A diamond is underwater; A light ray enters one face of the diamond, then travels at an angle of 30 degrees with respect to the normal. What was the ray's angle of incidence on the diamond? The index of refraction of water is nwater=1.33, and the index of refraction of diamond is ndiamond=2.42. Finally, apply Snell's law and find the ray's angle of incidence θ1 on the diamond.

Question

A diamond is underwater; A light ray enters one face of the diamond, then travels at an angle of 30 degrees with respect to the normal. What was the ray's angle of incidence on the diamond? The index of refraction of water is nwater=1.33, and the index of refraction of diamond is ndiamond=2.42.    Finally, apply Snell's law and find the ray's angle of incidence θ1 on the diamond.

anonymous · Answer

I really don't understand what they are asking. the answer came out to be 65.5 degrees, but I don't see how. please advise

jt950 · Answer

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anonymous · Answer

yup i got that part. but why 30 degrees? based on math i get 33 degrees sin-1(1.33/2.42)

anonymous · Answer

so other angle would be 60, but I don't see the 65. using trig analysis, we need to complete the triangle.

jt950 · Answer

just use the snell's law 

n(water)sinx=n(diamond)sin30

1.33 sin x = 2.42 sin 30

x = 65

anonymous · Answer

ok. so i would 30 is my angle of incidence

jt950 · Answer

Ray is incident from water to diamond.

30 is angle of refraction

anonymous · Answer

oh ok thanks for the clarification. that helps a lot.