Lecture 21. Physics 1202: Lecture 22 Today s Agenda

Transcription

1 Physics 1202: Lecture 22 Today s Agenda Announcements: Team problems today Team 16: Navia Hall, Laura Irwin, Eric Kaufman Team 18: Charles Crilly Jr, Kyle Eline, Alexandra Vail Team 19: Erica Allen, Shana Morel, Daniel Shoushani Team 20: Ugbad Awale, Jenny Ma, Randi Szabo Team 21: Erin Flajnik, Katerina Koch, Cindy Li Homework #10: due Friday Chapter 28: Physical Optics Interference Two-slit experiment Interference in Reflected Waves Diffraction/resolution 28- Physical Optics Interference Diffraction 1

2 28-1 Superposition and Interference If two waves occupy the same space, their amplitudes add at each point. They may interfere either constructively or destructively. Superposition and Interference Light waves interfere with each other much like mechanical waves do All interference associated with light waves arises when the electromagnetic fields that constitute the individual waves combine (superposition) Constructive interference Destructive interference 2

3 Conditions for Interference Incoherent light beams pass through each other, with no obvious interference. Although white-light interference is observed under certain conditions, it s easier to see interference when sources are monochromatic. For sustained interference between two sources of light to be observed, there are two conditions which must be met The sources must be coherent They must maintain a constant phase with respect to each other The waves must have identical wavelengths A wave through a slit Wavefronts: slit acts like point source Rays 3

4 A wave through two slits (two coherent point sources) Intensity What happens when two light waves are present at the same point in space and time? What will we see? Intensity! Add Amplitudes! (electric fields or magnetic fields) Brightness ~ <Amplitude 2 > ~ ½ E 0 2 4

5 Lecture 22 Act 1 Suppose laser light of wavelength l is incident on the two-slit apparatus as shown below. Which of the following statements are true? (A) There are new patterns of light and dark. (B) The light at all points on the screen is increased (compared to one slit). (C) The light at all points on the screen is decreased (compareed to two slits). A wave through two slits l 1 q d q l 2 L Assume L is large, Rays are parallel Screen 5

6 A wave through two slits In Phase, i.e. Maxima when DP= l 2 - l 1 = d sinq = ml Out of Phase, i.e. Minima when DP = d sinq = (m+1/2)l d q DP= l 2 -l 1 = d sinq Screen A wave through two slits In Phase, i.e. Maxima when DP = d sinq = ml + Out of Phase, i.e. Minima when DP = d sinq = (m+1/2)l + 6

7 Waves and Interference Note that you could derive the reflectance equation (q i =q R ) using a particle model for light. Bouncing balls. You could also derive Snell s Law for particles. n 1 sin (q i )=n 2 sin(q 2 ) The particles change speed in different media (Newton did just this) You cannot get a particle model for these interference effects. You would have to magically create particles at the bright spots and annihilate them at the dark spots. Interference effects mean that light must be made up of waves Young s Two-Slit Experiment In this experiment, the original light source need not be coherent it becomes so after passing through the first very narrow single slit Thomas Young first demonstrated interference in light waves from two sources in 1801 Light is incident on a screen with a narrow slit, S o The light waves then pass through two narrow, parallel slits, S 1 and S 2 The bright areas : constructive interference The dark areas : destructive interference 2017 Pearson Education, Inc. 7

8 28-2 Young s Two-Slit Experiment The light on the screen has alternating light and dark fringes, corresponding to constructive and destructive interference. The path difference is Therefore, the condition for bright fringes (constructive interference) is: 28-2 Young s Two-Slit Experiment This diagram illustrates the numbering of the fringes. The dark fringes are between the bright fringes; the condition for dark fringes is: 2017 Pearson Education, Inc. 8

9 28-3 Interference in Reflected Waves Reflected waves can interfere due to path length differences, and to phase changes upon reflection. No phase change if reflected from a lower n Half-wavelength phase change if reflected from a higher n, or from a solid surface. There is also no phase change in the refracted wave. Phase Change upon Reflection from a Surface/Interface Reflection from Optically Denser Medium (larger n) Reflection from Optically Lighter Medium (smaller n) 180 o Phase Change No Phase Change by analogy to reflection of traveling wave in mechanics 9

10 Interference in Thin Films Air phase change 1 no phase change 2 A wave traveling from air toward film undergoes phase change upon reflection. The wavelength of light l n in the medium with refraction index n is Film Air t The ray 1 is out of phase with ray 2 which is equivalent to a path difference l n /2. The ray 2 also travels extra distance 2t. Constructive interference Destructive interference constructive: 2t = (m +1/2) l n destructive: 2t = m l n Examples : constructive: 2t = m l n destructive: 2t = (m +1/2) l n 10

11 28-3 Interference in Reflected Waves Constructive interference: Destructive interference: Interference if light refracts and reflects from both surfaces of a thin film This accounts for the colors we see in oil slicks and soap bubbles. Newton s Rings Placing a curved piece of glass on flat glass gives interference patterns called Newton s rings. Monochromatic light is not necessary: white light is OK. Newton s rings can provide precision measurements of shapes and radii of lenses. If the bottom glass isn t flat, or the top glass isn t spherical or some other symmetric shape, Newton s rings appear irregular, not circular. 11

12 Problem Solving with Thin Films Equation 1 phase reversal 0 or 2 phase reversals 2t = (m + ½) l/n constructive destructive 2t = m l/n destructive constructive 28-3 Interference in Reflected Waves Wavelength of light in a medium of index of refraction n: Destructive interference (t= film thickness) Constructive interference: The rainbow of colors we see is due to the different wavelengths of light 12

13 28-3 Interference in Reflected Waves DVDs depend on interference for their functioning. The signal is encoded as tiny bumps in the surface, and the reflected laser varies in intensity depending on whether it is reflecting from a bump or not. Change of Phase Due to Reflection S Lloyd s mirror P 2 The reflected ray (red) can be considered as an original from the image source at point I. Thus we can think of an arrangement S and I as a double-slit source separated by the distance between points S and I. I L P 1 Mirror An interference pattern for this experimental setting is really observed.. but dark and bright fringes are reversed in order This mean that the sources S and I are different in phase by An electromagnetic wave undergoes a phase change by upon reflecting from the medium that has a higher index of refraction than that one in which the wave is traveling. 13

14 28-4 Diffraction A wave passing through a small opening will diffract, (e.g., water waves) After the opening, there are waves traveling in directions other than the direction of the original wave Diffraction is the bending of waves as they pass by an obstacle or through an opening. This is why we hear sound even if not in a straight line from the source sound waves will diffract around corners and other barriers The amount of diffraction depends on the wavelength, which is why we can hear around corners but not see around them. Experimental Observations: (pattern produced by a single slit?) 14

15 Huygen s Principle Huygen assumed that light is a form of wave motion rather than a stream of particles Christian Huygens ! Huygens Principle has two parts:! Each point on a wave front is the source of a spherical wavelet that spreads out at the wave speed.! At a later time, the shape of the wave front is the curve that is tangent to all the wavelets. How do we understand this pattern? First Destructive Interference: (a/2) sin Q = l/2 sin Q = l/a Second Destructive Interference: (a/4) sin Q = l/2 sin Q = 2 l/a m th Destructive Interference: sin Q = m l/a m= 1, 2, See Huygen s Principle 15

16 So we can calculate where the minima will be! sin Q = m l/a m= 1, 2, So, when the slit becomes smaller the central maximum becomes? Why is the central maximum so much stronger than the others? Other Examples Light from a small source passes by the edge of an opaque object and continues on to a screen. A diffraction pattern consisting of bright and dark fringes appears on the screen in the region above the edge of the object. What type of an object would create a diffraction pattern shown on the left, when positioned midway between screen and light source? A penny, Note the bright spot at the center. 16

17 28-5 Resolution Diffraction through a small circular aperture results in a circular pattern of fringes. This limits our ability to distinguish one object from another when they are very close. The location of the first dark fringe determines the size of the central spot: 2017 Pearson Education, Inc. Resolution (single-slit aperture) Rayleigh s criterion: two images are just resolved WHEN: When central maximum of one image falls on the first minimum of another image sin Q = l / a Q min ~ l / a 17

18 Resolution (circular aperture) Diffraction patterns of two point sources for various angular separation of the sources Rayleigh s criterion for circular aperture: Q min = 1.22 ( l / a) Recap of Today s Topic : Announcements: Team problems today Team 16: Navia Hall, Laura Irwin, Eric Kaufman Team 18: Charles Crilly Jr, Kyle Eline, Alexandra Vail Team 19: Erica Allen, Shana Morel, Daniel Shoushani Team 20: Ugbad Awale, Jenny Ma, Randi Szabo Team 21: Erin Flajnik, Katerina Koch, Cindy Li Homework #10: due Friday Chapter 28: Physical Optics Interference Two-slit experiment Interference in Reflected Waves Diffraction/resolution 18