Physics 102: Lecture 20, Slide 1 Hour Exam 3 Monday, Apr. 18 (two weeks from today!) –Lectures 14 – 21 –Homework through HW 11 –Discussions through Disc 11 Review session –Sunday, Apr. 17, 3pm, 141 Loomis –Will cover Fall ‘10 exam 3 Sign up for conflict exam!

Physics 102: Lecture 20, Slide 2 Physics 102: Lecture 20 Interference

Physics 102: Lecture 20, Slide 3 Phys 102 recent lectures Lecture 14 – EM waves Lecture 15 – Polarization Lecture 20 & 21 – Interference & diffraction Lecture 16 – Reflection Lecture 17 – Spherical mirrors & refraction Lecture 18 – Refraction & lenses Lecture 19 – Lenses & your eye Light as a wave Light as a ray

Physics 102: Lecture 20, Slide 4 Superposition t +1 t +1 t Constructive Interference In Phase

Physics 102: Lecture 20, Slide 5 Superposition t +1 t +1 t Destructive Interference Out of Phase 180 degrees

Physics 102: Lecture 20, Slide 6 ACT: Superposition + Different f 1) Constructive2) Destructive3) Neither

Physics 102: Lecture 20, Slide 7 Interference Requirements Need two (or more) waves Must have same frequency Must be coherent (i.e. waves must have definite phase relation)

Physics 102: Lecture 20, Slide 8 Demo: Interference for Sound … For example, a pair of speakers, driven in phase, producing a tone of a single f and : l1l1 l2l2 But this won’t work for light—can’t get coherent sources hmmm… I’m just far enough away that l 2 - l 1 = /2, and I hear no sound at all!

Physics 102: Lecture 20, Slide 9 Interference for Light … Can’t produce coherent light from separate sources. (f  Hz) Need two waves from single source taking two different paths –Two slits –Reflection (thin films) –Diffraction * Today’s lecture Next lecture

Physics 102: Lecture 20, Slide 10 Young’s double slit/rays Bright spots Shadow This is not what is actually seen! Monochromatic light travels through 2 slits onto a screen What pattern emerges on the screen?

Physics 102: Lecture 20, Slide 11 Young’s double slit/Huygens Recall Huygens’ principle: Every point on a wave front acts as a source of tiny wavelets that move forward. Wave crests in phase = constructive interference Bright and dark spots on screen! Constructive = bright Destructive = dark

Physics 102: Lecture 20, Slide 12 Young’s double slit: Key idea Key for interference is this small extra distance. Consider two rays traveling at an angle  : θ Bottom ray travels a little further (2 in this case)

Physics 102: Lecture 20, Slide 13 Young’s double slit: Quantitative Constructive: dsin(  ) = m Destructive: dsin(  ) = (m+1/2) Consider two rays traveling at an angle  Assume screen is very far away (L>>d): θ ≈ ≈ ≈ d L Path length difference = dsin(  ) θ where m = 0,  1,  2 m = +2 Need  < d

Physics 102: Lecture 20, Slide 14 Young’s double slit: Quantitative Constructive: dsin(  ) = m Destructive: dsin(  ) = (m+1/2) Assume screen is very far away (L>>d), angles  are small: θ d L dsin(  ) θ m = 0,  1,  2 m = 0 m = +1 m = -1 m = -2 m = +2 y sin(  )  tan(  ) = y/L y ≈ m L/d y ≈ (m+1/2) L/d

Physics 102: Lecture 20, Slide 15 ACT: Young’s Double Slit Screen a distance L from slits Single source of monochromatic light d 2 slits- separated by d A.Constructive B.Destructive C.Depends on L The rays start in phase, and travel the same distance, so they will arrive in phase. L Light waves from a single source travel through 2 slits before meeting on a screen. The interference will be:

Physics 102: Lecture 20, Slide 16 Preflight 20.1 Screen a distance L from slits Single source of monochromatic light d 2 slits- separated by d 1)Constructive 2)Destructive 3)Depends on L The rays start out of phase, and travel the same distance, so they will arrive out of phase. L ½ shift The experiment is modified so that one of the waves has its phase shifted by ½. Now, the interference will be:

Physics 102: Lecture 20, Slide 17 ACT: Preflight ) increases 2) same3) decreases Under water decreases so y decreases! 19%26%55% θ d L dsin(  ) θ m = 0 m = +1 m = -1 m = -2 m = +2 y When this Young’s double slit experiment is placed under water, the separation y between minima and maxima:

Physics 102: Lecture 20, Slide 18 Preflight 20.2 In the Young double slit experiment, is it possible to see interference maxima when the distance between slits is smaller than the wavelength of light? 1) Yes2) No Need: d sin  = m => sin  = m   d If   d then   d > 1 so sin  > 1 Not possible!

Physics 102: Lecture 20, Slide 19 Thin Film Interference n 1 (thin film) n2n2 n 0 =1.0 (air) t 1 2 Get two waves by reflection off two different interfaces: interference! Ray 2 travels approximately 2t further than ray 1. Light is incident normal to a thin film Note: angles exaggerated for clarity Why stop at 2 reflections?

Physics 102: Lecture 20, Slide 20 Reflection & Phase Shifts n1n1 n2n2 Upon reflection from a boundary between two transparent materials, the phase of the reflected light may change. If n 1 > n 2 – no phase change upon reflection If n 1 < n 2 – 180º phase change upon reflection (shift by /2) Incident Reflected n1n1 n2n2 Refracted Incident Reflected Refracted

Physics 102: Lecture 20, Slide 21 Thin Film Summary n 1 (thin film) n2n2 n = 1.0 (air) t 1 2 Ray 1:  1 = 0 or ½ Determine , number of extra wavelengths for each ray. If |  2 –  1 | = ½, 1 ½, 2 ½ …. (m + ½) destructive If |  2 –  1 | = 0, 1, 2, 3 …. (m) constructive Note: this is wavelength in film! ( film = o /n 1 ) + 2 t/ film ReflectionDistance Ray 2:  2 = 0 or ½ + 0 This is important!

Physics 102: Lecture 20, Slide 22 ACT: Thin Film Practice n 1 (thin film) n2n2 n = 1.0 (air) t 1 2 Blue light ( 0 = 500 nm) incident on a glass (n 1 = 1.5) cover slip (t = 167 nm) floating on top of water (n 2 = 1.3). A)  1 = 0B)  1 = ½C)  1 = 1 What is  1, the total phase shift for ray 1

Physics 102: Lecture 20, Slide 23 Is the interference constructive or destructive or neither? Thin Film Practice n 1 (thin film) n2n2 n = 1.0 (air) t 1 2 Blue light ( 0 = 500 nm) incident on a glass (n 1 = 1.5) cover slip (t = 167 nm) floating on top of water (n 2 = 1.3).  1 = ½  2 = 0 + 2t / glass = 2t n glass / 0 = (2)(167)(1.5)/500) = 1 Phase shift = |  2 –  1 | = ½ wavelength

Physics 102: Lecture 20, Slide 24 ACT: Thin Film Practice II n 1 (thin film) n2n2 n = 1.0 (air) t 1 2 Blue light ( 0 = 500 nm) incident on a glass (n 1 = 1.5) cover slip (t = 167 nm) floating on top of plastic (n 2 = 1.8).  1 = ½  2 = ½ + 2t / glass = ½ + 2t n glass / 0 = (2)(167)(1.5)/500) = 3/2 Phase shift = |  2 –  1 | = 1 wavelength Is the interference : 1) constructive 2) destructive 3) neither?

Physics 102: Lecture 20, Slide 25 See you Wednesday!