Physics 2 - Lecture 5: Interference 2 - Pham Tan Thi

Nội dung text: Physics 2 - Lecture 5: Interference 2 - 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. 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
3. 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.
4. 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
5. 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
6. 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
7. 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
8. Review: Lights of Different Colors Long wavelength (Low frequency) Short wavelength (High frequency) Violet light has the highest energy in visible region
9. Light Polarization
10. 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.
11. 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 
12. 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.
13. 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 .
14. 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 + )
15. 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
16. 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
17. (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.
18. Interference of Thin Film (inverse phase) Incoming wave Transmitted wave Reflected wave Incoming wave (same phase) Transmitted wave Reflected wave
19. 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
20. 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
21. 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✓ ◆