Feras_Awartani_ECE 313_lab1

ECE 313: Optoelectronics Lab Section 01 Experiment 01 Introduction to the Principles of Optical Waveguiding Submitted by: Feras Awartani Alexander Sharp, Wan Ediwan, Jack Shields Feb 6 th , 2024

ABSTRACT This lab focused on exploring the principles of total internal reflection, critical angles, and numerical aperture within optical fibers. Initially, we observed laser behavior under three conditions: a 45-degree angle of incidence directed at an empty notched slab, a large incidence angle aimed at a water- filled notched slab, and a similar angle directed at a mineral-oil filled notched slab. Utilizing Snell’s law of refraction, we derived crucial angles essential for understanding how light is guided through optical fibers via total internal reflection. Additionally, we conducted experiments with different mode patch cords (single mode and multimode) by adjusting the angle of incidence and monitoring voltage readings. Our results indicated that multimode fibers exhibit a higher numerical aperture at a larger range when compared to single mode fibers. THEORETICAL BACKGROUND This lab delves into the fundamentals of optics, particularly focusing on the behavior of light as it traverses various materials and propagates through optical fibers. By investigating Snell’s Law of refraction, expressed as n1sin θ 1=n2sin θ 2, we gain insight into how light interacts with materials of different refractive indices, n. Examination of Snell’s law reveals that when n1>n2, the refracted angle θ 2 must exceed the incident angle θ 1, a phenomenon known as total internal reflection. Total internal reflection occurs when the light is incident at a specific angle known as the critical angle ( θ c). The critical angle is the minimum angle θ 1 needs to be so that θ 2 is 90˚ and all incident light is reflected. sin θ c = n2/n1 (1.1) This concept serves as the central theme of the lab, particularly as it pertains to fiber optics. Optical fibers consist of a solid cylindrical core surrounded by a cladding layer, with the cladding has a lower refractive index than the core to facilitate total internal reflection. Two primary types of optical fibers exist: single mode and multimode. The former features a refractive index profile peaking at the core center and is zero on the boundaries, making it ideal for high-speed, long-distance transmissions. In contrast, multimode fibers have a larger core capable of accommodating various light profiles and shapes.

Figure 2 Diagram of experimental apparatus. Position (a) of the HeNe laser is used for the study of total internal reflection; position (b) is used to measure NA of a fiber. To set up the experiment we mounted the HeNe laser in position (a) as shown above in Fig. 2, we then set θa to 45˚ as shown in Fig. 3. After we had that set up, we were able to use a protractor to measure θ1 shown in Fig. 3 which we measured to be 25˚. Having θa and θ1 and knowing that. n air = 1 allowed us to use Snell’s law to find the index of refraction of the plastic. n air sin θa = n pl sin θ. 1 sin (45) = n pl sin (25) n pl = sin (45)/sin (25) = 1.673 Figure 3 Ray diagram of refraction through model waveguide. Next, we filled both notches of the plexiglass ¾ full of water, we started with a large incidence angle θa and decreased till a convenient angle, then we measured θ1 which for us was 90-34 = 56˚. Next, we measured θ2 at θ1= 56 to be θ2 = 90-21 = 69˚, knowing that n water = 1.33 along with these angle measurements we were able to use Snell’s law to find n pl: n pl sin(θ1) = n water sin(θ2) n pl = n water sin(θ2) / sin(θ1) n pl = 1.33 sin (69) / sin (56) n pl = 1.4977

Then, we changed the angle θ1 until the angle θ2 was 90˚ meaning that all the light hitting the water was being reflected and none of it was transmitted through, we used the protractor to find that angle to be θc =54˚. Next, we used Eqtn. (1.3) to find the NA NA = (n1 2 – n2 2 ) 1/2 NA= (1.4977 2 -1.33 2 ) 1/2 ~ 0.688 Then, to check our work, we looked up the index of refraction of our plastic to compare it to our answer, we got that n pl (expected) = 1.49, so: θc = Sin -1 (1.33 / 1.49) ~ 63.2˚ and NA= (1.49 2 -1.33 2 ) 1/2 ~ 0.672 These results show that our NA was actually very close to expected, however our critical angle was slightly off, this is probably due to human error. After that we removed the water from our notches and replaced it with some mineral oil. Using our calculated n pl = 1.4977, we redid the steps for water, but trying to find the index of the mineral oil now that we have the index of plastic. We measured our θ1 to be 90-24 = 66˚ and our θ2 to be 90-14 = 76˚. Using Snell’s law, we have that: n pl sin(θ1) = n Oil sin(θ2) n oil = n pl sin(θ1) / sin(θ2) n oil = 1.4977 sin (66) / sin(76) n oil = 1.41 afterwards we changed the angle until we observed total internal reflection which we measured to be at θc= 70˚. Next, we used Eqtn. (1.3) to find the NA NA = (n1 2 – n2 2 ) 1/2 NA= (1.4977 2 -1.41 2 ) 1/2 ~ 0.505

NA = 1 sin 16 = 0.275 This number is close to the expected NA for multimode fibers which is around 0.3 Next, we did the same thing but with a single mode patchcord. Fig. 5 below shows the measurements we got as we slowly adjusted the angle. Figure 5 a graph of the dB reading of the power meter vs the angle of a single mode patchcord Our maximum here was -7.4 at an angle 0˚ from the center. The angles that measure 3db under that are -3˚ and 3˚, so the acceptance angle is 3-(-3) = 6. Using this we can find the NA using Eqtn. (1.3). NA = n a sin θ a NA = 1 sin 6 = 0.1 This number is exactly what we expect for single mode fiber. CONCLUSION This lab provided me with an insightful introduction to the mechanics of fiber optics, offering a deeper understanding of the foundational principles behind various technologies and services. Exploring how light behaves differently across different mediums offered valuable insights into the practical

applications of optical phenomena. Moreover, delving into the distinctions between the two different types of optical fibers expanded my comprehension of their respective functionalities. Reflecting on our experimental results, I found that our measurements and calculations, for the most part, closely aligned with theoretical expectations, reaffirming the effectiveness of our learning process. Overall, this hands-on experience has enriched my understanding of fiber optics and its real-world implications. POST-LAB QUESTIONS For these questions n1 = n core and n2 = n cladding 1. Corning multimode fiber has a numerical aperture of 0.2 and a cladding index of 1.46. Calculate a) the core index, and the acceptance angle in air, NA = (n1 2 – n2 2 ) 1/2 n1= (NA 2 + n2 2 ) 1/2 = (0.2 2 + 1.46 2 ) 1/2 =1.4736 NA = n a sin θ a; na (air) = 1 θ a = sin -1 (NA/na) = sin -1 (0.2/1) = 11.54˚ b) the acceptance angle in water(nw=1.33) NA = n a sin θ a θ a = sin -1 (NA/na) = sin -1 (0.2/1.33) = 8.65˚ c) the critical angle at the core-cladding interface. sin θ c = n2/n1 θ c =sin -1 (n2/n1) = sin -1 (1.46/1.4736) = 82.21˚ 2. Corning single mode fiber has a cladding index of 1.46 and a core index of 1.46*1.0036. Calculate a) the core index, the numerical aperture, and the acceptance angle in air, core index = 1.46*1.0036 = 1.465256

NA = 1 sin (8) = 0 .1392 NA = (n1 2 – n2 2 ) 1/2 n2 = (n1 2 – NA 2 ) 1/2 = (1.52 2 – 0.1392 2 ) 1/2 = 1.5136 5. Calculate the percentage error produced in the measurement of the numerical aperture of the fiber in problem 2 above, when the cladding index is given as 1.45 instead of 1.46. The core index should remain equal to 1.46*1.0036 NA = (n1 2 – n2 2 ) 1/2 = (1.465256 2 – 1.45 2 ) 1/2 = 0.21089 %error = [(actual – expected)/(expected)] * 100 %error = [(0.21089 – 0.124)/(0.124)] * 100 = 70% 6. The velocity of light in the core of a step index fiber is 2.01 x 10^8 m/s, and the critical angle at the core-cladding interface is 80˚ . Determine the numerical aperture and the acceptance angle for the fiber in air. Assume that the velocity of light in vacuum is 3 x10^8 m/s. n1 = c/v = 3 x10^8 / 2.01 x 10^8 = 1.49254 sin θ c = n2/n1 n2 = n1 sin θ c = 1.49254 sin (80) = 1.47 NA = (n1 2 – n2 2 ) 1/2 = (1.49254 2 – 1.47 2 ) 1/2 = 0.2584 NA = n a sin θ a; n a (air) = 1 θ a = sin -1 NA = sin -1 0.2584 = 14.975˚