How do the angle of incidence relate mathematically with the angle of reflection after the critical angle, that is, in a TIR?(adsbygoogle = window.adsbygoogle || []).push({});

Say we have a person standing at the bottom of a swimming pool, if we take the n for water as 1.33 and for air as 1, we get the critical angle of 48.75 degrees. When the person sees the surface of the water that's further away, the angle of incidence (eye to surface) increases, but how does the other angle change?

The equation n1 sin(theta1) = n2 sin(theta2) doesn't seem to work anymore when the criticle angle is passed...

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# Total internal reflection question

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