Reflection and Interference from Thin Films

1 PHY 2049: Chapter 3614 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 dn1n2n0= 1(Angle shown, but actuallynormally incident)PHY 2049: Chapter 3615 Standard analysis of Thin Film Interference ()11122 Max ( constructive )2 Min (destructive)nndmdm ==+dn1n2(Angle shown, but actuallynormally incident)n0= 111nn =Wavelength inside film!

2 PHY 2049: Chapter 3616 Example of Thin Film , 362 nm, 543 nm,dm ==.. ==Let = 500 n1= (MgF2) n2= (glass)Max intensityMin intensity() nm, 272 nm, 453 nm,dm =+ =..Must be careful about phase shift at boundary: Reflection for nin< nouthas phase shift, 0 if nin> nout Since n0< n1and 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 3617 Thin Film Interference for Soap Bubble()122 Min (destructive)2 Max ( constructive )nndmdm ==+nn =Wavelength inside soap!ndSimilar analysis, Phase shift of for air soap Reflection No phase shift for soap air Reflection Net phase shift switchesmax minPHY 2049: Chapter 3618 Example of Soap Bubble nm, 379 nm, 568 nm,dm ==.

3 ==Let = 500 n = (soap + water)Min intensityMax intensity() nm, 284 nm, 473 nm,dm =+ =..PHY 2049: Chapter 3619 Quiz What is the condition for destructive Interference for light reflecting from a soap bubble of thickness d? (1) (2) 2ndm =()122ndm =+Only 1 Reflection has a phase shift,so this switches min and maxPHY 2049: Chapter 3620 Quiz Consider an oil film (thickness d, n = ) on top of water (n = ). Light of = 600 nm is normally incident. Which value of d corresponds to destructive Interference ? (1) 300 nm (2) 150 nm (3) 200 nmdn1n2n0= 1 Only 1 Reflection has a phase shift,so d = m * 200 nmPHY 2049: Chapter 3621 Quiz Consider a soap bubble of thickness d and n = Light of = 600 nm is incident on the bubble.

4 Which value of d corresponds to destructive Interference ? (1) 300 nm (2) 150 nm (3) 200 nmndOnly 1 Reflection has a phase shift,so d = m * 200 nmPHY 2049: Chapter 3622 Diffraction Grating: 1000s of Slits!PHY 2049: Chapter 3623 Diffraction Grating Analysis similar to double slit Many slits instead of 2 Slits still separated by distance d Maxima again occur only for Maxima are much sharpersindm = 2 3 4 LWavesGratingN = 4m = 1ddddPHY 2049: Chapter 3624 Intensity Pattern for Diffraction Grating Calculation of intensity vs Basically a product of single slit and multiple slit formula Let a= slit width, d= slit separation, N= # of slits When N is large, maxima are extremely sharp For central peak, angular half-width Imagine N 10000 100000!

5 22maxsinsinsinNIIN = sin /sin /ad ==/hwNd PHY 2049: Chapter 36252 Slits: d = 4 PHY 2049: Chapter 36264 Slits: d = 4 PHY 2049: Chapter 362710 Slits: d = 4 PHY 2049: Chapter 362820 Slits: d = 4 PHY 2049: Chapter 3629 Diffraction Gratings in Astronomy Use to determine wavelength ( measured) Use to determine wavelength ( measured) Typically have N= 60,000 100,000!! Sharp peaks allow closely spaced wavelengths to be resolved Accuracy better than nm, nm Important for distinguishing element signatures Find all the elements in a stellar spectrumsindm =PHY 2049: Chapter 3630 Example: Separate = 600, = nm Poorly resolved with 15,000 slit grating Too few slits lines too wide to tell apartPHY 2049: Chapter 3631 Example (cont.)

6 Lines easily resolved with 60,000 slit gratingPHY 2049: Chapter 3632 Example: Resolution of Sodium D1, D2 Lines High resolution solar spectrum near sodium absorption linesSodium lines DD12589 54588 94==..nmnm = nm