PHY2049ch35B%284-14-10%29

PHY2049ch35B%284-14-10%29 - Wave Interference and...

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Unformatted text preview: Wave Interference and Diffraction: I Introduction, thin films PHY 2049 Physics 2 with Calculus PHY 2049: Chapter 35 1 Quiz Three beams of light, a, b and c, of the same wavelength are sent through 3 layers of plastic with the indices of refraction as shown. Which material has the most number of wavelengths inside the material? 1. 2. 3. 4. a Shortest wavelength in material, so fits most # of waves b c Same for all a b c PHY 2049: Chapter 35 2 Need to Understand Light as Wave! (You already have read this material) Index of refraction Speed of EM wave in medium: Wavelength of light: Propagation of light: Huygens principle (36-2) Explains reflection and refraction Explains interference (from superposition) Explains diffraction (spreading of light around barrier) PHY 2049: Chapter 35 3 Interference as a Wave Phenomenon Interference of light waves Caused by superposition of waves Intensity can increase or decrease! Contrast with particle model of light Effects and applications Double slit Single slit Diffraction gratings Anti-reflective coatings on lenses Highly reflective coatings for mirrors Iridescent coatings on insects Colors on thin bubbles Interferometry with multiple telescopes PHY 2049: Chapter 35 4 Interference from Wave Superposition Basic rule: Add displacement at every point Wave 1 Wave 2 Sum PHY 2049: Chapter 35 5 Constructive Interference Same wavelength, phase difference = 0° Amplitude larger: Higher intensity Sum PHY 2049: Chapter 35 6 Destructive Interference Same wavelength, phase difference = 180° (1/2 ) Amplitude smaller: Lower intensity Sum PHY 2049: Chapter 35 7 Examples Two waves, same No interference Combined intensity: Inew = 4I + I = 5I , with amplitudes 2A and A Initial intensities 4I and I, respectively (I =kA2) Maximum constructive interference ( New amplitude: New intensity: New amplitude: New intensity: Anew = 2A + A = 3A Inew = 9I = 0) Maximum destructive interference ( Anew = 2A – A = A Inew = I =) PHY 2049: Chapter 35 8 General Treatment of Interference Most interference is partial Amplitudes for 2 waves are generally different Phase difference : 0 < < 180° Additional considerations Wavelengths can be different Multiple waves may interfere (e.g., diffraction grating) But easy to accommodate: just sum over all waves PHY 2049: Chapter 35 9 Interference and Path Length Two sources, spaced 3 wavelengths apart, emit waves with the same wavelength and phase. In how many places on the circle will the net intensity be a relative maximum? Answer = 12 Can you see why? Hint: Start at far right and move counterclockwise towards top, noting path length changes. Key idea: Path difference leads to phase difference PHY 2049: Chapter 35 10 Interference and Path Length Two sources, separated by 4 , emit waves at same wavelength and phase. Find relative minima on +x axis. Solution: path difference must be a half-multiple of 4 4 values n=0 x = 15.8 x = 4.58 x = 1.95 x = 0.54 L = /2 L = 3 /2 L = 5 /2 L = 7 /2 11 x n=1 n=2 n=3 PHY 2049: Chapter 35 Reflection and Interference from Thin Films Normal-incidence light strikes surface covered by a thin film Some rays reflect from film surface Some rays reflect from substrate surface (distance d further) Path length difference = 2d causes interference From full constructive to full destructive, depending on d n0 = 1 n1 = 1.2 n2 = 1.5 PHY 2049: Chapter 36 (Angle shown, but actually normally incident) 12 Standard analysis of Thin Film Interference Wavelength inside film! d n0 = 1 n1 = 1.2 n2 = 1.5 PHY 2049: Chapter 36 (Angle shown, but actually normally incident) 13 Example of Thin Film Interference Let = 500 n1 = 1.38 (MgF2) n2 = 1.5 (glass) Max intensity Min intensity Must be careful about phase shift at boundary: Reflection for nin < nout has phase shift, 0 if nin > nout Since n0 < n1 and n1 < n2, the phase shift has no effect here But for other cases, there can be an extra phase shift Soap bubble (next slide) PHY 2049: Chapter 36 14 Thin Film Interference for Soap Bubble Wavelength inside soap! Similar analysis, but… Phase shift of for air–soap reflection No phase shift for soap–air reflection Net phase shift switches max min n d PHY 2049: Chapter 36 15 Example of Soap Bubble Interference Let = 500 nm n = 1.32 (soap + water) Min intensity Max intensity PHY 2049: Chapter 36 16 Quiz What is the condition for destructive interference for light reflecting from a soap bubble of thickness d? (1) (2) (3) (4) d=m d = m+ 1 2 ( n ) n PHY 2049: Chapter 36 17 Quiz What is the condition for destructive interference for light reflecting from a soap bubble of thickness d? (1) (2) (3) (4) Only 1 reflection has a phase shift, so this switches min and max d=m d = m+ 1 2 ( n ) n PHY 2049: Chapter 36 18 Quiz Consider an oil film (thickness d, n = 1.5) on top of water (n = 1.3). Light of = 600 nm is normally incident. Which value of d corresponds to destructive interference? (1) 300 nm (2) 150 nm (3) 200 nm d n0 = 1 n1 = 1.5 n2 = 1.3 PHY 2049: Chapter 36 19 Quiz Consider an oil film (thickness d, n = 1.5) on top of water (n = 1.3). Light of = 600 nm is normally incident. Which value of d corresponds to destructive interference? (1) 300 nm (2) 150 nm (3) 200 nm Only 1 reflection has a phase shift, so d = m * 200 nm corresponds to min d n0 = 1 n1 = 1.5 n2 = 1.3 PHY 2049: Chapter 36 20 Quiz Consider a soap bubble of thickness d and n = 1.5. Light of = 600 nm is incident on the bubble. Which value of d corresponds to constructive interference? (1) 300 nm (2) 150 nm (3) 200 nm n d PHY 2049: Chapter 36 21 Quiz Consider a soap bubble of thickness d and n = 1.5. Light of = 600 nm is incident on the bubble. Which value of d corresponds to constructive interference? (1) 300 nm (2) 150 nm (3) 200 nm Only 1 reflection has a phase shift, so d = (m + 1/2) * 200 nm corresponds to max n d PHY 2049: Chapter 36 22 ...
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