Lab11_VanLe

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Jan 9, 2024

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Van Thin Le Professor Xiang Song Phys-2426-51700 22 November 2023 LAB 11: WAVE INTERFERENCE Part A Theory Please study Interference concept to answer the following questions. 1. Discuss interference, constructive and destructive interference. List the phase and path difference equations for constructive and destructive conditions. Interference occurs when two or more waves superpose, combining their amplitudes and resulting in a new wave pattern. This phenomenon can be categorized into constructive and destructive interference . Constructive Interference: This occurs when waves combine in such a way that their crests align with each other, resulting in an amplified wave with a larger amplitude. o Phase difference of two wave for CI: ∆ ϕ = ϕ 2 ϕ 1 = 2 m π o The path difference for Bright: ∆r = r 2 r 1 = m λ Destructive Interference: This occurs when waves combine in such a way that the crest of one wave aligns with the trough of another, leading to a reduced or canceled-out wave. o Condition for out of phase (Dark): ∆ ϕ = ϕ 2 ϕ 1 =( 2 m + 1 ) π o The path difference for Dark: ∆r = r 2 r 1 =( 2 m + 1 ) λ 2 2. Discuss the Double Slits (Young’s) interference. Give path difference equation for the angle condition of interference in Young’s interference. Define the meaning of each term in your equation. In Young's double-slit interference experiment, a coherent light source passes through two closely spaced slits, creating two coherent wavefronts that overlap and interfere with each other on a screen placed some distance away. This produces a pattern of light and dark fringes known as interference fringes. Path difference: ∆r = r 2 r 1 = dsinθ Constructive interference (bright) at point P: ∆r = dsin θ bright = m λ Destructive interference (dark) at point P: ∆r = dsin θ dark =( 2 m + 1 ) λ 2
3. Double slits interference shown in graph: slits separation d = 1900 nm, distance between slits and screen D = 4436 nm, central bright y o = 0, first bright position on screen y 1 = 1555 nm. Determine angle θ and wavelength λ using the given information. tanθ = ∆ y D = 1555 4436 →θ = tan 1 ( 1555 4436 ) = 19.3 o d y D = m λ d y 1 D = 1 × λ → λ = d y 1 D = 1900 × 1555 4436 = 666 nm 4. Discuss the phase change condition due to reflection of light from a surface. Summarize equations of interference for thin film. When light reflects from a denser medium to a rarer one, it undergoes a phase change of π (or 2λ/2). In interference within thin films, like soap bubbles, Equations dictate constructive interference when: 2 n 1 t =( 2 m + 1 ) λ 2 Equations dictate destructive interference when: 2 n 1 t = m λ Where: t - the film's thickness n1 - the surrounding medium's refractive index, m - an integer representing fringe order, λ - the light's wavelength in the film.
Part B Lab Go to PhET website. Click on Simulation/Physics. Under Physics, choose Light & Radiation. Under Light & Radiation, Wave Interference is the 2nd of last simulation (Location might change. A to Z search.). Click to run the Wave Interference . I. Complete Data Table with Wave Interference/Slits Simulation: Basic Operation for Slits Simulation: Keep the default on Frequency (green) and Amplitude (Max). Check Light Source (below Amplitude). Place check mark on Screen and Intensity. Choose Two Slits. Adjust Slit Width a = 500 nm (default) and Slits Separation d = 1900 nm. Press the Green Button on Light (left side screen). Drag Slits to the far left (by ◄► on bottom of Slits). Drag Ruler out of Toolbox to measure the distance, D, between Slits and Screen. Move Ruler to Screen and measure the centerto-center distance Δy (as shown in Figure). Test and understand all functional tools on screen. You must practice figuring out what is the best way to complete the measurement. Data Table 1: Measure Wavelength λ of Light Sorce Adjust Slit Width a = 500 nm and Slits Separation d = 1900 nm. Measure the distance, D, between Slits and Screen and measure the center-to-center distance Δy by Ruler (or peak-peak distance on Intensity) . Referring to Young’s Interference, calculate wavelength λ=dsinθ. Drag Stopwatch out of Toolbox to measure 10 waves time as t 10 and calculate average period time T = t 10 /10. Calculate wavelength by λ T =cT, where c = 2.998 x 10 8 m/s. 1 nm = 10 −9 m, 1fs = 10 −15 s. Measured the distance, D , between Slits and Screen : D = 4465.6 Attach a Screenshot of Red Data from the Simulation to Lab Report . Color Δy (nm) θ=tan -1 (Δy/D) λ=dsinθ (nm) T (fs) λ T =cT (nm)
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Red 1700 20.84 676.1 2.25 674.6 Green 1261 15.77 516.4 1.70 509.7 Violet 1015 12.8 421.2 1.40 419.7 Question: If you decrease the Slits Separation d, what do you observe? The intensity of the remaining fringes reduces while the brightness of the bright fringe amplifies. If you decrease the distance between Slits and Screen D, what do you observe? The intensity of the bright fringe grows, and this increase extends to the fringes nearing the center. II. Complete Data Table with Wave Interference/Interference Simulation: Basic Operation for Interference Simulation: Check Light Source (below Amplitude). Keep the default value on Frequency (green), Amplitude (Max), and Separation distance d = 1500 nm. Place check mark on Screen and Intensity. Press the Green Button on both Lights. Drag Ruler
out of Toolbox to measure the distance, D, between Light and Screen. Move Ruler to Screen (or Intensity) and measure the center-to-center distance Δy as shown in Figure of Table 1). Test and understand all functional tools on screen. You must practice figuring out what is the best way to complete the measurement. Data Table 2: Measure Wavelength λ of Light Source Measure the distance, D, between Light and Screen and measure the center-to-center distance Δy by Ruler . Referring to Young’s Interference, calculate wavelength λ=dsinθ. Drag Stopwatch out of Toolbox to measure 10 waves time as t 10 and calculate average period time T = t 10 /10. Calculate wavelength by λ T =cT, where c = 2.998 x 10 8 m/s. Separation distance: d = 1500 nm. 1 nm = 10 −9 m, 1fs = 10 −15 s. Measured the distance, D , between Slits and Screen: D = 4460.8nm Attach a Screenshot of Green Data from the Simulation to Lab Report . Color Δy (nm) θ=tan -1 (Δy/D) λ=dsinθ (nm) T (fs) λ T =cT (nm) Green 1710.3 21 537 1.85 554.6 Blue 1585 19.6 502.2 1.69 506.7 Violet 1371.3 17.1 440.76 1.45 434.7 Question: Calculate the percent difference between λ and λ T of Green Data: % difference = { Difference in Values / [Average Value]}•100%
={ Value1 – Value2 / [ (Value1 + Value2)/2]} •100% 506.7 502.2 ( 506 . 7 + 502.2 )/ 2 100% = 0.89% If you increase the Separation d, what do you observe? The bright fringes approach the center. Part C Discovery Beyond what you have done with Wave Interference Simulation, make few new findings using other functions in this simulation , or give real life examples for interference. You can express your findings by data table, word, equations, and/or screenshot graph/Video, but do giving physics explanation. You could also design meaningful experiment and show it here. In Part C minimum 10 sentences are required. The rainbow effect is often observed in thin films, akin to the colors seen in soap bubbles. Interference plays a pivotal role here, showcasing captivating spectrums due to the constructive and destructive interference of light waves reflecting off the top and bottom surfaces of the film. Consider a soap bubble, which acts as a thin film of water sandwiched between two layers of air. When light hits the soap film, some of it reflects off the outer surface, while another portion reflects off the inner surface. These reflected waves interfere with each other when they recombine, creating an array of colors. Interference occurs as light waves of different wavelengths travel through the film and undergo phase shifts. The varying thicknesses of the film lead to different path differences for different colors of light, causing them to interfere constructively or destructively. For instance, at certain thicknesses, the path difference for red light might result in constructive interference, intensifying the red color. Meanwhile, blue light might experience destructive interference, causing some colors to cancel out. This leads to the mesmerizing display of iridescent colors seen in soap bubbles or oil slicks, showcasing interference's ability to create vibrant, shifting hues in everyday phenomena.
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