Reflection and Interference from Thin Films
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Reflection and Interference from Thin Films ÎNormal-incidence light strikes surface covered by a thin film Some rays reflect from film surface Some rays reflect from substrate surface (distance d further) ÎPath length difference = 2d causes interference From d full constructive to full destructive, depending on λ (Angle shown, but actually normally incident) n0 = 1 n1 n2 PHY 2049: Chapter 36 14 Standard analysis of Thin Film Interference 2d = mλn1 ( 2d = m + d Max (constructive) 1 2 ) λn 1 Min (destructive) λn1 = λ n1 Wavelength inside film! (Angle shown, but actually normally incident) n0 = 1 n1 n2 PHY 2049: Chapter 36 15 Example of Thin Film Interference Let λ = 500 λ1.38 n1 = 1.38 (MgF2) n2 = 1.5 (glass) 500 = = 362.3nm 1.38 Max intensity d = 12 mλ1.38 = 181nm, 362 nm, 543nm, … Min intensity d= 1 2 ( m + 12 ) λ1.38 = 90.6 nm, 272 nm, 453nm, … Must be careful about phase shift at boundary: Reflection for nin < nout has ½ λ phase shift, 0 if nin > nout ¾ Since n0 < n1 and n1 < n2, the phase shift has no effect here ¾ But for other cases, there can be an extra ½ λ phase shift ¾ Soap bubble (next slide) PHY 2049: Chapter 36 ¾ 16 Thin Film Interference for Soap Bubble λ 2d = mλn Min (destructive) λn = ( 2d = m + 1 2 ) λn n Max (constructive) Wavelength inside soap! Similar analysis, but… ¾ Phase shift of ½ λ for air–soap reflection ¾ No phase shift for soap–air reflection ¾ Net ½ λ phase shift switches max ↔ min n d PHY 2049: Chapter 36 17 Example of Soap Bubble Interference Let λ = 500 λ1.32 n = 1.32 (soap + water) 500 = = 379 nm 1.32 Min intensity d = 12 mλ1.32 = 189 nm, 379 nm, 568nm, … Max intensity d= 1 2 ( m + 12 ) λ1.32 = 94.7 nm, 284 nm, 473nm, … PHY 2049: Chapter 36 18 Quiz ÎWhat is the condition for destructive interference for light reflecting from a soap bubble of thickness d? (1) 2d = mλn (2) 2d = m + 12 λn ( ) Only 1 reflection has a phase shift, so this switches min and max PHY 2049: Chapter 36 19 Quiz ÎConsider an oil film (thickness d, n = 1.5) on top of water (n = 1.3). Light of λ = 600 nm is normally incident. Which value of d corresponds to destructive interference? (1) 300 nm (2) 150 nm (3) 200 nm d Only 1 reflection has a phase shift, so d = m * 200 nm n0 = 1 n1 n2 PHY 2049: Chapter 36 20 Quiz ÎConsider a soap bubble of thickness d and n = 1.5. Light of λ = 600 nm is incident on the bubble. 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 n d PHY 2049: Chapter 36 21 Diffraction Grating: 1000s of Slits! PHY 2049: Chapter 36 22 Diffraction Grating ÎAnalysis similar to double slit Many slits instead of 2 Slits still separated by distance d Maxima again occur only for d sin θ Maxima are much sharper = mλ Grating N=4 d d m=1 λ 2λ d 3λ d 4λ θ L Waves PHY 2049: Chapter 36 23 Intensity Pattern for Diffraction Grating ÎCalculation of intensity vs θ Basically a product of single slit and multiple slit formula Let a = slit width, d = slit separation, N = # of slits 2 ⎛ sin α ⎞ ⎛ sin N β ⎞ I = I max ⎜ ⎟ ⎜ ⎟ α β N sin ⎝ ⎠ ⎝ ⎠ 2 α = π a sin θ / λ β = π d sin θ / λ ÎWhen N is large, maxima are extremely sharp For central peak, angular half-width Δθ hw λ / Nd Imagine N ≅ 10000 – 100000! PHY 2049: Chapter 36 24 2 Slits: d = 4λ PHY 2049: Chapter 36 25 4 Slits: d = 4λ PHY 2049: Chapter 36 26 10 Slits: d = 4λ PHY 2049: Chapter 36 27 20 Slits: d = 4λ PHY 2049: Chapter 36 28 Diffraction Gratings in Astronomy to determine wavelength (θ measured) Use d sin θ = mλ to determine wavelength (θ measured) ÎUse have N = 60,000 – 100,000!! Sharp peaks allow closely spaced wavelengths to be resolved Accuracy better than 0.001 nm, e.g. 589.605 ± 0.001 nm Typically ÎImportant Find for distinguishing element “signatures” all the elements in a stellar spectrum PHY 2049: Chapter 36 29 Example: Separate λ = 600, λ = 600.05 nm ÎPoorly Too resolved with 15,000 slit grating few slits ⇒ lines too wide to tell apart PHY 2049: Chapter 36 30 Example (cont.) ÎLines easily resolved with 60,000 slit grating PHY 2049: Chapter 36 31 Example: Resolution of Sodium D1, D2 Lines ÎHigh resolution solar spectrum near sodium absorption lines Sodium lines λ D1 = 589.54 nm λ D2 = 588.94 nm Δλ = 0.60 nm PHY 2049: Chapter 36 32
