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Chapter 35 Interference 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. 35- 1 Light as a Wave 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. Fig. 35-2 35- 2 Law of Refraction 1 2 1 v1 t v1 v2 2 v2 1 sin 1 (for triangle hce) hc sin 2 2 hc (for triangle hcg) c sin 1 1 v1 Index of Refraction: n v sin 2 2 v2 c c n1 and n2 v1 v2 sin 1 c n1 n2 sin 2 c n2 n1 Fig. 35-3 Law of Refraction: n1 sin1 n2 sin2 35- 3 Wavelength and Index of Refraction n v v n n c c n v cn c fn f n n 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 L L Ln1 Number of wavelengths in n1: N1 Fig. 35-4 n1 n1 L L Ln2 Number of wavelengths in n2 : N 2 n 2 n2 Ln2 Ln2 L Assuming n2 n1: N 2 N1 n2 n1 N2 N1 1/2 wavelength destructive interference 35- 4 Rainbows and Optical Interference Fig. 35-5 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 35- 5 Diffraction For plane waves entering a single slit, the waves emerging from the slit start spreading out, diffracting. Fig. 35-7 35- 6 Young’s Experiment For waves entering a two slit, the emerging waves interfere and form an interference (diffraction) pattern. Fig. 35-8 35- 7 Locating Fringes The phase difference between two waves can change if the waves travel paths of different lengths. What appears at each point on the screen is determined by the path length difference DL of the rays reaching that point. Path Length Difference: DL d sin Fig. 35-10 35- 8 Locating Fringes if DL d sin integer bright fringe Maxima-bright fringes: d sin m for m 0,1, 2, Fig. 35-10 if DL d sin odd number dark fringe Minima-dark fringes: d sin m 12 for m 0,1,2, 2 1 1.5 m 1 dark fringe at: sin m 2 bright fringe at: sin 1 35- 9 d d Intensity in Double-Slit Interference E1 E0 sin t and E2 E0 sin t 2 d I 4 I 0 cos2 12 sin maxima when: 12 m for m 0,1, 2, d sin m for m 0,1, 2, 2m E 2 E 2 d sin (maxima) minima when: 12 m 12 d sin m 12 for m 0,1,2, (minima) I avg 2 I 0 Fig. 35-12 35-10 Proof of Eqs. 35-22 and 35-23 Eq. 35-22 E t E0 sin t E0 sin t ? E 2 E0 cos 2 E0 cos 12 E 2 4 E02 cos2 12 I E2 2 4cos2 12 I 4 I 0 cos2 12 I 0 E0 Eq. 35-23 Fig. 35-13 phase path length difference difference 2 phase 2 path length difference difference 2 d sin 35- 11 Combining More Than Two Waves In general, we may want to combine more than two waves. For eaxample, there may be more than two slits. Prodedure: 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. E4 E3 E E1 E2 35-12 Interference from Thin Films 12 ? 0 Fig. 35-15 35-13 Equations for Thin-Film Interference Three effects can contribute to the phase difference between r1 and r2. 1. Differences in reflection conditions 2 2. Difference in path length traveled. 0 3. Differences in the media in which the waves travel. One must use the wavelength in each medium ( / n), to calculate the phase. Fig. 35-17 ½ wavelength phase difference to difference in reflection of r1 and r2 odd number odd number 2L wavelength = n 2 (in-phase waves) 2 2 2 L integer wavelength = integer n 2 (out-of-phase waves) n 2 n2 2L m 2L m n2 1 2 n2 for m 0,1, 2, for m 0,1, 2, (maxima-- bright film in air) (minima-- dark film in air) 35-14 Film Thickness Much Less Than r1 r2 If L much less than , for example L < 0.1, than phase difference due to the path difference 2L can be neglected. Phase difference between r1 and r2 will always be ½ wavelength destructive interference film will appear dark when viewed from illuminated side. 35-15 Problem Solving Tactic 1: Thin-Film Equations Equations 35-36 and 35-37 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? L r2 r1 n1=1 n2=1.5 n3=1.7 35-16 Michelson Interferometer DL 2d1 2d2 (interferometer) DLm 2 L (slab of material of thickness L placed in front of M 1 ) 2 L 2 Ln Nm = = (number of wavelengths m in slab of material) Na = 2L (number of wavelengths in same thickness of air) Fig. 35-20 N m -N a = 2 Ln 2L 2L = n-1 (difference in wavelengths For each change in path by 1, the interference pattern shifts by one fringe at T. By counting the fringe change, one determines Nm- Na and can then solve for L in terms of and n. for paths with and without thin slab) 35-17