PHYS_Lab_Report_6

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Dec 6, 2023

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Shannon Coffey & Kiersten Roberts 11/1/2023 Diffraction PHYS 202
Results: d = 5 mm m y θ Sin θ 3 1.4cm 0.617 0.01077 5 2.4cm 1.057 0.01844 6 2.7cm 1.189 0.02077 Fig 1. This graph represents the m values observes through our experiment along the x-axis, while the y-axis is calculated by dSinθ. The slope of this graph is λ, as dSinθ = m ( λ). The data reported in this figure is derived from the 5 mm slits. d = 25 mm m y θ Sin θ 2 0.7cm 0.308 0.00537 3 1.0cm 0.441 0.00769 7 3.0cm 1.321 0.02305 y = 0.0017x + 0.0004 R² = 0.988 0 0.002 0.004 0.006 0.008 0.01 0.012 0 1 2 3 4 5 6 7 dSinθ m 5mm Slit
Fig 2. This graph represents the m values observes through our experiment along the x-axis, while the y-axis is calculated by dSinθ. The slope of this graph is λ, as dSinθ = m ( λ). The data reported in this figure is derived from the 25 mm slits. D = 130 cm Calculations: dSinθ d= 5mm -> 0.005m 0.005 Sin ( 0.617) = 0.005384 0.005 Sin ( 1.057) = 0.009223 0.005 Sin ( 1.189) = 0.010385 d= 25mm -> 0.025m 0.025 Sin ( 0.308) = 0.01343 0.025 Sin ( 0.441) = 0.01924 0.025 Sin ( 1.321) = 0.05763 θ = tan -1 (y/D) 5mm slit θ = tan -1 (1.4/ 130) = 0.617 θ = tan -1 (2.4/ 130) = 1.057 θ = tan -1 (2.7/ 130) = 1.189 25mm slit θ = tan -1 (0.7/ 130) = 0.308 θ = tan -1 (1.0/ 130) = 0.441 θ = tan -1 (3.0/ 130) = 1.321 y = 0.0091x - 0.0061 R² = 0.9953 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0 1 2 3 4 5 6 7 8 dsin θ m 25mm Slit
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25mm Slit Calculated λ Value: 0.0091m 5 mm Slit Calculated λ Value: 0.0017m Questions: Part I: 1. Using your interference pattern, compute the best estimate of the distance between two maxima(∆y). 25mm ∆y = 0.7cm => 0.007m/2 = 0.0035m 5mm ∆y = 1.4cm => 0.014m/3 = 0.0046m 2. Using the previous result and the distance from the slits to the wall(D), compute the angle between the central maximum (order 0) and the next maximum (or order 1), Compute sin(θ) also. 5mm Y= 0.0017(1) + 0.004 = 0.0057= dsin(θ) 0.0057 = (0.005) sin(θ) Sin -1 (1.14) = θ 25mm Y= 0.0091(1)– 0.0061 =0.003 = dsin(θ) 0.003= (0.025) sin(θ) Sin -1 (0.12) = θ = 0.002094 sin(θ) : 5mm Sin ( 0.617) = 0.01077 Sin ( 1.057) = 0.01844 Sin ( 1.189) = 0.02077 25mm Sin ( 0.308) = 0.00537 Sin ( 0.441) = 0.00769 Sin ( 1.321) = 0.02305 3. Using answers from Q1, Q2 and n equals 1, calculate the wavelength(λ) of the laser light. Compare with the given value. (λ) = (d ∆y)/D 25mm
= ((0.25e-3m)( 0.0035m))/1.3m = 6.73e-7m 5mm = ((0.05e-3m)(0.0035m))/1.3m = 1.769e-7m Given value = 633nm or 6.33e-7m 25mm = (633-673)/673 = 0.0594 = 5.94% error 5mm = (633-177)/177 = 2.576 = 257.6% error 4. What happened when you rotated the slits? Explain. When we switched from the slit at 5mm to the slit at 25mm the wavelength (found through calculations of observed information) shifted upwards slightly from 0.0017m to 0.0091m. 5. Why do maxima (bright spots) and minima (dark areas) appear when light is passed through the slits? Why don’t we observe just two bright spots on the wall separated by the same distance as the slits? We see the bright and dark spots (maxima and minima) because of constructive interference and deconstructive interference. The dark spots or minima are where the waves are not in phase and “cancel out”. The bright spots, or maxima are where the waves are in phase with each other and “add up”. 6. Thomas Young used a single slit for the light to pass through before it hit the double slits, why don’t we use a single slit before the double in our experiment? Hint: What is the difference between the light sources? Thomas Young used sunlight and in our experiment, we used a laser. The laser creates a thin stream of light that we can chose where it is directed, this way we could aim the light at each slit we desired to use. Part II: 7. Using your graph(sin(θ) vs 1/w) deduce the wavelength of the laser light. Compare with the given value. Wavelength = slope of figures 25mm Slit Calculated λ Value: 0.0091m 5 mm Slit Calculated λ Value: 0.0017m 8. What is the difference between interference and diffraction? Interference are waves that come from two separate sources while diffraction is secondary waves that emerge from different segments from the wave of origin. When observing the intensity of
waves from interference it is around the same but with diffraction there is a great deal of peaks and valleys in intensity. 9. What does this experiment tell us about light? This experiment teaches us about diffraction, interference, the difference between single and double-slit experiments, and maxima/minima. This experiment showed how light waves behave as other waves do, more so than participles.
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