1}{n} = \frac{1}{{1.33}}\]
anonymous
  • anonymous
okay! so that gets 0.751879 right? what happens next?
Michele_Laino
  • Michele_Laino
we have to compute this value: \[\Large \arcsin \left( {0.7518} \right) = ...\]

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anonymous
  • anonymous
0.850787638?
Michele_Laino
  • Michele_Laino
no, it is: 48.753
anonymous
  • anonymous
ohh :/ oops :P okay! so that means our solution is 49º?
Michele_Laino
  • Michele_Laino
yes! that's right!
anonymous
  • anonymous
yay!! thank you!
Michele_Laino
  • Michele_Laino
:)

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