wk3-lab Alex Lei Claudia Nguyen and Grace Amour

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Dec 6, 2023

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Week 3 Lab Report – Reflection Date: 7/19/2023 Section: A3 Lab Partners: 1. Alex Lei 2. Grace Amour 3. Claudia Nguyen 3.3.2 Specular Reflection (Total Points: 20) Goal: Verify Eq. 3.1 ( θ mirror = ½ θ bench or equivalently θ i = θ r ) for specular reflection. ** θ bench and θ mirror are defined in Fig. 3.7. Schematic Diagram (2 points) : 1

Background voltage reading (1 point): ● 4.4 mV Table #1 – Measurements (6 points) : Trial # θ bench, initial direct compass reading (degrees) Distance from laser to mirror (cm) Distance from laser to optical bench (cm) θ mirror , direct compass or protractor reading (degrees) Flux/voltage recorded from photometer (mV) Flux minus background (mV) 1 50° 30 cm 19 cm 20° 115.5 mV 111.1 mV 2 60° 38 cm 27.5 cm 29° 214.6 mV 210.2 mV 3 70° 36.5 cm 30 cm 26° 146 mV 141.6 mV 4 80° 35 cm 30 cm 40° 168 mV 163.6 mV Sample calculation for flux – background (1 point) : ● Voltage Reading - Background Value =VR - BV 115.53 mV - 4.4 mV = 111.13 mV 2

Table #2 – Calculations (4 points): Trial # θ bench from distances/using trigonometry (degrees) θ mirror from distances/using trig; should = ½ * θ bench (degrees) θ mirror theoretical, from direct compass reading of θ bench (degrees) Discrepancy in θ mirror (% error) 1 39.3° 19.7° 25° 21.2% 2 46.4° 23.2° 30° 22.6% 3 55.3° 27.7° 35° 20.8% 4 58.9° 29.5° 40° 26.2% Which of the two methods for measuring θ mirror experimentally (from distances or from the compass/protractor) did you use when calculating the discrepancy? Why? (Your explanation should only be a sentence or two.) (1 point) ● The methods used for measuring θ mirror included the distances between the laser to optical and the laser to mirror. This method was used because it was determined that the values recorded by the distances could show a more accurate reference of the readings when compared to the protractor. Sample calculations for θ bench from distances, θ mirror from distances/using trig, θ mirror theoretical, and % error (2 points, 0.5 each): ● Arcsin (19 cm / 30 cm ) = 39.3° Is the discrepancy/% error of θ mirror comparable to the % error in your angular measurements, or is there an additional source of systematic error? (1 point) ● = 100 = 21.2% 𝐸𝑥?𝑒𝑟𝑖?𝑒?𝑡𝑎? − 𝑡ℎ𝑒?𝑟𝑒𝑡𝑖𝑐𝑎? 𝑇ℎ𝑒?𝑟𝑒𝑡𝑖𝑐𝑎? | 19.7−25 25 | × Is there an angle for which the reflected light is maximized? (1 point) ● No, there is also no direct correlation associated with the angle and the maximum mV that is reflected. There is also evidence of fluctuations that occur with the voltages due to light from the windows and many sources outside what was recorded. Using your answers to the previous two questions, can you conclude from your results whether the mirror is a perfect specular reflector? (1 point) 3

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● Base d on discrepancy in θ mirror , the % error of our results (from 20%-26%) is greater than precision error (~10%), so I conclude that the mirror is not a perfect specular reﬂector. ● Theta mirror also does not perfectly represent ½ of the theta bench. 3.3.3 Diffuse Reflector (Total points: 15) Goal: Measure how the amount of light reflected by a diffuse reflector varies with the angle of reflection. *For this section, keep the laser fixed at a given θ bench and rotate the white card. It is much easier than trying to change the position of the laser. θ bench value, should be between 0 and 90 degrees for simplicity (1 point) : ● 40° What is the measured angle θ white card when the white card is perpendicular to the optical bench? (1 point) ● 0° The angle θ white card should be corrected to 0 degrees when the white card is perpendicular to the optical bench. What is the correction (in degrees) between your measured angle and 0 degrees? Show your calculation. (1 point) ● 360° - θ white card , direct compass = θ white card corrected value 360° - 350° = 10° Calculate the expected θ white card for the expected intensity maximum corresponding to specular reflection, based on your θ bench value and your correction from the previous part. This is t he angle at which you expect to see the maximum flux. (1 point) ● ½ q bench = ½ (40) = 20 degrees for maximum flux expectation. **For the table below, you may do more than 10 angles if you wish, but only 10 are required. You may also want to measure the background for each individual angle if you are in a lab station by a window/to make sure you have no V – bkg values; if you keep it the same or only measure it once, you can fill out the “background” column below with the same value all the way down. For the corrected angle, set it to zero degrees when the white card is perpendicular to the optical 4

bench. Take angles to be positive when the white card is rotated toward the laser and negative away from the laser. Table 3.3.3 (5 points): θ white card , direct compass reading (degrees) θ white card , corrected value (degrees) V (mV) Background (mV) V – background (mV) 0° 0° 2.0 mV 4.4 mV 6.4 mV 350° 10° 3.4 mV 4.4 mV 7.8 mV 340° 20° 2.4 mV 4.4 mV 6.8 mV 330° 30° 2.2 mV 4.4 mV 6.6 mV 320° 40° 1.7 mV 4.4 mV 6.1 mV 310° 50° 1.3 mV 4.4 mV 5.7 mV 300° 60° 1.1 mV 4.4 mV 5.5 mV 290° 70° 0.7 mV 4.4 mV 5.1 mV 280° 80° 0.3 mV 4.4 mV 4.7 mV 270° 90° 0.1 mV 4.4 mV 4.5 mV Sample calculations for θ white card (corrected value) and V – background (1 point): ● Voltage Reading - Background Value =VR - BV 2.0 mV - (-4.4 mV) = 6.4 mV Plot: θ white card (corrected angle) vs V – background (2 points) : 5

Calculate the discrepancy between the expected θ white card for the expected intensity maximum and the angle from your data that gives you an intensity maximum. (1 point) ● = 100 = 50% 𝐸𝑥?𝑒𝑟𝑖?𝑒?𝑡𝑎? − 𝑡ℎ𝑒?𝑟𝑒𝑡𝑖𝑐𝑎? 𝑇ℎ𝑒?𝑟𝑒𝑡𝑖𝑐𝑎? | 10−20 20 | × An ideal diffuse reflector scatters light uniformly in all directions. Is the white card an ideal diffuse reflector? (1 point) ● No, it is not an ideal diffuse reflector because it scatters the light everywhere equally with a specific angle shown to represent the maximum. How are the results for the reflector different from the mirror? (1 point) ● The results are different because the reflector has a different voltage represented by the angle, whereas the mirror shows no inherent correlations associated with the voltage fluctuations from the angles used. 3.3.4 Internal Reflection (Total Points: 15) Goal: Find the critical angle for propagation in an optical fiber by two methods. ● Critical angle value = 30.3° Background reading (the rubber stopper should be in the photometer, but the fiber optic cable has not yet been put into it) (1 point) : ● -0.15 mV 6

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Measure and write down the signal from a single loop of the fiber optic and a double loop of the fiber optic. Can you conclude that there is no loss of intensity from curving the fiber, or at least that the difference is not significant? (2 points) ● 3.00 mV at the single loop. ● 3.04 mV at the double loop. ● Yes, based on the values obtained, it can be concluded that there is no loss of intensity when curving the fiber optic cable because even when bending or curving the fiber for twice the amount, the value of the light’s voltage does not change. **Note: The corrected reading in the table below has α = 0 when the face of the optical fiber is perpendicular to the laser beam. Your correction for your angle should be the same as in 3.3.3. Table 3.3.4 (4 points) : α ( Compass Reading) (degrees) α (Corrected Reading) (degrees) V (mV) V − background (mV) 340° 20° 3.6 mV 3.75 mV 336° 24° 3.9 mV 4.05 mV 334° 26° 3.6 mV 3.75 mV 332° 28° 2.95 mV 3.10 mV 330° 30° 2.99 mV 3.14 mV 328° 32° 2.99 mV 3.14 mV 326° 34° 3.00 mV 3.15 mV 324° 36° 2.7 mV 2.85 mV 322° 38° 2.2 mV 2.35 mV 320° 40° 1.5 mV 1.65 mV 318° 42° 1.3 mV 1.45 mV 316° 44° 0.77 mV 0.92 mV Sample calculations for α (corrected value) and V – background (1 point): 7

Voltage Reading - Background Value =VR - BV ● 1.60 mV -(-0.15 mV) = 1.75 mV Based on your data above, what is the critical angle for propagation, α critical ? How did you determine this? (You do not need to estimate the error in your measurement.) (2 points) ● 34°, this is where the voltage peaked. After this, there was a decrease in the voltage, which indicates that 34°, or the peak is where the critical angle is, according to the data. Show your calculation for the theoretical critical angle, using the data provided to you in the manual. (1 points): = 1.5 𝑛 ?𝑜𝑟? = 1.4 𝑛 ?𝑙𝑎? = = = 30.4° α ?𝑟𝑖𝑡 𝑠𝑖𝑛 −1 (𝑛 ?𝑜𝑟? ) 2 − (𝑛 ?𝑙𝑎? ) 2 𝑠𝑖𝑛 −1 (1. 5) 2 − (1. 4) 2 Calculate the discrepancy between your theoretical critical angle and your experimental critical angle for propagation from above. Show your calculation below. Taking any error of less than 20% to be “reasonable,” is this error reasonable? (2 points) : ● [(30.4 - 34 ) / (34)] x 100 = 10.6% ● Taking any error of less than 20% is reasonable because it is below the precision error of 20% that we were given. Place one end of your fiber optic at the hole in the metal plate, and take the other end and shine it onto a white surface (piece of paper, white card from 3.3.3, etc.). Observe the image patterns projected onto the white surface. Sketch the image patterns observed at α ≈ 0 o and α ≈ 30 o , and briefly explain what causes these patterns. (2 points) 8

● There's a donut that appears from the light on the white surface at 24 degrees. A mirror of the fiber op±c causes the image to appear and distort at an angle of 24–30 degrees. This distor±on occurs at roughly 24, but cannot be visualized at an angle of 0 because at 0, the light is shot out directly with no angular displacement. 9

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