Physics A2 - Lecture 2: Interference - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 2: Interference - Huynh Quang Linh

1. Lecture 2: Interference P y S3 d S2 S1 L Incident wave (wavelength l) Content:  Interference of Sound waves  Two-Slit Interference of Light: Young Interference  Phasors  Multiple-Slit Interference
2. Interference of Waves:  When two waves are present at the same point in space and time (single w), they lead to interference:  Add amplitudes (e.g., pressures or electric fields).  What we observe however is Intensity (absorbed power). I = A2 Stereo speakers: Listener: y1 = A1cos(kx - wt) y2 = A1cos(kx - wt + ) y1 +y 2 = 2A 1 cos( / 2) cos(kx w t  / 2) A 2 2  = 0:waves add “in phase” (“constructive”) I = |2 A1| = 4|A1| = 4I1  = p:waves add “out of phase” (“destructive”) I = |2 A1*0|2 = 0
3. Interference Exercise The relative phase of two waves also depends on the relative distances to the sources: The two waves at this point are “out of phase”. Their phase r1 difference  depends on the path difference d  r2 - r1. Each fraction Path Phase of a wavelength of difference difference path difference d  A = I gives that fraction 2A cos(/2) of 360º (or 2p) of 0 1 phase difference: l/4  d l/2 = 2p l l
4. Solution The relative phase of two waves also depends on the relative distances to the sources: The two waves at this point are “out of phase”. Their phase r1 difference  depends on the path difference d  r2 - r1. Each fraction Path Phase Reminder: A can be negative. of a wavelength of difference difference “Amplitude” is the absolute value. path difference gives that fraction d  A = 2A1cos(/2) I of 360º (or 2p) of 0 2A1 phase difference: 04I1 p/2 2A1 2I1 l/4  d p 0 0 = l/2 2p l 2p 2A1 4I1 l
5. Amplitude vs. Intensity (for 2 interfering waves) 2 cos(/2) cos (/2) Plot here as a function of . A = 2A1cos(/2) 2A1  Constructive 2 2 I = 4A1 cos (/2) Interference 2 4A1 Destructive Interference  0 2p 4p 6p 8p 10p d 0 l 2l 3l 4l 5l Note: What is the average intensity? Iave = 4I1*0.5 = 2I1
6. Sound wave example: 2 Each speaker alone produces intensity I1 = 1 W/m at the listener, and f = 900 Hz. Drive the speakers in phase. Compute the intensity I at the listener: Sound velocity in air: v = 330 m/s  d = 2p l r1 3 m  = 2p(d/l) with d = r2 – r1 4 m Procedure: 1) Compute path-length difference: d = r2 - r1 = 2) Compute wavelength: l = 3) Compute phase difference (in degrees):  = 4) Write a formula for the resultant amplitude: 5) Compute the resultant intensity, I =
7. Sound wave example: Each speaker alone produces intensity I1 = 1 W/m2 at the listener, and f = 900 Hz. Drive the speakers in phase. Compute the intensity I at the listener: Sound velocity: v = 330 m/s  d = 2p l r1 3 m  = 2p(d/l) with d = r2 – r1 4 m Procedure: 1) Compute path-length difference: d = r2 - r1 = 1 m 2) Compute wavelength: l = v/f = (330 m/s)/(900 Hz) = 0.367 m 3) Compute phase difference (in degrees):  = 360 (d/l) = 360(1/0.367) =981 4) Write a formula for the resultant amplitude: A = 2A1cos(/2), A1 = I1 2 2 2 5) Compute the resultant intensity, I = 4 I1cos (/2) = 4 (1 W/m ) (0.655) =1.72W/m2
8. Exercise 2 : Speaker interference r1 What happens to the intensity at the listener if we decrease the frequency f? (Recall the phase shift was 981.) a. decrease b. stay the same c. increase
9. Exercise 2 : Speaker interference r1 What happens to the intensity at the listener if we decrease the frequency f? (Recall the phase shift was 981.) a. decrease I b. stay the same c. increase  = 981 f decreases l increases d/l decreases 0 360 720 900   decreases I decreases (see figure)
10. Summary  The resultant intensity of two equal-intensity waves at the same point in space is: 2 I = 4 I1cos (/2)  For nonequal intensities, the maximum and minimum intensities are 2 2 Imax = |A1 + A2| Imin = |A1 - A2|  The phase difference between the two waves may be due to a difference in their source phases or a difference in the path lengths to the observer. In the latter case:  = 2p(d/l) with d = r2 – r1
11. Huygen’s principle All points on wavefront are point sources for ‘spherical secondary Wavefront Wavefront at t=0 at time t wavelets’ with speed, frequency equal to initial wave. What happens when a plane wave meets a small aperture? Answer: The result depends on the ratio of the wavelength l to the size of the aperture a : The transmitted wave is still concentrated in the forward l > a “Diffraction”: Interference of waves from objects or apertures
12. Light: Particle or Wave? Diffraction of light played an important historical role.  1818: French Academy held a science competition  Fresnel proposed the diffraction of light.  One judge, Poisson, knew light was made of particles, and thought Fresnel’s ideas ridiculous; he argued that if Fresnel ideas were correct, one would see a bright spot in the middle of the shadow of a disk.  Another judge, Arago, decided to actually do the experiment Conclusion (at the time): Light must be a wave, since particles don’t diffract!
13. Transmission of light through narrow slits the usual setup: Monochromatic light source at a great distance, Observation Slit pattern or a laser. screen
14. Double-slit interference  Light (wavelength l) is incident on a two-slit (two narrow, rectangular openings) apparatus: I1  If either one of the slits is closed, S a diffuse image of the other slit will 1 appear on the screen. (The image will be “diffuse” due to diffraction. We will discuss this effect in more detail S later.) 2 Diffraction Monochromatic light profile (wavelength l) screen  If both slits are now open, we see interference “fringes” (light and dark S bands), corresponding to constructive 1 and destructive interference of the electric-field amplitudes from both slits. S2 I
15. It’s just like sound waves! l Observer r1 Sound d Observer l S1 Light d S2 2 In both cases, I = 4 I1cos (/2) with  = 2p(d/l), d = r2 - r1 However, for light the observer distance is generally >> d.
16. Simple formula for the path difference, d , when the observer is far from sources.  Assume 2 sources radiating in phase: Observer q r d Normal to d When observer distance >> slit spacing q (r >> d) : d d d = dsinq d  = 2p(d/l) = 2p(d sinq/l)
17. Two-Slit Interference q Basic result: 2l/d l d = dsinq = ml Constructive r /d y Interference q d 0 m = 0, ±1, ±2, I Destructive d = dsinq = (m + 1/ )l -l/d 2 Interference L “lines” of m=2 constructive Usually we care about the linear m=1 interference: (as opposed to angular) q displacement y of the pattern: q = sin-1(ml/d) m=0 y = L tanq m=-1 m=-2
18. Two-Slit Interference, small angles: The slit-spacing d is often large compared to l, so that q is small. Then we can use the small angle approximations to simplify our results: For small angles: (q << 1 radian) sin q q tan q (only in radians!) y = L tan q L q (in radians) Y Lq Constructive q m(l/d) Interference: y m(l/d)L 2lL/d m = 0, ±1, ±2, lL/d q d 0 I -lL/d Destructive 1 q (m + /2)(l/d) Interference: L 1 y (m + /2)(l/d)L
19. Exercise 3: 2-slit interference A laser of wavelength 633 nm is S1 incident on two slits separated Dy by 0.125 mm. S2 I 1. What is the spacing Dy between fringe maxima on a screen 2m away? a. 1 mm b. 1 mm c. 1 cm 2. If we increase the spacing between the slits, what will happen to Dy? a. decrease b. stay the same c. increase 3. If we instead use a green laser (smaller l), Dy will? a. decrease b. stay the same c. increase
20. exercise 3: 2-slit interference A laser of wavelength 633 nm is S1 incident on two slits separated Dy by 0.125 mm. S2 1. What is the spacing Dy between fringe maxima on a screen 2m away? a. 1 mm b. 1 mm c. 1 cm 2. If we increase the spacing between the slits, what will happen to Dy? a. decrease b. stay the same c. increase Since Dy ~ 1/d, the spacing decreases. Note: This is a general phenomenon – the “far-field” interference pattern varies inversely with slit dimensions. 3. If we instead use a green laser (smaller l), Dy will? a. decrease b. stay the same c. increase Since Dy ~ l, the spacing decreases.
21. Phasors  We now want to introduce a new way of solving interference problems, using phasors to represent the interfering amplitudes (this will make it easier to solve other problems later on). A Represent a wave by a A1 A = A1 cos A1 cos vector with magnitude  (A1) and direction (). = 2A1 cos One wave has  = 0. A1  Now = /2 A = 2A1 cos 2 To get the intensity, we simply square this amplitude:  2  2 2 2 where I1= A1 is the intensity I = 4A1 cos I = 4I1 cos 2 2 when only one slit is open This is identical to our previous result ! More generally, if the phasors have C Here  is B different amplitudes A and B,  the external C2 = A2 + B2 + 2AB cos  A angle.
22. Multi-Slit Interference P What changes if we increase the number of slits (e.g., N = 3, 4, 1000?) Incident wave (wavelength l) y  First look at the “principle maxima”: S3  If slit 1 and 2 are in phase with each other, than slit 3 will also be in phase. d Conclusion: Position of “principle S2 interference maxima” are the same! (i.e., d sinq = m l) S1 L • What about amplitude of principle maxima? Draw phasor diagram: A = 3 A For N slits tot 1 2 A A A Itot = N I1 1 1 1 Itot = 9 I1  For other directions, simple geometry can tell us the resultant amplitude A Note: phasor A 1 angle is with in terms of A1 and . A  respect to  Use this to find interference minima A1 adjacent slit!  A1
23. exercise 4 I =? 1. In 2-slit interference, the first minimum 9I9 1 corresponds to  = p. For 3-slits, we have a secondary maximum at  = p (see g(Ix) 5 diagram). What is the intensity of this secondary maximum? (Hint: Use phasors.) 0 0 10 2p 00 2p 10 (a) I1 (b) 1.5 I1 (c) 3 I1 10 x 10  =? 2. What value of  corresponds to the first zero of the 3-slit interference pattern? (a) =p/2 (b) =2p/3 (c) =3p/4 3. What value of  corresponds to the first zero of the 4-slit interference pattern? (a) =p/2 (b) =2p/3 (c) =3p/4
24. exercise 4 I =? 1. In 2-slit interference, the first minimum 9I9 1 corresponds to  = p. For 3-slits, we have a secondary maximum at  = p (see g(Ix) 5 diagram). What is the intensity of this secondary maximum? (Hint: Use phasors.) 0 0 (c) 3 I 0 2p 0 2p (a) I1 (b) 1.5 I1 1 10 0 10 10 x 10  =? 2. What value of  corresponds to the first zero of the 3-slit interference pattern? (a) =p/2 (b) =2p/3 (c) =3p/4   =3p/4  p/3 A =p/2 =2 A No. A is not zero. Yes! Equilateral No, triangle does not close. triangle gives A = 0.
25. exercise 4 I =?  2. What value of corresponds to the first zero 9I9 1 of the 3-slit interference pattern? (a) =p/2 (b) =2p/3 (c) =3p/4 g(Ix) 5 0 0 3. What value of  corresponds to the first zero 10 2p 00 2p 10  of the 4-slit interference pattern? 10 x 10 (a) =p/2 (b) =2p/3 (c) =3p/4  =?  A  =3p/4 =p/2 =2p/3 Yes. Square No. With 4 slits we No. A 0 gives A = 0. start to “wrap around” again (A = A1) For N slits the first zero is at 2p/N.
26. General properties of N-Slit Interference 9I9 1 1616I1 25I1 N=3 N=4 25 N=5 20 10 5 g(Ix) h(x) I h5(Ix) 10 0 0 0 00 0 0 10 2p 00 10  10 0 10  0 2p 2p 0 2p 10 2p 00 2p 10  10 l/d 0x l/d 10 q 10 l/ 0x l/ 10 q d d 10 l/d 0x l/d 10 q • The positions of the principal maxima of the intensity patterns always occur at  = 0, 2p, 4p, [ is the phase between adjacent slits] (i.e., dsinq = ml, m = 0, 1, 2, ). • The principal maxima become taller and narrower as N increases. • The intensity of a principal maximum is equal to N2 times the maximum intensity from one slit. The width of a principal maximum goes as 1/N. • The # of zeroes between adjacent principal maxima is equal to N-1. The # of secondary maxima between adjacent principal maxima is N-2.
27. Questions  Is Young’s experiment an interference experiment or a diffraction experiment, or both?  What changes, if any, occur in the pattern of interference fringes if the apparatus if Young’s experiment is placed under water?  Do interference effects occur for sound waves?