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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 d far < To correct, produce virtual image of far object d 0 = at the far point (d i = d far ). Nearsightedness.
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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 d0 = at the far point (di = dfar)
Nearsightedness • Far point dfar < • To correct, produce virtual image of far object d0 = 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 d0 = 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
+1 t -1 +1 t -1 +2 t -2 Superposition ConstructiveInterference + In Phase 5
+2 t -2 Superposition Destructive Interference +1 t -1 + +1 Out of Phase 180 degrees t -1 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
hmmm… I’m just far enough away that l2-l1=l/2, and I hear no sound at all! l1 l2 Interference for Sound … For example, a pair of speakers, driven in phase, producing a tone of a single f and l: But this won’t work for light--can’t get coherent sources 15
Two different paths Single source Interference possible here 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* 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: • Constructive • Destructive • Depends on L d The rays start in phase, and travel the same distance, so they will arrive in phase. Single source of monochromatic light L 2 slits-separated by d Screen a distance L from slits 23
½ l shift 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: • Constructive • Destructive • Depends on L d The rays start out of phase, and travel the same distance, so they will arrive out of phase. Single source of monochromatic light L 2 slits-separated by d Screen a distance L from slits 25
At points where the difference in path length is the screen is dark. (destructive) 2 slits-separated by d Young’s Double Slit Concept At points where the difference in path length is 0, l,2l, …, the screen is bright. (constructive) d Single source of monochromatic light L 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
Constructiveinterference Destructive interference where m = 0, or 1, or 2, ... Young’s Double Slit Quantitative d d Path length difference = d sin q Need l < d 32
y Young’s Double Slit Quantitative L 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
Need: d sin q = ml => sin q = ml / d Not possible! 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 2) No If l > d then l / d > 1 sosin q > 1 35
1 2 Thin Film Interference n0=1.0 (air) n1 (thin film) t n2 Get two waves by reflection off two different interfaces. Ray 2 travels approximately2t further than ray 1. 37
Reflected wave Incident wave n1 n2 Reflection + Phase Shifts Upon reflection from a boundary between two transparent materials, the phase of the reflected light may change. • If n1 > n2 - no phase change upon reflection. • If n1 < n2 - phase change of 180º upon reflection. (equivalent to the wave shifting by l/2.) 39
This is important! Distance Reflection Thin Film Summary Determine d, number of extra wavelengths for each ray. 1 2 n = 1.0 (air) n1 (thin film) t n2 Note: this is wavelength in film! (lfilm= lo/n1) Ray 1: d1 = 0 or ½ + 0 Ray 2: d2 = 0 or ½ + 2 t/ lfilm If |(d2 – d1)| = 0, 1, 2, 3 …. (m) constructive If |(d2 – d1)| = ½ , 1 ½, 2 ½ …. (m + ½) destructive 42
What is d1, the total phase shift for ray 1 A) d1 = 0 B) d1 = ½ C) d1 = 1 Example Thin Film Practice (ACT) 1 2 n = 1.0 (air) nglass = 1.5 t 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? 45
Example Thin Film Practice 1 2 n = 1.0 (air) nglass = 1.5 t 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 d1 = ½ d2 = 0 + 2t / lglass = 2t nglass/ l0= (2)(167)(1.5)/500) =1 Phase shift = d2 – d1 = 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 (B) destructive (C) neither 1 2 n=1 (air) nglass =1.5 t nplastic=1.8 Reflection at both interfaces! d1 = ½ d2 = ½ + 2t / lglass = ½ + 2t nglass/ l0= ½ + 1 Phase shift = d2 – d1 = 1 wavelength 48
nair=1.0 t =l noil=1.45 ngas=1.20 nwater=1.3 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 (nwater=1.33). When the thickness of the two films is exactly one wavelength… • The gas looks: • bright • dark • The oil looks: • bright • dark d1,gas = ½ d2,gas = ½ + 2 d1,oil = ½ d2,oil = 2 | d2,oil – d1,oil | = 3/2 | d2,gas – d1,gas | = 2 constructive destructive 50