On the flat surface of an optical fiber by the diameter of d = 2 cm and refractive index n = Ö2, a ray of light falls at an angle of incidence i = 45°, entering the optical axis of the fiber. What is the distance traveled along the optical fiber at the N reflection on the cylinder? Find out also for N = 10 reflections.
The ray of light penetrates through fiberglass in point A such that the sin i = n sinr, and r = 30o
The angle limit l = 45o, because sinl = 1 / n
So, light rays that are reflected in P1, P2, P3, suffers total reflection because the incident angle in these points are at 60o, at this point is greater than the limit angle.
On first reflection, we have:
The second reflection the distance on fiber direction is 2AP1, and after N reflections we have:
For N = 10 reflections, we get D = 32,9 cm.