Physics 102 Lecture 20 Interference But firstMore on

Physics 102: Lecture 20 Interference

But first…More on the Eye • Recall the lens formula: • Normal eyes – Far point = – Near point = 25 cm • Nearsightedness – Far point dfar < – To correct, produce virtual image of far object d 0 = at the far point (di = dfar)

Nearsightedness • Far point dfar < • To correct, produce virtual image of far object d 0 = at the far point (di = dfar) flens = - dfar 1/flens = -1/dfar Example: • My prescription reads -6. 5 dipoters • flens = -1/6. 5 = -0. 154 m = -15. 4 cm (a diverging lens) • dfar = 15. 4 cm (!)

Farsightedness • Near point dnear > 25 cm • To correct, produce virtual image of object at d 0 = 25 cm to the near point (di = dnear) Example: • My near prescription reads +2. 5 dipoters • flens = +1/2. 5 = 0. 4 m = 40 cm • therefore dnear = 67 cm (with my far correction)

Magnifying Glasses • Will not cover in class • Read on your own – Lecture 19 at the very end – Text Ch. 24. 4

Superposition Constructive Interference +1 -1 t + +1 -1 In Phase t +2 t -2 5

Superposition Destructive Interference +1 -1 t + +1 -1 Out of Phase t 180 degrees +2 t -2 7

Superposition ACT + Different f 1) Constructive 2) Destructive 3) Neither 10

Interference Requirements • Need two (or more) waves • Must have same frequency • Must be coherent (i. e. waves must have definite phase relation) 12

Interference for Sound … For example, a pair of speakers, driven in phase, producing a tone of a single f and : hmmm… I’m just far enough away that l 2 l 1= /2, and I hear no sound at all! l 1 l 2 But this won’t work for light--can’t get coherent sources 15

Interference for Light … • Can’t produce coherent light from separate sources. (f 1014 Hz) • Need two waves from single source taking two different paths – Two slits – Reflection (thin films) – Diffraction* Single source Two different paths Interference possible here 18

ACT: Young’s Double Slit Light waves from a single source travel through 2 slits before meeting on a screen. The interference will be: A. Constructive B. Destructive C. Depends on L d Single source of monochromatic light L 2 slitsseparated by d The rays start in phase, and travel the same distance, so they will arrive in phase. Screen a distance L from slits 23

Preflight 20. 1 The experiment is modified so that one of the waves has its phase shifted by ½ l. Now, the interference will be: 1) Constructive ½ l shift 2) Destructive 3) Depends on L d Single source of monochromatic light L 2 slitsseparated by d The rays start out of phase, and travel the same distance, so they will arrive out of phase. Screen a distance L from slits 25

Young’s Double Slit Concept At points where the difference in path length is 0, , 2 , …, the screen is bright. (constructive) d Single source of monochromatic light At points where the difference in path length is L 2 slitsseparated by d the screen is dark. (destructive) Screen a distance L from slits 27

Young’s Double Slit Key Idea L Two rays travel almost exactly the same distance. (screen must be very far away: L >> d) Bottom ray travels a little further. Key for interference is this small extra distance. 30

Young’s Double Slit Quantitative d d Path length difference = d sin q Constructive interference Destructive interference where m = 0, or 1, or 2, . . . Need l < d 32

Young’s Double Slit Quantitative L y d A little geometry… sin(q) tan(q) = y/L Constructive interference Destructive interference where m = 0, or 1, or 2, . . . 33

Preflight 20. 3 L y d When this Young’s double slit experiment is placed under water. The separation y between minima and maxima 1) increases 2) same 3) decreases 21% 23% 53% Under water l decreases so y decreases 35

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) Yes Need: d sin q = m l If l>d 2) No => sin q = m l / d then l/d>1 so sin q > 1 Not possible! 35

Thin Film Interference 1 2 t n 0=1. 0 (air) n 1 (thin film) n 2 Get two waves by reflection off two different interfaces. Ray 2 travels approximately 2 t further than ray 1. 37

Reflection + Phase Shifts Incident wave Reflected wave n 1 n 2 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 - phase change of 180º upon reflection. (equivalent to the wave shifting by /2. ) 39

Thin Film Summary Determine d, number of extra wavelengths for each ray. 1 2 n = 1. 0 (air) n 1 (thin film) t This is important! n 2 Reflection Distance Ray 1: d 1 = 0 or ½ + 0 Ray 2: d 2 = 0 or ½ + 2 t/ lfilm If |(d 2 – d 1)| = 0, 1, 2, 3 …. If |(d 2 – d 1)| = ½ , 1 ½, 2 ½ …. Note: this is wavelength in film! (lfilm= lo/n 1) (m) constructive (m + ½) destructive 42

Thin Film Practice (ACT) 1 2 t n = 1. 0 (air) nglass = 1. 5 nwater= 1. 3 Blue light (lo = 500 nm) incident on a glass (nglass = 1. 5) cover slip (t = 167 nm) floating on top of water (nwater = 1. 3). Is the interference constructive or destructive or neither? What is d 1, the total phase shift for ray 1 A) d 1 = 0 B) d 1 = ½ C) d 1 = 1 45

Thin Film Practice 1 2 t n = 1. 0 (air) nglass = 1. 5 nwater= 1. 3 Blue light (lo = 500 nm) incident on a glass (nglass = 1. 5) cover slip (t = 167 nm) floating on top of water (nwater = 1. 3). Is the interference constructive or destructive or neither? Reflection at air-film interface only d 1 = ½ d 2 = 0 + 2 t / lglass = 2 t nglass/ l 0= (2)(167)(1. 5)/500) =1 Phase shift = d 2 – d 1 = ½ wavelength 45

ACT: Thin Film Blue light l = 500 nm incident on a thin film (t = 167 nm) of glass on top of plastic. The interference is: (A) constructive 1 2 t (B) destructive n=1 (air) nglass =1. 5 nplastic=1. 8 (C) neither d 1 = ½ Reflection at both interfaces! d 2 = ½ + 2 t / lglass = ½ + 2 t nglass/ l 0= ½ + 1 Phase shift = d 2 – d 1 = 1 wavelength 48

Preflights 20. 4, 20. 5 A thin film of gasoline (ngas=1. 20) and a thin film of oil (noil=1. 45) are floating on water t= (nwater=1. 33). When the thickness of the two films is exactly one wavelength… The gas looks: • bright • dark d 1, gas = ½ d 2, gas = ½ + 2 | d 2, gas – d 1, gas | = 2 constructive nair=1. 0 ngas=1. 20 noil=1. 45 nwater=1. 3 The oil looks: • bright • dark d 1, oil = ½ d 2, oil = 2 | d 2, oil – d 1, oil | = 3/2 destructive 50

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