PHYS-2020: General Physics IICourse Lecture NotesSection XIIIDr. Donald G. LuttermoserEast Tennessee State UniversityEdition 3.3AbstractThese class notes are designed for use of the instructor and students of the course PHYS-2020:General Physics II taught by Dr. Donald Luttermoser at East Tennessee State University. Thesenotes make reference to the College Physics, 9th Edition (2012) textbook by Serway and Vuille.XIII. Wave OpticsA. Wave Interference and Young’s Double Slit Experiment.1. We have already discussed wave interference (see VII.E.2 on PageVII-15 and VIII.E on Page VIII-12). In this section we will ex-pand upon t he p hysics of interference.2. In ord er to carry out experiments concerning constructive or de-struct ive interference of waves of two or more sources, the wavesfrom the sources must be coherent.a) This means that the waves they emit must maintain aconstant phase with respect to each other.b) In addition to this, the waves must have identical wave-lengths to be coherent.i) Both of these conditions require the sources toemit a single emission-line spectrum.ii) A laser (light amplification of st imulated emissionof radiation) is a good source of coherent rad iation.c) Most ordinary light sources p roduce a continuous spec-trum, hence produce incoherent light. Incoherent lightcan be converted to coherent light through the use of two“pinholes” or “slits” at a specific separation as describedbelow.3. In 1801, Thomas Young first demonstrated interference in lightwaves from two sources produced from a sin gle source of lightpassing through two pinholes in a screen as shown in Figure 24.1XIII–1XIII–2 PHYS-2020: General Physics I Iin your textbook. (From this point forward, we will describe thisexperiment with slits in stead of pinh oles.)a) This experiment showed that the resultin g wavefronts fromthe two slits produced as series of bright and dark parallelbands called fringes.b) In order for the wavefronts (i.e., wavecrests) of the lightfrom the two different slits to arrive on a proj ection screenat the same time at t he same place, the following equationmust be trueδ = r2− r1= d sin θ , (XIII-1)where r1is the distance that the wavefront traveled fromslit 1, r2the distance traveled from slit 2, δ is the pathdifference between r2and r1, d the distance between t heslits, and θ is the angle shown in the figure below.S2QS1dslit planeSourceLviewingscreenOPyr2r1δ = r2 − r1 = d sin θθθDonald G. Luttermoser, ETSU XIII–3c) The “bright fringes” correspond to regions of construc-tive interference. This will occur at a position on theviewing screen when δ = ±mλ:δ = d sin θbright= mλ , m = 0, ±1, ±2, . . . (XIII-2)i) The number m is called the order number.ii) The central bright fringe at θbright= 0 is called thezeroth-order maximum.iii) The first m aximum on either side, where m =±1, is called the first-order maximum, and so forth.d) The “dark fringes” correspond to regions of destructiveinte rference. This will occur at a position on the viewingscreen when the waves are 180◦out of phase (i.e. , δ is anod d multiple of λ/2):δ = d sin θdark= m +12!λ , m = 0, ±1, ±2, . . .(XIII-3)i) Here, the first dark fringe occurs at m = 0 givinga path difference of λ/2.ii) The second dark fringe occurs at m = 1 giving apath difference of 3λ/2, and so on.e) Let L be the distance between the “slit plane” and th eviewing screen an d y be the distance that a frin ge is awayfrom the zeroth-order maximum on the viewing scr een.Then, if L  d and d  λ, which is typically the casein these types of experiments, we note fr om trigonometrythaty = L tan θ ≈ L sin θ . (XIII-4)XIII–4 PHYS-2020: General Physics I Ii) Using this approximation in Eq . (XIII-2), we findthat th e positions of the bright fri nges measu redfrom the zeroth-order b right f ringe isybright=λLdm, m = 0, ±1, ±2, . . . (XIII-5)ii) Now, using this approximation in Eq. (XIII-3)givesydark=λLd m +12!, m = 0, ±1, ±2, . . .(XIII-6)for the location of the dark fringes as measuredfrom the zeroth-order b right f ringe.4. Young’s double-slit experiment provides a meth od for measuringthe wavelength of light. In addition, h is experiment gave thewave model of light a great d eal of credibility.Example XIII–1. Problems 24.3 (Page 852) from the Ser-way & Vuille textbook: A pair of narrow, parallel slits separatedby 0.250 mm is illuminated by the green component from a mercuryvapor lamp (λ = 546.1 nm). The interference pattern is observedon a screen 1.20 m from the plane of the parallel slits. Calculate thedistance ( a) f r om the central maximum to the first bright region oneither side of the central maximum and (b) between the first andsecond dark bands in the interference pattern.Solution (a):Here, we have λ = 546.1 nm = 546.1×10−9m, L = 1.20 m, and d= 0.250 mm = 0.250×10−3m. The distance between the centralmaximum and the first-order b right fringe can be d eterminedwith the use of Eq. (XIII-5):∆y = ybright|m=1− ybright|m=0=λLd· 1 − 0Donald G. Luttermoser, ETSU XIII–5=λLd=546.1 × 10−9m(1.20 m)0.250 × 10−3m= 2.62 × 10−3m = 2.62 mm .Solution (b):The distance between the first and second dark bands can bedetermined with Eq. (XIII-6):∆y = ydark|m=1− ydark|m=0="λLd 1 +12!#−"λLd 0 +12!#=3λL2d−λL2d=λLd= 2.62 × 10−3m = 2.62 mm .B. Interference in Thin Films.1. Consider light reflected off of a surface. Let nibe the in dex ofrefraction of the incident side of the surface and nobe the index ofrefraction on the opposite side (i . e., the “other” side). With thisset up, we can write the Rules of Phase Change of ReflectedLight:a) If no> ni, any light reflected off of this surface will un-dergo a 180◦phase change upon reflection.b) If no< ni, any light reflected off of t his surface will expe-rience no phase change upon reflection.2. Consider a film of uniform thickness ` (note that your tex tbookuses t for this variable) with an index of refraction n (as shownXIII–6 PHYS-2020: General Physics I Iin Figure 24.7 in the textbook). We will label the top side of thefilm surface A and the bottom side surface B. On either side ofthis film, we will assume that air is the medium and hence setnoutside= nair= 1.a) An incident beam passes through air and interacts withsurface A. Part of the beam gets reflected (which we willcall ray ‘1’), where the reflected ray has an 180◦phasechange since no> ni, and part of th e ray is