At what angle must a ray be incident as it passes from a material which has index of refraction 1.74 to a material with index of refraction is 1.13 in order to be totally internally reflected?
When a ray passes from a material to another with a higher index of refraction, there are incident angles `thetaIn for which Snell's Law sin(`thetaRef) / sin(`thetaIn) = n2 / n1 has no solution for `thetaRef. This occurs when sin(`thetaIn) * n2 / n1 is greater than 1. This is possible because n2 > n1 and therefore n2 / n1 > 1; we only need for `thetaIn to be close enough to 90 deg to make sin(`thetaIn) * n2 / n1 > 1.
For such an angle, we will have
Since the sine of any angle can be at most 1, sin(`thetaRef) cannot be > 1 and we therefore have no solution.
The largest `thetaIn which will result in a solution to the equation will then be the one for which sin(`thetaIn) * n2 / n1 = 1; for this angle `thetaRef will be 90 degrees. This angle is called the critical angle.
We can find the specific `thetaIn that satisfies this condition.
We easily solve this for `thetaIn:
This is called the critical angle; we will denote it `thetaCrit:
For the present case we have
For incident angles greater than `thetaCrit, we will have no solution at all for `thetaRef, meaning that no refraction takes place. In this case the ray is totally reflected at the interface between the materials. None of its energy is transmitted and therefore none of its energy is lost at the interface. Since the ray remains in the original medium, we call this phenomenon total internal reflection.
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