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 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
3. 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
4. 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
5. 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)
6. 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
7. Transmission of light through narrow slits the usual setup: Monochromatic light source at a great distance, Observation Slit pattern or a laser. screen
8. 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.
9. 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
10. 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
11. 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.
12. exercise 4 I =? 1. In 2-slit interference, the first minimum 99I1 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
13. exercise 4 I =?  2. What value of corresponds to the first zero 99I1 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.
14. 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?