PS 13 Fall 14, Spendier Lecture 3/Page 1 Lecture today: Chapter 35 Interference 1) Intensity in Double-Slit Interference ) Thin Film Interference Announcements: - Shortened office hours this Thursday (1-1:3am). Please e-mail me to find another time in case you planned on stopping by my office between 1:3-11am Last lecture: started diffraction & interference of light Waves can interfere: that is, combine together to form complex wave patterns Interference is best demonstrated using coherent waves. Coherence means that waves have a phase relationship that is maintained for many cycles - Laser light is coherent - light from slits S 1 and S in the double slit experiment is said to be completely coherent - direct sunlight/light bulb is partially coherent - Young s two slit Interference xperiment: example of the interference of light waves Constructive interference, bright fringes (two slits) d sinθ = mλ (m =, ±1, ±, ±3, ) Destructive interference, dark fringes (two slits) d sinθ = (m+1/) λ (m =, ±1, ±, ±3, )
PS 13 Fall 14, Spendier Lecture 3/Page 1) Intensity in Double-Slit Interference Goal is to find the intensity at any point in the fringe pattern of double slit experiment A) Superposition of two waves: combine the two sinusoidally varying fields from two sources at a point P, taking proper account of the phase difference ϕ of the two waves at point P total (x,t) = 1 (x,t) + (x,t) = cos(kx-ωt) + cos(kx-ωt+ ϕ) total (x,t) = cos (ϕ/)cos(kx-ωt+ ϕ/) B) The intensity is then proportional to the square of the resultant electric-field amplitude I max c amplitude: max = cos (ϕ/) Intensity of interfering waves as a function of phase constant ϕ I cos ( /) 4 cos ( /) = 4 cos ( /) max I c c c I I 4 cos / Here we used Intensity for original wave: I c Now recall that recall that a phase difference of π rad corresponds to one wavelength L L d sin Fully constructive interference (bright fringes) occurs when / m, for m =,1,,... Fully destructive interference (dark fringes) occurs when / ( m 1/ ) Therefore at a bright fringe I 4I cos / 4I cos m since m is an integer I 4I at,, 4,... bright and dark fringe: I 4Icos / 4Icos m1/ I at 1, 3, 5,... dark
PS 13 Fall 14, Spendier Lecture 3/Page 3 Since we used any integer in this proof, it does not matter which intensity peak we choose, the maximum intensity is the same for each bright fringe. This is true only if d << λ (We will see in chapter 36 that is d << λ, then the intensity profile of light hitting the screen will have an envelope and fall off. Also, in reality, the waves get less intense as they propagate. So the further away you are, the less intense the pattern. On the screen, the intensity maxima wil be dimmer the wider out you go. ) Thin Film Interference The colors we see when sunlight illuminates a soap bubble or an oil slick are caused by the interference of light waves reflected from the front and back surfaces of a thin transparent film. To understand, why soap bubbles show vibrant color patterns, even though soapy water is colorless, we need to reconsider the change of phase of an M wave due to reflection.
PS 13 Fall 14, Spendier Lecture 3/Page 4 Thin and Thick Films: We emphasized thin films, since for two waves to cause a steady interference pattern, the waves must be coherent! Sun light and a light bulb emits light in a stream of short bursts, each of which is only a few micrometers long. - if light reflects from two surfaces of a thin film, the two reflected waves are part of the same burst and they are coherent ==> steady interference pattern - if the film is two thick, the two reflected waves will belong to different bursts and they are not coherent ==> no steady interference pattern Phase shifts during reflection: Depending on the differences in index of refraction, an M wave can undergo or not undergo a phase change upon reflection: Recall wave on string: a)free boundary c)fixed boundary Hence: An electromagnetic wave undergoes a phase change of 18 o upon reflection from a medium that has a higher index of refraction than the one in which the wave is traveling.
PS 13 Fall 14, Spendier Lecture 3/Page 5 For simplicity, we assume that the incident light ray is almost perpendicular to the film (θ ) Case 1: Reflection with no phase shift: Consider a thin film in a medium that has a larger index of refraction than the film. There are reflections at surfaces A and B Now recall that a phase difference of π rad corresponds to one wavelength L b λ b : Light wavelength in the film. The path difference between waves 1 and is L for normal incidence. Fully constructive interference occurs when / m, for m =,1,,... = L π = mπ λ b or L = mλ b Fully destructive interference occurs when / ( m 1/ ) L m 1/ or b L m1/ b NOT: there relations also hold both waves have a π phase shift. Case : Reflection with phase shift: Consider a thin film in a air. Since n air < n film, phase shift of π, or halfwavelength occurs upon reflection. Hence in this case: Fully constructive interference occurs when Lm 1/ film Fully destructive interference occurs when L m film since λ film = λ air /n
PS 13 Fall 14, Spendier Lecture 3/Page 6 Fully constructive interference (maxima - bright film in air) air Lm 1/ n film Fully destructive interference (minima - dark film in air) air L m n film SUMMARY: Normal incidence No phase shift or both have π-shift One of the two wave has π- shift L =mλ film Constructive reflection Destructive reflection L = (m+1/) λ film Destructive reflection Constructive reflection m=,1,,3...
PS 13 Fall 14, Spendier Lecture 3/Page 7 xample: White light in air shines on an oil film (n = 1.5) that floats on water (n = 1.33). When looking straight down at the film, the reflected light is red, with a wavelength of 636 nm. What is the minimum possible thickness of the film? If 16 nm gives constructive interference for red light, what about the other colors? They are not completely cancelled out, because 16 nm is not the right thickness to give completely destructive interference for any wavelength in the visible spectrum. The other colors do not reflect as intensely as red light, so the film looks red.