Hi, i'm having a bit of trouble with a question here. I have a diagram of a beam of light entering a medium with a high refractive index from a lower refractive index medium. The beam however doesn't leave the second medium because of total internal reflection, so it's angle to the normal when it enters the second medium is it's critical angle. The critical angle i have is 41.1 deg, and the refractive index of the second medium is 1.52, and the refractive index of the first medium is 1.33. can anyone give me any hints on how to find the angle to the normal that the beam of light enters the second medium is ?
Thanks.
2007-08-01 10:21:05 · 1 answers · asked by David M 1 in Science & Mathematics ➔ Physics
The critical angle for light traveling from a material with a higher (n1) to a lower (n2) refractive index is:
theta = arcsin(n2/n1)
If your angle of incidence is greater than the critical angle, there will be ZERO transmission into the second medium. That's what total internal reflection means. All the light stays inside the more optically dense medium. It is totally internally reflected.
2007-08-01 10:27:55 · answer #1 · answered by lithiumdeuteride 7 · 4⤊ 1⤋