Presentation on theme: "1 Chapter 35 The concept of optical interference is critical to understanding many natural phenomena, ranging from color shifting in butterfly wings to."— Presentation transcript:
1 Chapter 35 The concept of optical interference is critical to understanding many natural phenomena, ranging from color shifting in butterfly wings to intensity patterns formed by small apertures. These phenomena cannot be explained using simple geometrical optics, and are based on the wave nature of light. In this chapter we explore the wave nature of light and examine several key optical interference phenomena. Interference 35-
2 Huygen’s Principle: All points on a wavefront serve as point sources of spherical secondary wavelets. After time t, the new position of the wavefront will be that of a surface tangent to these secondary wavelets. Light as a Wave 35- Fig. 35-2
3 Law of Refraction 35- Index of Refraction: Fig Law of Refraction:
4 Wavelength and Index of Refraction 35- Fig The frequency of light in a medium is the same as it is in vacuum Since wavelengths in n1 and n2 are different, the two beams may no longer be in phase
hitt The image of an erect candle, formed using a convex mirror (f < 0), is always: A. virtual, inverted, and smaller than the candle B. virtual, inverted, and larger than the candle C. virtual, erect, and larger than the candle D. virtual, erect, and smaller than the candle E. real, erect, and smaller than the candle 5
hitt A 5:0-ft woman wishes to see a full length image of herself in a plane mirror. The minimum length mirror required is: A. 5 ft B. 10 ft C. 2.5 ft D ft E. variable: the farther away she stands the smaller the required mirror length 6
question Two thin lenses (focal lengths f 1 and f 2 ) are in contact. Their equivalent focal length is: A. f 1 + f 2 B. f 1 f 2 /(f 1 + f 2 ) C. 1=f 1 + 1=f 2 D. f 1 /f 2 E. f 1 (f 1 f 2 )=f 2 7
8 The geometrical explanation of rainbows given in Ch. 34 is incomplete. Interference, constructive for some colors at certain angles, destructive for other colors at the same angles is an important component of rainbows Rainbows and Optical Interference 35- Fig. 35-5
9 Diffraction 35- Fig For plane waves entering a single slit, the waves emerging from the slit start spreading out, diffracting.
10 Young’s Experiment 35- Fig For waves entering a two slit, the emerging waves interfere and form an interference (diffraction) pattern.
11 The phase difference between two waves can change if the waves travel paths of different lengths. Locating Fringes 35- Fig What appears at each point on the screen is determined by the path length difference L of the rays reaching that point. Path Length Difference:
13 Coherence 35- Two sources to produce an interference that is stable over time, if their light has a phase relationship that does not change with time: E(t)=E 0 cos( t+ ) Coherent sources : Phase must be well defined and constant. When waves from coherent sources meet, stable interference can occur. Sunlight is coherent over a short length and time range. Since laser light is produced by cooperative behavior of atoms, it is coherent of long length and time ranges Incoherent sources : jitters randomly in time, no stable interference occurs
14 Intensity in Double-Slit Interference 35- Fig E1E1 E2E2
15 Fig Proof of Eqs and Eq Eq
16 hitt In a Young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in the distance they travel by a multiple of: A. a fourth of a wavelength B. a half a wavelength C. a wavelength D. three-fourths of a wavelength E. none of the above
17 In general, we may want to combine more than two waves. For example, there may be more than two slits. Procedure: 1.Construct a series of phasors representing the waves to be combined. Draw them end to end, maintaining proper phase relationships between adjacent phasors. 2.Construct the sum of this array. The length of this vector sum gives the amplitude of the resulting phasor. The angle between the vector sum and the first phasor is the phase of the resultant with respect to the first. The projection of this vector sum phasor on the vertical axis gives the time variation of the resultant wave. Combining More Than Two Waves 35- E1E1 E2E2 E3E3 E4E4 E
18 Interference from Thin Films 35- Fig
19 Reflection Phase Shifts 35- Fig n1n1 n2n2 n 1 > n 2 n1n1 n2n2 n 1 < n 2 Reflection Reflection Phase Shift Off lower index 0 Off higher index 0.5 wavelength
20 Equations for Thin-Film Interference 35- Fig Three effects can contribute to the phase difference between r 1 and r 2. 1.Differences in reflection conditions 2.Difference in path length traveled. 3.Differences in the media in which the waves travel. One must use the wavelength in each medium ( / n ), to calculate the phase. ½ wavelength phase difference to difference in reflection of r 1 and r 2
21 Film Thickness Much Less Than 35- If L much less than l, for example L < 0.1, than phase difference due to the path difference 2 L can be neglected. Phase difference between r 1 and r 2 will always be ½ wavelength destructive interference film will appear dark when viewed from illuminated side. r2r2 r1r1
22 Color Shifting by Morpho Butterflies and Paper Currencies 35- Fig For the same path difference, different wavelengths (colors) of light will interfere differently. For example, 2 L could be an integer number of wavelengths for red light but a half integer wavelengths for blue. Furthermore, the path difference 2 L will change when light strikes the surface at different angles, again changing the interference condition for the different wavelengths of light.
23 Problem Solving Tactic 1: Thin-Film Equations 35- Equations and are for the special case of a higher index film flanked by air on both sides. For multilayer systems, this is not always the case and these equations are not appropriate. What happens to these equations for the following system? n 1 =1n 2 =1.5n 3 =1.7 r1r1 r2r2 L
24 Fig Michelson Interferometer 35- For each change in path by 1, the interference pattern shifts by one fringe at T. By counting the fringe change, one determines N m - N a and can then solve for L in terms of and n.