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Homework B on Optical Fibers (solutions) |
Feel free to work on the homework in groups. The work you hand in, however, should reflect your understanding of the material. You should show all of your calculations (neatly) and justify all of your answers for full credit.
In your previous homework assignment, you considerd a step-index fiber having a core index of refraction of n1 = 1.55. Light of vacuum wavelength 633 nm entered the fiber from air (n0=1) at an angle q0 as shown in the picture below. The critical angle at the core-cladding interface was 75.4o, and you should have found that n2 must then equal 1.50. The path taken by the light if it is totally internally reflected looks something like the diagram below. q1a is the angle the light makes with the normal upon entering the fiber, and q1b is the angle light makes with the normal upon hitting the upper edge of the core. In your previous homework, you should have identified this angle q1b as the one which might correspond to the critical angle.
| 1. | What is the numerical aperture of this fiber? Show your work. |
| The numerical aperture is given by sqrt(n12 - n22). For this fiber, 1.42 - 1.22 = 0.52. The square root of that gives a NA of0.721. | |
| 2. | Will light entering the fiber at an angle q0 = 30o be trapped in the fiber? How do you know this? |
| Yes. The numerical aperture is the largest value that n0 sin q0 can have and still have light trapped inside the fiber. Since n0 = 1 and the NA is 0.721, sin q0 < 0.721. This means that q0 must be less than 46.14o. | |
| 3. | Use Snell's Law to determine the angle of refraction q1a as the light enters the fiber if q0 = 30o. |
| n0 sin q0 = n1 sin q1. (1) (sin 30o) = (1.4) (sin q1) gives a value for q1 of 20.92o. | |
| 4. | Use geometry to determine the angle of incidence q1b on the upper edge of the core. |
| 90o - 20.92o = 69.08o | |
| 5. | Compare your answer to the previous question to the critical angle for the core-cladding interface. Will light be trapped inside the fiber? |
| 69.08o > 59.0o. Yes, light will be trapped in the fiber. This agrees with the answer that was obtained using the numerical aperture. |
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