A. Consider a light ray traveling from medium 1 to medium 2 with medium 2 extend

Question

A.

Consider a light ray traveling from medium 1 to medium 2 with medium 2 extending downward indefinitely. Let the index of refraction of medium 1 be 1.0. If the index of refraction of medium 2 1.92 and theta is nearly 90 degree, what is theta 2? Answer in units of degree What is the condition that the ray will totally reflect at interface B? At what angles of theta 1 will total internal reflection at interface B take place, for the values of n1and n2 given above? Now suppose the index of refraction medium 2 is 1.24. What will be the maximum value of 6 that will have total internal reflection?

Explanation / Answer

A. Snell's law: (Sin 1)/(Sin 2) = n2/n1

Sin 90 / Sin 2 = 1.92/1

2 = 31.4 deg

B. For total internal reflection at face B, Sin 2 / Sin 90 = 1/1.92

2 > 31.4 deg

C. Sin 1 / Sin 31.4 = 1.92/1

1 = 90 deg

D. Sin 2/ Sin 90 = 1/1.24

2 = 53.8 deg

Sin 1 / Sin 53.8 = 1.24/1

1 = 90 deg

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