Interference - Pham Tan Thi

Nội dung text: Interference - Pham Tan Thi

1. Interference Pham Tan Thi, Ph.D. Department of Biomedical Engineering Faculty of Applied Science Ho Chi Minh University of Technology
2. A Single Oscillating Wave The formula y(x, t)=Acos(kx !t) describes a harmonic plane wave of amplitude A moving in the +x direction For a wave on a string, each point on the wave oscillates in the x direction with simple harmonic motion of angular frequency ω 2⇡ ! The wavelength = The speed/velocity v = f = k k The intensity is proportional to I A2 the square of the amplitude /
3. Multiple Waves: Superposition The principle of superposition states that when two or more waves of the same type cross at a point, the resultant displacement at that point is equal to the sum of the displacements due to each individual wave. For inequal intensities, the maximum and minimum intensities are: Imax = |A1 + A2|2 Imin = |A1 - A2|2
4. Multiple Waves: Superposition Constructive “Superposition” Destructive “Superposition”
5. Superposing sine waves If you added the two sinusoidal waves shown, what would theSuperposing result look Sinusoidal like? Waves If we added the two sinusoidal waves shown, what would the result look like? 1.00 0.50 0.00 0 100 200 300 400 500 600 700 800 900 1000 -0 .5 0 -1 .0 0 The sum of two sines having the same frequency is another sine with the same frequency. Its amplitude depends on their relative phases. 2.00 1.50 1.00 0.50 0.00 -0.50 0 100 200 300 400 500 600 700 800 900 -1.00 1000 -1.50 -2.00 Let’s see how this works.
6. Superposing sine waves If you added the two sinusoidal waves shown, whatSuperposing would the result look sine like? waves 1.00 If you added0.50 the two sinusoidal waves shown, 0.00 0 100 200 300 400 500 600 700 800 900 Superposing Sinusoidal Waves 1000 what would-0 .5 0 the result look like? If-1 we .0 0 added the two sinusoidal waves shown, what would the result look like? 1.00 0.50 0.00 0 100 200 300 400 500 600 700 800 900 1000 The sum of two-0 .5sines 0 having the same frequency is another sine -1 .0 0 with the same frequency. The sum of two sines having the same frequency is another sine Its amplitude dependswith the same on frequency their → relativeIts amplitude phases. depends on their relative phases The sum of two2.00 sines having the same frequency is another sine 1.50 with the same frequency.1.00 0.50 0.00 Its amplitude depends-0.50 0 on their relative phases. 100 200 300 400 500 600 700 800 900 -1.00 1000 -1.50 2.00 -2.00 1.50 1.00 0.50 0.00 -0.50 0 100 200 300 400 500 600 700 800 900 Let’s see how -1.00this works. 1000 -1.50 -2.00 Let’s see how this works.
7. Adding Sine Waves with Different Phases Suppose we have two sinusoidal waves with the same A1, ω and k: y = A cos(kx !t) and y = A cos(kx !t + ) 1 1 2 1 One starts at phase � after the other Spatial dependence of 2 waves at t = 0 Resulting wave: y = y1 + y2 ↵ + ↵ A (cos↵ + cos)=2A cos 1 1 2 2 y1 + y2 ✓ ◆✓ ◆ y =2A cos((/2)2)cos(kx !t + /2) 1 y =2A cos(/2)cos(kx !t + /2) 1 Amplitude Oscillation
8. Interference of Waves What happens when two waves are present at the same place? Interference of Waves Always add amplitudes (pressures or electric fields). WhatHowever, happens whenwe observe two waves intensity are present(power). at the same place? ForAlways equal Aadd and amplitude ω: (e.g. pressures or electric fields) However, we observe intensity (i.e. power) 2 For equal AA and==φ ω2A: 11 cos(φ /2) ⇒ I 4I cos ( /2) A = 2A1cos(�/2) ⇒ I = 4I1cos2(�/2) Example: Terminology: Stereo speakers: Constructive interference: Stereo speakers: Listener: Terminology: waves are “in phase” Constructive(φ = 0, interference:2π, 4π, ) Listener: wavesDestructive are “in interference: phase” (� = 0,waves 2π, 4π are, ) “out of phase” Destructive(φ = π interference:, 3π, 5π, ) waves are “out of phase” Of course,(� φ= canπ, 3 πtake, 5π , )on an infinite number of values. We won’t use terms like “mostly constructive” or “slightly destructive”. Lecture 2, p.6
9. Quiz Each speaker alone produces an intensity of I1 = 1 W/m2 at the Example: Changing phase of the Source listener:Each speaker alone produces an intensity of I = 1 W/m2 at the listener: Example: Changing phase1 of the Source 2 Each speaker alone produces an intensity of I1 = 1 W/m at the listener: I = I = A 2 = 1 W/m2 2 2 Example: Changing phase1 of1 the SourceI = I1 = A1 = 1 W/m 2 2 2 Each speaker alone produces an intensity of I1 =I =1 IW/m1 = A1 at= the 1 W/mlistener: Drive the speakers in phase. What is the intensity I at the listener? 2 2 DriveDrive the the speakers speakers in phase. in Whatphase. is the intensity IWhat = I1 = A I1 at is =the 1 the W/mlistener? intensity I at the listener? I = Drive the speakers in phase. What is the intensityI = I at the listener?I =? Now shift phase of one speaker by 90o.What is the intensity I at the listener? I = Now shift phase of one speaker by 90o.What is the intensity I at the listener? Now shift phase of one speaker by 90°. What is the intensity I at the I = listener? o Now shift phase of one speaker by 90 .What is Ithe = intensity I at the listener? φφφ φφφ Lecture 2, p.7 Lecture 2, p.7 I = I =? φφφ Lecture 2, p.7
10. Quiz Each speaker alone produces an intensity of I1 = 1 W/m2 at the Example: Changing phase of the Source listener:Each speaker alone produces an intensity of I = 1 W/m2 at the listener: Example: Changing phase1 of the Source 2 Each speaker alone produces an intensity of I1 = 1 W/m at the listener: I = I = A 2 = 1 W/m2 2 2 Example: Changing phase1 of1 I the= I1 Source = A1 = 1 W/m 2 2 2 Each speaker alone produces an intensity of I1 =I =1 IW/m1 = A1 at= the 1 W/mlistener: Drive the speakers in phase. What is the intensity I at the listener? 2 2 DriveDrive the the speakers speakers in phase. in Whatphase. is the intensity IWhat = I1 = A I1 at is =the 1 the W/mlistener? intensity I at the listener? I = Drive the speakers in phase. What is the intensity I at the listener? I = I = (2A1)2 = 4I1 = 4 W/m2 Now shift phase of one speaker by 90o.What is the intensity I at the listener? I = Now shift phase of one speaker by 90o.What is the intensity I at the listener? Now shift phase of one speaker by 90°. What is the intensity I at the I = listener? o Now shift phase of one speaker by 90 .What is Ithe = intensity I at the listener? φφφ φφφ Lecture 2, p.7 Lecture 2, p.7 I = I =? φφφ Lecture 2, p.7
11. Quiz Each speaker alone produces an intensity of I1 = 1 W/m2 at the Example: Changing phase of the Source listener:Each speaker alone produces an intensity of I = 1 W/m2 at the listener: Example: Changing phase1 of the Source 2 Each speaker alone produces an intensity of I1 = 1 W/m at the listener: I = I = A 2 = 1 W/m2 2 2 Example: Changing phase1 of1 I the= I1 Source = A1 = 1 W/m 2 2 2 Each speaker alone produces an intensity of I1 =I =1 IW/m1 = A1 at= the 1 W/mlistener: Drive the speakers in phase. What is the intensity I at the listener? 2 2 DriveDrive the the speakers speakers in phase. in Whatphase. is the intensity IWhat = I1 = A I1 at is =the 1 the W/mlistener? intensity I at the listener? I = Drive the speakers in phase. What is the intensity I at the listener? I = I = (2A1)2 = 4I1 = 4 W/m2 Now shift phase of one speaker by 90o.What is the intensity I at the listener? I = Now shift phase of one speaker by 90o.What is the intensity I at the listener? Now shift phase of one speaker by 90°. What is the intensity I at the I = listener? o Now shift phase of one speaker by 90 .What is Ithe = intensity I at the listener? φφφ φφφ Lecture 2, p.7 Lecture 2, p.7 I = I = 4I1cos2(45°) = 2 I1 = 2 W/m2 φφφ Lecture 2, p.7
12. ACT 1: Noise CancellingNoise-cancelling Headphones Headphones Noise-cancelingNoise-canceling headphones headphones working work using using interference.interference. A microphone A microphone on the on earpiece the monitorsearpiece the instantaneous monitors the instantaneousamplitude of the externalamplitude sound wave, of the andexternal a speaker sound on wave, the inside of theand earpiece a speaker produces on the a soundinside ofwave the cancel it. earpiece produces a sound wave to cancel it. 1. What1. mustWhat be must the bephase the phaseof the signalof the signalfrom the from speaker the speak relativeer relative to the to the externalexternal noise? noise? a. 0a. 0 b. 90b. 90˚ c. πc. π d. d.180-180˚ e. e.2 π2π 2. What must the intensity Is of the signal from the speaker relative to the external2. Whatnoise? must be the intensity Is of the signal from the speaker relative to the a. Is = In b. Is In external noise In? a. Is = In b. Is In Lecture 2, p.10
13. ACT 1: Noise CancellingNoise-cancelling Headphones Headphones Noise-cancelingNoise-canceling headphones headphones working work using using interference.interference. A microphone A microphone on the on earpiece the monitorsearpiece the instantaneous monitors the instantaneousamplitude of the externalamplitude sound wave, of the andexternal a speaker sound on wave, the inside of theand earpiece a speaker produces on the a soundinside ofwave the cancel it. earpiece produces a sound wave to cancel it. 1. What1. mustWhat be must the bephase the phaseof the signalof the signalfrom the from speaker the speak relativeer relative to the to the externalexternal noise? noise? a. 0a. 0 b. 90°b. 90˚ c. πc. π d. d.180°-180˚ e. e.2 π2π Destructive interference occurs when the waves are ±180° out of phase (=π radians) 2. What must the intensity Is of the signal from the speaker relative to the external2. Whatnoise? must be the intensity Is of the signal from the speaker relative to the a. Is = In b. Is In external noise In? We want A = As - An = 0. Note that I is never negative. a. Is = In b. Is In Lecture 2, p.10
14. Interference of Light
15. Review: Lights of Different Colors Long wavelength (Low frequency) Short wavelength (High frequency) Violet light has the highest energy in visible region
16. Interference of Light Interference is a phenomenon in which two coherent light waves superpose to form a resultant wave of greater or lower amplitude (a) Constructive interference: If a crest of one wave meets a crest of another wave, the resultant intensity increases. (a) Destructive interference: If a crest of one wave meets a trough of another wave, the resultant intensity decreases.
17. Light Polarization
18. Conditions for Interference ✓ Two interfering waves should be coherent, i.e. the phase difference between them must remain constant with time. ✓ Two waves should have same frequency. ✓ If interfering waves are polarized, they must be in same state of polarization. ✓ The separation between the light sources should be as small as possible. ✓ The distance of the screen from the sources should be quite large. ✓ The amplitude of the interfering waves should be equal or at least very nearly equal. ✓ The two sources should be narrow. ✓ The two sources should give monochromatic or very nearly monochromatic.
19. Young’s Double Split Experiment • In 1801, Young admitted the sunlight through a single pinhole and then directed the emerging light onto two pinholes. • The spherical waves emerging from the pinholes interfered with each other and a few colored fringes were observed on the screen.
20. Calculation of Optical Path Difference between Two Waves As slits (S1 and S2) are equidistant from source (S), the phase of the wave at S1 will be same as the phase of the wave at S2 and therefore S1, S2 act as coherent sources. The waves leaving from S1 and S interfere and produce alternative bright and dark bands on the screen. 2d is the distance between S1 and S2 θ is the angle between MO and MP x is the distance between O and P Let P is an arbitrary point on screen, which is at a distance D from source S1N is the normal on to the line S2P S1M = S2M = GO = OH = d
21. Calculation of Optical Path Difference between Two Waves The optical paths are identical with the geometrical paths, if the experiment is carried out in air The path difference between two waves is S2P - S1P = S2N Let S1G and S2H are perpendicular on the screen and S2HP forms triangle (S2P)2 = (S2H)2 + (HP)2 = D2 + (x + d)2 (x + d)2 (S P)2 = D2 1+ 2 D2 
22. Calculation of OpticalDivergence Path Diff erenceOperator between Two Waves (x + d)2 (S P)2 = D2 1+ 2 D2  (x + d)2 1/2 S P=D 1+ 2 D2  Since D >> (x + d), (x + d)2/D2 is very small. After expansion, 1 (x + d)2 S P=D 1+ 2 2 D2  1 (x + d)2 S P= D + 2 2 D  1 (x d)2 Similarly, S P= D + 1 2 D  1 2xd Path difference, S P S P= (x + d)2 (x d)2 = 2 1 2D D ⇥ ⇤
23. Conditions for Observing Fringes The nature of the inference of the two waves at P depends simply on what waves are contained in the length path difference (S2N). • If the path difference (S2N) contains an integral number of wavelengths, then the two waves interfere constructively producing a maximum in the intensity of light on the screen at P. • If the path difference (S2N) contains an odd number of half- wavelengths, then the two waves interfere destructively and produce minimum intensity of light on the screen at P.
24. Conditions for Observing Fringes For bright fringes (maxima), 2xd S N= = n where n = 0, 1, 2, 2 D D x = n 2d For dark fringes (minima), 2xd S N= =(2n + 1) where n = 0, 1, 2, 2 D 2 D x = (2n + 1) 2d 2 Let xn and xn+1 denote the distances of nth and (n+1)th bright fringes Then, D D D x = n ⇒ x(n+1) xn = (n + 1) n n 2d 2d 2d
25. Continued . Spacing between nth and (n+1)th bright fringe is Let x and x denote the distances of the nth and (n+1)th bright fringes. n n+1 D D x(n+1) xn = (n + 1) n Then the distance between(n+1)th and2d nth brightfringes2d is D = 2d * This is independent of n Hence, the spacing between any consecutive bright fringe is the same. * The distanceSimilarly,between the spacingany betweentwo consecutive two dark fringesbright is fringe(Dλ/2dis). the same i.e. Dλ/2d. ` The spacing between the fringes is independent of n. * Similarly, the distance between any two consecutive dark fringes is the same i.eThe. D λspacing/2d. between any two consecutive bright or dark fringe is called * The distancethe “fringeDλ /width2d is”called and is thedenoted“ Fringe by X-width” and is denoted by D X= 2d The wavelength of unknown light (λ) can be calculated by measuring the values of D, 2d, and .
26. Resultant Intensity due to superposition of two interfering waves • Let S be a narrow slit illuminated by a monochromatic (single wavelength) source. • S1 and S2 are two similar parallel slits (S1 and S2 are equidistant from S and very close to each other) • Suppose the waves from S reach S1 and S2 in the same phase (coherent). • Beyond S1 and S2, the wave proceed as if they started from S1 and S2.
27. Resultant Intensity due to superposition of two interfering waves • Let us calculate the resultant intensity of light of wavelength (λ) at point P on a screen placed parallel to S1 and S2. • Let A1 and A2 be the amplitude at P due to the waves from S1 and S2, respectively. • The waves arrive at P, having transversed different paths S1P and S2P. • Hence, they are superposed at P with a phase difference δ 2⇡ = path di↵erence ⇥ 2⇡ = (S P-SP) ⇥ 2 1 The displacement at P due to the simple harmonic waves from S1 and S2 can be represented by y1 = A1sin!t y2 = A2sin(!t + )
28. Resultant Intensity due to superposition of two interfering waves By the principle of superposition, when two or more waves simultaneously reach at point, the resultant displacement is equal to the sum of the displacements of all the waves. Hence the resultant displacement is: y = y1 + y2 = A1sin!t + A2sin(!t + ) = A1sin!t + A2sin!tcos + A2cos!tsin =sin!t(A1 + A2cos) + cos!t(A2sin) Let us define: A1 + A2cos = Rcos✓ A2sin = Rsin✓
29. Resultant Intensity due to superposition of two interfering waves The resultant displacement is y =sin!t(Rcos✓) + cos!t(Rsin✓) y = Rsin(!t + ✓) Hence, the resultant displacement at P is simple harmonic and of amplitude R A1 + A2cos = Rcos✓ A2sin = Rsin✓ 2 2 2 (A1 + A2cos) = R cos ✓ 2 2 2 (A2cos) = R sin ✓ Adding the above two equations, 2 2 2 2 2 2 R cos ✓ + R sin ✓ =(A1 + A2cos) +(A2sin) 2 2 2 R = A1 + A2 +2A1A2cos
30. Resultant Intensity due to superposition of two interfering waves 2 2 2 R = A1 + A2 +2A1A2cos The resultant intensity I at point P, which is proportional to the square of the resultant amplitude, is given by I = R2 By assuming proportionality constant = 1, 2 2 1=A1 + A2 +2A1A2cos 2⇡ 2⇡ where, = path di↵erence = (S P-SP) ⇥ ⇥ 2 1 In the case of interference, the resultant intensity at P is not just the sum of the intensities of the individual waves.
31. Conditions for Maxima and Minima Intensity The resultant intensity I at point P, 2 2 1=I = A1 + A2 +2A1A2cos The intensity I is a maximum, when cosδ = +1 or δ = 2nπ; n = 0, 1, 2, Phase difference δ = 2nπ; n = 0, 1, 2, Path difference (S2P - S1P)= 2nπ x λ/2π = nλ 2 2 2 Imax = A1 + A2 +2A1A2 =(A1 + A2) The maximum intensity is greater than the sum of two separate intensities The intensity I is a minimum, when cosδ = -1 or δ = (2n + 1)π; n = 0, 1, 2, Phase difference δ = (2n + 1) π; n = 0, 1, 2, Path difference (S2P - S1P)= (2n + 1)π x λ/2π = (2n + 1)λ I = A2 + A2 2A A =(A A )2 min 1 2 1 2 1 2 The minimum intensity is less than the sum of two separate intensities
32. Techniques for Producing Interference The phase relation between the waves emitted by two independent light sources rapidly changes with time and therefore they can never be coherent, even though sources are identical in all respects. • If two sources are derived from a single source by some devices, then any phase change occurring in single source is simultaneously accompanied by the same phase change in the other source. Therefore, the phase difference between waves emerging from the two sources remains constant and the sources are coherent. • The techniques used for creating coherent sources of light are (a) Wavefront splitting (b) Amplitude splitting
33. (a) Wavefront Splitting One of the method consists of dividing a light wavefront, emerging from a narrow slit, by passing it through two slits closely spaced side by side. The two parts of the same wavefront travel through different paths and reunite on a screen to produce fringe pattern. This is known as interference due to the division of wavefront. This method is useful only with narrow sources. Example: Young’s double split; Fresnel’s double mirror; Fresnel biprism; Lloyd’s mirror; etc. Fresnel’s Biprism BP A S1 B 2d S C S2 E D * S is a narrow vertical slit illuminated by a monochromatic light. * The light from S is allowed to fall symmetrically on the biprism BP. * The light beams emerging from the upper and lower halves of the prism appears to start from two virtual images of S, namely S1 and S2. * S1 and S2 act as coherent sources. * The cones of light BS1E and AS2C, diverging from S1 and S2 are superposed and the interference fringes are formed in the overlapping region BC of the screen.
34. (b) Amplitude Splitting The amplitude (intensity) of a light wave is divided into two parts, namely reflected and transmitted components, by partial reflection at a surface. The two parts travel through different paths and reunite to produce interference fringes. This is known as interference due to division of amplitude. Optical elements such as beam splitters, mirrors are used for achieving amplitude division. Examples: Newton’s ring experiment; Michelson’s interferometer; etc.
35. Interference of Thin Film (inverse phase) Incoming wave Transmitted wave Reflected wave Incoming wave (same phase) Transmitted wave Reflected wave
36. Interference of Thin Film Assuming, Incident ray y(x, t)=Acos(kx !t) Phase change, Reflected(∆Φ = ray π) Reflected(∆Φ = ray 0) = k(2h) ⇡ For zero reflection, =(2n 1)⇡ For maximum reflection, =2n⇡
37. Interference of Thin Film For zero reflection, Incident ray =2hk ⇡ =(2n 1)⇡ 4⇡h Reflected(∆Φ = ray π) ⇡ =(2n 1)⇡ Reflected(∆Φ = ray 0) 2⇡ where k = ✓ ◆ Cancellation of rays n 2h h = OR = 2 n
38. Interference of Thin Film For maximum reflection, Incident ray =2hk ⇡ =2n⇡ 4⇡h Reflected(∆Φ = ray π) ⇡ =2n⇡ Reflected(∆Φ = ray 0) 4h =(2n + 1)⇡ Maxima of rays (2n + 1) h = n = 0, 1, 2, 4 (2n 1) h = n = 1, 2, 4 The color of soap bubble is due to maximum reflected rays
39. Interference of Newton’s Ring OY2 + XY2 = 4r2 ON2 + NY2 = OY2 (x-t)2 + x2 = OY2 r NX2 + NY2 = XY2 t2 + x2 = XY2 (2r-t)2 + x2 + t2 + x2 = 4r2 4r2 - 4rt + t2 + x2 + t2 + x2 = 4r2 ∟ ∟ t2 + x2 = 2rt
40. Interference of Newton’s Ring t2 + x2 = 2rt Thin lens approximation: “a thin lens is a lens with a thickness (distance along the optical axis between the two surfaces of the lens) that is negligible r compared to the radii of curvature of the lens surfaces.” r >> t ⇒ neglect t2 x2 x2 ≈ 2rt ⇒ t ⇡ 2r Condition for Destructive Interference: ∟ n ∟ t = ⇒ x = pnr 2 Always dark fringes at the center Condition for Constructive Interference: (2n + 1) 1 t = ⇒ x = n + r 2 4 s✓ ◆
41. Interference of Newton’s Ring t2 + x2 = 2rt Thin lens approximation: “a thin lens is a lens with a thickness (distance along the optical axis between the two surfaces of the lens) that is negligible r compared to the radii of curvature of the lens surfaces.” r >> t ⇒ neglect t2 x2 x2 ≈ 2rt ⇒ t ⇡ 2r Interference of Newton’s ring Condition for Destructive Interference: n t = ⇒ x = pnr 2 Always dark fringes at the center Condition for Constructive Interference: (2n + 1) 1 t = ⇒ x = n + r 2 4 s✓ ◆