An optical fiber is made of a clear plastic with index of refraction n=1.6, For what angles with the surface the light remain contained within the plastic guide?
To determine the angle of incidence at which light remains contained within the plastic guide, we can use the concept of total internal reflection.
Total internal reflection occurs when light travels from a medium with a higher refractive index to a medium with a lower refractive index, and the angle of incidence exceeds the critical angle.
The critical angle can be calculated using the equation:
sin(critical angle) = (refractive index of lower medium) / (refractive index of higher medium)
In this case, the refractive index of the lower medium is 1 (air) and the refractive index of the higher medium is 1.6 (plastic).
sin(critical angle) = 1 / 1.6
Taking the inverse sine (sin^-1) of both sides, we can find the critical angle:
critical angle = sin^-1(1 / 1.6) ≈ 39.81 degrees
Therefore, for angles of incidence less than 39.81 degrees, the light will remain contained within the plastic guide of the optical fiber.