Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is virtual (since i 2 < 0). 30 The magnification is m = (-i 1 /p 1 ) x (-i 2 /p 2 ) = (-40/40)x(30/-30) =+1, so the image has the same size orientation as the object. f1f1 f1f1 Lens 1Lens 2 f2f2 f2f Actual ray diagram purple. Dashed lines are virtual rays

Lecture 15 Interference Chp. 35 Topics –Interference is due to the wave nature of light –Huygen’s principle, Coherence –Young’s Interference Experiment and demo –Intensity in double slit experiment –Change in wavelength and phase change in a medium - soap bubble –Interference from thin films Demos Polling

Huygen’s Principle, Wavefronts and Coherence k Examples of coherence are: Laser light Small spot on tungsten filament Wavefront Most light is incoherent: Two separate light bulbs Two headlight beams on a car Sun is basically incoherent

In order to form an interference pattern, the incident light must satisfy two conditions: (i) The light sources must be coherent. This means that the plane waves from the sources must maintain a constant phase relation. For example, if two waves are completely out of phase with φ = π, this phase difference must not change with time. (ii) The light must be monochromatic. This means that the light consists of just one wavelength λ = 2π/k. Light emitted from an incandescent lightbulb is incoherent because the light consists of waves of different wavelengths and they do not maintain a constant phase relationship. Thus, no interference pattern is observed.

An important question to consider is once light is in phase what are the ways it can get out of phase 1. The selected rays travel different distances. 2. Rays go through different material with different index of refraction 3. Reflection from a medium with greater index of refraction First what do we mean by out of phase or in phase

In Phase Out of Phase by 180 degrees or  radians or  /2 In between Destructive interferenceConstructive interference

1) Lets consider the first case. How does light get out of phase when selected rays travel different distances. This leads to the famous Young’s double slit experiment

Young’s Double Slit Interference Experiment m=0 m=1 m=2  D y

Constructive interference Destructive interference If you now send the light from the two openings onto a screen, an interference pattern appears, due to differing path lengths from each source we have constructive interference if paths differ by any number of full wavelengths destructive interference if difference is half a wavelength longer or shorter

Constructive interference Destructive interference We want to find the pathlength difference

How do we locate the vertical position of the fringes on the screen? We want to relate y to m, L and the angle 1) L >> d 2) d >> λ These tell us that θ is small Therefore, y is the distance to the m’th fringe

m+/- y m L /d 2L /d 3L /d Maxima m+/- y m L /2d 3L /2d 5L /2d 7L /2d Minima ymym

Young’s Double Slit Interference Experiment m=0 m=1 m=2  D y

Problem 13E. Suppose that Young’s experiment is performed with blue-green light of 500 nm. The slits are 1.2mm apart, and the viewing screen is 5.4 m from the slits. How far apart the bright fringes? From the table on the previous slide we see that the separation between bright fringes is How far off the axis is the m=3 bright fringe? y =3L /d = 3S=6.75 mm

What is the Intensity Distribution along the screen The electric field at P is sum of E 1 and E 2. The intensity is proportional to the square of the net amplitude. represents the correlation between the two waves. For incoherent light, as there is no definite phase relation between E 1 and E 2 and cross term vanishes and incoherent sum is I

For coherent sources, the cross term is non-zero. In fact, for constructive interference E 1 = E 2 and I = 4I 1 For destructive interference E 1 = -E 2 and and correlation term = - I 1, the total intensity becomes I = I 1 -2I 1 + I 1 = 0 Suppose that the waves emerged from the slits are coherent sinusoidal plane waves. Let the electric field components of the wave from slits 1 and 2 at P be given by We have dropped the kx term by assuming that P is at the origin (x =0) and we have acknowledged that wave E 2 has traveled farther by giving it a phase shift ( ϕ ) relative to E 1 For constructive interference, with path difference of δ = λ would correspond to a phase shift of ϕ =2π. This then implies

E1E1 E2E2 E0E0 E0E0 E tt 

The intensity I is proportional to the square of the amplitude of the total electric field Then..

What about the intensity of light along the screen?

2) How does a wave get out of phase when passing through different media with different indices of refraction? You must consider the thickness of the two materials in numbers of wavelength N 1 and N 2. Note that the velocity and wavelength changes but not the frequency. Path length difference = c = f Vacuum v n = f n  f /n n=c/v

Concept of path length difference, phase and index of refraction c = f Vacuum v n = f n  f /n n=c/v Rays are in phase if  where m  Rays are out of phase if where m=1, 2, 3

n2n2 n1n1 air water E Explain using wave machine Demonstrate using rope 3) The wave changes phase by 180 degrees when reflection from a medium with greater index of refraction. See cartoon below. E stands for the Electric field of the wave

air 1.0 air 1.00 soap 1.30 L eye 2 1 Reflection 180 deg phase change for Ray 1when reflecting from the soap film surface but not ray 2 What are the conditions for Constructive Interference in reflection from a soap bubble? We must add a half wavelength to account for the 180 deg phase change for constructive interference. Path difference must = integral number of wavelengths plus 1/2 a wavelength. Now consider the path length differences First consider phase change upon reflection Answer

soap 1.30 What are the conditions for Constructive Interference in transmission from a soap bubble? air 1.0 air 1.00 L eye Transmission 2 1 m = 1,2,3,4…. For constructive interference path difference must = integral number of wavelengths No phase changes upon reflection n =1.30 Demonstrate with soap bubble Why does it look black on top? What are the conditions for destructive interference?

39. A disabled tanker leaks kerosene (n = 1.20) into the Persian Gulf, creating a large slick on top of the water (n = 1.30). (a) If you are looking straight down from an airplane while the Sun is overhead at a region of the slick where its thickness is L=460 nm, for which wavelength(s) of visible light is the reflection brightest because of constructive interference? air 1.0 Water 1.30 Kerosene 1.20 L deg phase change

(b) If you are scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest? (Hint: use figure (a) with appropriate indices of refraction.) Scuba diver air 1.0 Water 1.30 Kerosene 1.20 L 2 1

We note that only the nm wavelength (blue) is in the visible range, Visible spectrum is 430 nm nm

27. S 1 and S 2 in Fig are point sources of electromagnetic waves of wavelength 1.00 m. They are in phase and separated by d = 4.00 m, and they emit at the same power. (a) If a detector is moved to the right along the x-axis from source S 1, at what distances from S 1 are the first three interference maxima detected? x detector Consider what the path difference is between the detector and S 1 and the detector and S 2

The solution for x of this equation is For constructive interference we have x detector

Solve for x

m=3 What about m = 4 ? This corresponds to x=0. Path difference =4 meters. m=2 m=1

Although the amplitudes are the same at the sources, the waves travel different distances to get to the points of minimum intensity and each amplitude decreases in inverse proportion to the square of the distance traveled. The intensity is not zero at the minima positions. m=3m=2m=1 Where do the minima occur? m=0 x=15.75 m m=1 x =4.55 m m=2 x=1.95 m m=3 x= 0.55 m m=0 m=1

Demo with speakers using sound waves Set oscillator frequency to 1372 Hz, Then wavelength of sound is 343/1372 =0.25 m Set speakers apart by 1m. Then maxima occur at