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Unformatted text preview: University of Alabama Department of Physics and Astronomy Department of Electrical and Computer Engineering PH 495/ECE 493 LeClair & Kung Spring 2011 Problem Set 5 Instructions: 1. Answer all questions below. All questions have equal weight. 2. Show your work for full credit. 3. All problems are due Friday 25 March 2011 by 11:59pm. 4. You may collaborate, but everyone must turn in their own work. Problems 9 and 10 involve a take-home experiment, for which you will need 2 glass slides and a plano-convex lens. These items will be provided in lecture on 10 March, or you may pick them up from Dr. LeClair by appointment after that time. 1. Bekefi & Barrett 8.2; Hecht 9.24 A radar antenna operating on a wavelength of 0.10 m is located 8 m above the water line of a torpedo boat. Treat the reflected beam from the water as originating in a source 8 m below the water directly under the radar antenna. The dipole antenna is oriented perpendicular to the plane of the page. 8m θ x R (a) What is the altitude x of an airplane 12 km from the boat if it is to be in the first interference minimum of the radar signal? (b) What is the total number of minima one observes as one scans the sky in the vertical plane as a function of the angle θ, from θ = 0 to θ = π, keeping the distance R fixed? 2. Bekefi & Barrett 8.3 Two dipole radiators (e.g., the oscillating current segments we discussed in class) are separated by a distance λ/2 along the x axis (half-wave dipole antenna). The dipoles are oriented along z, as in the problem we worked in class. Assume the distance to the observation point r satisfies r ￿ λ. (a) Plot the intensity of radiation in the x−y plane. Note the values of intensity at θ = 0, π/3, π/2, π if the oscillators are in phase.i (b) Repeat (a) if the oscillators are 180◦ out of phase. (c) The oscillators are now spaced by a distance λ/4 and are 90◦ out of phase. Repeat (a). Note that this configuration would be very useful for a broadcast station in a coastal city, for example ... 3. Bekefi & Barrett 8.5 We desire to superpose the oscillations of several simple harmonic oscillators having the same frequency ω and amplitude A, but differing from one another by constant phase increments α; that is, E(t) = A cos ωt + A cos (ωt + α) + A cos (ωt + 2α) + A cos (ωt + 3α) + · · · (1) (a) Using graphical phasor addition, find E(t); that is, writing E(t) = Ao cos (ωt + ϕ), find Ao and ϕ for the case when there are five oscillators with A = 3 units and α = π/9 radians. (b) Study the polygon you obtained in part (a) and, using purely geometrical considerations, show that for N oscillators ￿ ￿ ￿￿ sin Nα/2 N−1 E(t) = (NA) cos ωt + α N sin α/2 2 (2) (c) Sketch the amplitude of E(t) as a function of α. The above calculation is the basis of finding radiation from antenna arrays and diffraction gratings. 4. Hecht 9.10 White light falling on two narrow slits emerges and is observed on a distant screen. If red light (λo = 780 nm) in the first-order fringe overlaps violet in the second-order fringe, what is the latter’s wavelength? 5. Hecht 9.26 A soap film surrounded by air has an index of refraction of 1.34. if a region of the film appears as bright red (λo = 633 nm) in normally reflected light, what is its minimum thickness there? 6. Hecht 9.36 One of the mirrors in a Michelson interferometer is moved, and 1000 fringe pairs i You want to make a polar plot with intensity as the radial distance and reference the angle from the midpoint between the two sources. Wolfram alpha is handy for this, http://wolframalpha.com.Try a query like “plot of r = 4cos^2(-(pi/4)*sin(theta) + pi/4)” shift past the hairline in a viewing telescope during the process. If the device is illuminated with 500 nm light, how far was the mirror moved? 7. Hecht 9.47 A glass camera lens with an index of refraction of 1.55 is to be coated with a cryolite film (n ≈ 1.30) to decrease the reflection of normally incident green light (λo = 500 nm). What thickness should be deposited on the lens? 8. Bekefi & Barrett 8.9 A plane electromagnetic wave of wavelength λo is incident on two long, narrow slits, each having width 2a and separated by a distance 2b, with b ￿ a. One of the slits is covered by a thin dielectric plate of thickness d, and dielectric coefficient κ, with d chosen so that √ ( κ − 1)d/λo = 5/2. The interference pattern due to the slits is observed in a plane a distance L from the slits, where L is large enough so that the far field approximations may be used, that is, the pattern depends only on the angle θ from the normal to the slits, as shown. (a) Consider effects due to interference only. What is the condition for a maximum in the pattern? Sketch the interference pattern. 2a θ 2b 2a dielectric slab (b) Now include effects due to both interference and diffraction. How is the intensity distribution modified from that obtained in (a)? Let b/a = 10, sketch the resulting interference-diffraction pattern. (Assume that all angles involved are small enough so that cos θ ≈ 1, and hence that the optical path through the dielectric is independent of angle.) For questions 9 and 10, download the procedure for the take-home experiment here:ii ii These experiments have been adapted from Mavalvala, Nergis, Walter Lewin, and Wolfgang Ketterle, “8.03 http://ocw.mit.edu/courses/physics/8-03-physics-iii-vibrations-and-waves-fall-2004/ assignments/take_home_exp6.pdf In these experiments, you will investigate two interference effects. The phenomena are not hard to observe, and the mathematical derivations and descriptions (found in your text) are not difficult. However, until you see these effects yourself and explore them a bit, you will not have a true understanding of interference. Plus, the interference patterns can be quite striking and colorful! 9. Perform the interference experiment with two glass slides. Take care in making sure your slides are clean and dust-free. Observing the fringes may be easier using a desk lamp close to the slides, position yourself such that you see the mirror-like reflection from the glass surface. If you do not see the fringes clearly, try gently pressing the slides together with a pencil so they make better contact. For a color filter, any colored glass or plastic will do (perhaps a bottle). If you can’t find anything, try using a colored light source, such as bright LED. An interesting variation of the fringe patterns can be seen if you press (carefully!) on the slides with a pencil to slightly bow the slides - you are essentially observing the strain field created in the glass. Briefly, in a paragraph or two, discuss your findings qualitatively, and make an order-of-magnitude estimate of the amount by which your slides deviate from being perfectly flat. 10. Perform the experiment on Newton’s rings using your plano-convex lens and one of the glass slides. Note that the rings are quite small and harder to observe than the fringes in the previous experiment – using a second lens or magnifying glass to more easily observe the rings may be a good idea. Briefly discuss your findings qualitatively. Note in particular why the rings have the size that they do, and how you might make them bigger. Use the rings to roughly estimate the radius of curvature of your lens. From an earlier lab, you should know how to verify the radius of curvature from the focal distance of the lens. Physics III: Vibrations and Waves,” Fall 2004. (Massachusetts Institute of Technology: MIT OpenCourseWare), http://ocw.mit.edu (Accessed 09 Mar, 2011). License: Creative Commons BY-NC-SA. University of Alabama Department of Physics and Astronomy Department of Electrical and Computer Engineering PH 495/ECE 493 LeClair & Kung Spring 2011 Problem Set 5: Solutions 1. Bekefi & Barrett 8.2; Hecht 9.24 A radar antenna operating on a wavelength of 0.10 m is located 8 m above the water line of a torpedo boat. Treat the reflected beam from the water as originating in a source 8 m below the water directly under the radar antenna. The dipole antenna is oriented perpendicular to the plane of the page. is there a m 8m θ x R (a) What is the altitude x of an airplane 12 km from the boat if it is to be in the first interference minimum of the radar signal? (b) What is the total number of minima one observes as one scans the sky in the vertical plane as a function of the angle θ, from θ = 0 to θ = π, keeping the distance R fixed? Solution: From the geometry of the figure, referencing the crudely modified figure below, you should be able to convince yourself that (1) d2 = (x − h)2 + R2 ￿2 d x-h = (x + h) + R d h h 2 θ θ θ d’ R (2) 2 x h The path difference between the direct ray (d) and the reflected ray d￿ is thus ￿ ￿ 2 2− (x − h)2 + R2 d − d = (x + h) + R ￿ (3) The phase difference δ between the two rays is this path difference multiplied by k = 2π/λ plus an extra phase shift of π for the reflected ray: δ= The intensity is then easily found: ￿ 2π ￿ ￿ d −d +π λ (4) ￿￿ ￿π ￿ ￿ ￿π ￿ ￿ ￿￿ δ I = 4Io cos = 4Io cos2 d￿ − d + π = 4Io sin2 d￿ − d 2 λ λ 2 (5) The minima in intensity occur when ￿ δ π￿ ￿ = d − d = nπ 2 λ ￿ nλ = d￿ − d = (x + h)2 + R2 − (6) ￿ (x − h)2 + R2 (7) The problem is then the “simple” matter of solving the above for x. It can be done, with some tedium. First square both sides and isolate the remaining radical. nλ = ￿ 2 (x + h) + R2 − ￿ (x − h)2 + R2 ￿ ￿ ￿ n2 λ2 = (x + h)2 + R2 + (x − h)2 + R2 − 2 (x + h)2 (x − h)2 + R4 + R2 (x + h)2 + (x − h)2 Square both sides and collect terms . . . ￿ ￿2 ￿ 2 2 22 2 x + h + R − n λ = 2 (x2 − h2 )2 + R4 + 2R2 (x2 + h2 ) ￿ ￿2 ￿ ￿ ￿ ￿ ￿￿ 4 x2 + h2 + R2 + n4 λ4 − 4n2 λ2 x2 + h2 + R2 = 4 x4 + h4 − 2x2 h2 + R4 + 2R2 x2 + h2 4x4 + 4h4 + 4R4 − 8x2 h2 + 8x2 R2 + 8h2 R2 = 4x4 + 4h4 + 4R4 + 8x2 h2 + 8x2 R2 ￿ ￿ + 2h2 R2 + n4 λ4 − 4n2 λ2 x2 + h2 + R2 ￿ ￿ ￿ ￿ 16x2 h2 − 4n2 λ2 x2 = 4n2 λ2 h2 + R2 − n4 λ4 = n2 λ2 4h2 + 4R2 − n2 λ2 ￿ ￿ n2 λ2 4h2 + 4R2 − n2 λ2 2 x= 16h2 − 4n2 λ2 ￿ 4R2 + 4h2 − λ2 n2 x = nλ ≈ 75.0 m (8) 16h2 − 4λ2 n2 If we first use the very reasonable approximation λ ￿ {R, h} and then the nearly as reasonable h ￿ R, we can come to something a bit simpler: ￿ x = nλ √ 4R2 + 4h2 − λ2 n2 R2 + h2 R ≈ nλ ≈ nλ 16h2 − 4λ2 n2 2h 2h (9) √ These latter two forms are readily derived by starting from Eq. 7, factoring out R2 + x2 + h2 from both terms, and approximating the resulting radicals for 2xh ￿ R2 + x2 + h2 using a binomial expansion (since x is at minimum h, this is ok). Using the numbers given, one finds the first minimum at x ≈ 75.0 m. How many minima must there be over the interval [0, π]? We have solved the problem of finding the minima only over the interval [0, π/2]. However, owing to the symmetry of the problem, we know that the interval [π/2, π] has the same number of minima. Unfortunately, we have no obvious restriction on magnitude x from the information given, and everything else is fixed. We do know that at θ = 0, x = h, but as θ approaches π/2, x increases without bound. What to do? The exact expression above does have an interesting condition built in, however: x must be real. For this to be true, the numerator and denominator in Eq. 8 must both be positive. That is, 2￿ 2 R + h2 λ 2h = λ 4R2 + 4h2 − λ2 n2 > 0 =⇒ nmax = (10) 16h2 − 4λ2 n2 > 0 =⇒ nmax (11) Given that R ￿ h, it is clear that the second expression is more restrictive. Using the numbers given, one finds n = 160. This is for the interval [0, π/2], the problem we have solved. The interval (π/2, π] thus contains 159 minima (not double counting the minima at π/2, so in total over θ ∈ [0, π] we have n = 319 minima. Alternate method The above is an exact method, though far from obvious. Given that both λ and h are tiny compared to R, we need not solve the problem exactly. We can use the same dipole approximation we employed previously to make things far simpler. The gist of the idea is that the source is very distant compared to the spacing of the real and virtual sources (h << R), and the geometry looks more like this: A Δ h x r γ θ θ R We assume the source is sufficiently distant that the rays approaching the source can be approximated as parallel. Construct a line running from the intersection of the reflected ray with the water to the direct ray, such that this line meets the direct ray at a right angle (point A in the figure). The parallel ray approximation means that from point A rightward, both direct and reflected rays traverse the same distance. The only geometrical phase difference is thus due to the path difference: the reflected ray travels a distance r, while the direct ray travels a distance ∆. If we can find the difference r − ∆, we are nearly done. From the figure above, the surface of the water is made up from angles θ, γ, and 90, such that γ = 180 − 90 − 2θ = 90 − 2θ (12) h sin θ ∆ = r sin γ (13) You can convince yourself that r= (14) The path difference is then h h r−∆= − r sin γ = − sin θ sin θ ￿ h sin θ ￿ sin (90 − 2θ) = h (1 − cos 2θ) sin θ (15) Accounting for the phase shift on reflection, our total phase shift is then δ= ￿ 2π 2π h 2π h ￿ 4πh (r − ∆) + π = (1 − cos 2θ) + π = 2 sin2 θ + π = sin θ + π λ λ sin θ λ sin θ λ (16) The intensity is then ￿￿ ￿ ￿ ￿ ￿ δ π 2 2πh 2 2πh I = 4Io cos = 4Io cos sin θ + = 4Io sin sin θ 2 λ 2 λ 2 (17) The condition for a minimum is 2πh sin θ = nπ λ or sin θ = nλ 2h (18) For a very distant source, the separation of the parallel rays is small compared to x, and we may approximate x ≈ R tan θ ≈ R sin θ = Rnλ/2h, which gives x ≈ 75 m in agreement with our previous result. As for the number of minima, we need only note from the above expression that sin θ returns values only in the interval [0, 1] over θ ∈ [0, π/2]. That is, since sin θ is at most 1, nλ = sin θ ￿ 1 2h or n￿ 2h = 160 λ (19) This is precisely the condition we derived earlier; accounting for the angles [π/2, π], we end up with a total of 2 × 160 − 1 = 319 minima. Whether you prefer the first or second method is really a matter of taste. I preferred the first since it gives an exact answer, though such precision is of limited utility - it is probably not worth the extra difficulty. The advantage of the second method is that it is the same problem we’ve solved many times already (it is just the double slit again, or Lloyd’s mirror) with a couple of small twists. 2. Bekefi & Barrett 8.3 Two dipole radiators (e.g., the oscillating current segments we discussed in class) are separated by a distance λ/2 along the x axis (half-wave dipole antenna). The dipoles are oriented along z, as in the problem we worked in class. Assume the distance to the observation point r satisfies r ￿ λ. (a) Plot the intensity of radiation in the x−y plane. Note the values of intensity at θ = 0, π/3, π/2, π if the oscillators are in phase.i (b) Repeat (a) if the oscillators are 180◦ out of phase. (c) The oscillators are now spaced by a distance λ/4 and are 90◦ out of phase. Repeat (a). Note that this configuration would be very useful for a broadcast station in a coastal city, for example ... Solution: The intensity is I = 4Io cos 2 ￿ ￿ ￿ ￿ πd 1 1 2π sin ψ − (ϕ1 − ϕ2 ) = 4Io cos sin ψ − (ϕ1 − ϕ2 ) λ 2 2 2 (20) with d = λ . For (a) and (b), we need only make polar plots with I/Io as the radial coordinate and 2 ψ as the angular coordinate. For (a) we have ϕ1 − ϕ2 = 0, and for (b) we have ϕ1 − ϕ2 = π. For (c), the spacing is now λ/4 and ϕ1 − ϕ2 = π/4. Can you see now why this would be useful for a transmitter in a coastal city? You want to make a polar plot with intensity as the radial distance and reference the angle from the midpoint between the two sources. Wolfram alpha is handy for this, http://wolframalpha.com.Try a query like “plot of r = 4cos^2(-(pi/4)*sin(theta) + pi/4)” i 4 2 3 2 1 1 0.5π 0.5π 0.75π 0.75π 0.25π 0.25π π -4 π -4 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1.25π 4 1 2 3 4 1.75π 1.5π -1 1.25π 1.75π 1.5π -1 -2 -3 -2 -4 Figure 1: (left) Intensity pattern for two oscillators with zero phase offset, spaced at d = λ/2. The oscillators are at y = ±λ/4. (right) The same oscillators with a phase difference of ϕ1 − ϕ2 = π. The radiation pattern is rotated by 90◦ . 4 3 2 1 0.5π 0.75π 0.25π π -4 -3 -2 -1 0 1.25π 1 2 3 4 1.75π 1.5π -1 Figure 2: Intensity pattern for two oscillators with phase offset ϕ1 − ϕ2 = π/2, spaced at d = λ/4. The oscillators are at y = ±λ/8. 3. Bekefi & Barrett 8.5 We desire to superpose the oscillations of several simple harmonic oscillators having the same frequency ω and amplitude A, but differing from one another by constant phase increments α; that is, E(t) = A cos ωt + A cos (ωt + α) + A cos (ωt + 2α) + A cos (ωt + 3α) + · · · (21) (a) Using graphical phasor addition, find E(t); that is, writing E(t) = Ao cos (ωt + ϕ), find Ao and ϕ for the case when there are five oscillators with A = 3 units and α = π/9 radians. (b) Study the polygon you obtained in part (a) and, using purely geometrical considerations, show that for N oscillators E(t) = (NA) ￿ ￿ ￿￿ sin Nα/2 N−1 cos ωt + α N sin α/2 2 (22) (c) Sketch the amplitude of E(t) as a function of α. The above calculation is the basis of finding radiation from antenna arrays and diffraction gratings. Solution: You can find a quick solution using a phasor diagram here: http://ocw.mit.edu/courses/physics/8-03-physics-iii-vibrations-and-waves-fall-2004/ assignments/soln10.pdf We will solve the N oscillator problem by more direct means. Each oscillator has a phase offset of α from its neighbor, so the Nth oscillator has a phase offset of (N − 1) α from the first. Since the total field from all N oscillators is the sum of their individual electric fields, and thusii The symbol ￿ is the operator that takes the real part of ￿ expression, and ￿ is the operator that takes the an imaginary part. Also recall the sum of a finite geometric series n−1 ark = a(1 − rn )/(1 − r). 0 ii Etot = N−1 ￿ A cos (ωt + nα) = A n=0 N−1￿ ￿ cos (ωt) cos (nα) − sin (ωt) sin (nα) n=0 = A cos (ωt) N−1 ￿ cos (nα) − A sin (ωt) n=0 N−1 ￿ ￿ (23) (24) sin (ωt) cos (nα) n=0 ￿− ￿ ￿ ￿￿ ￿− ￿ ￿ ￿￿ ￿ ￿ cos N2 1 α sin Nα sin N2 1 α sin Nα 2 2 = A cos ωt − A sin ωt sin α sin α 2 2 ￿ ￿￿ ￿ ￿ ￿ ￿￿ sin Nα (N − 1) α (N − 1) α 2 =A cos (ωt) cos − sin (ωt) sin sin α 2 2 2 ￿ ￿ Nα ￿ ￿ ￿ ￿ ￿ ￿ ￿￿ sin 2 (N − 1) α sin Nα/2 N−1 (NA) =A cos ωt + = cos ωt + α sin α 2 N sin α/2 2 2 (25) (26) (27) Thus, total field is that of a single oscillator of amplitude and phase ￿ ￿ sin Nα 2 Ao = A sin α 2 (N − 1) α ϕ= 2 (28) (29) One can also perform the sum using complex exponentials, which makes it a simple geometric series. Given N = 5, A = 3 and α = π/9, one finds Ao ≈ 13.23 and ϕ = 2π/9 = 40◦ . In order to sketch the amplitude as a function of α, we should recall that lim β →0 sin Nβ =N β (30) which means that the intensity has zeros at Nα/2 = nπ unless sin (α/2) = 0, in which case we have a maximum. Below are plots of I/NA versus α for N = 5 (red) and N = 6 (black) for ωt = 0. Note the primary maxima at α = 2nπ and the (N − 2) secondary maxima in between. 1 0.75 0.5 0.25 -10 -7.5 -5 -2.5 0 2.5 5 7.5 10 4. Hecht 9.10 White light falling on two narrow slits emerges and is observed on a distant screen. If red light (λo = 780 nm) in the first-order fringe overlaps violet in the second-order fringe, what is the latter’s wavelength? Solution: The interference condition for a fringe to occur is sin θm ≈ θm = mλ a (31) where a is the slit spacing. If θred,m=1 = violet,m=2 , this implies 1 λviolet = λred = 390 nm 2 (32) 5. Hecht 9.26 A soap film surrounded by air has an index of refraction of 1.34. if a region of the film appears as bright red (λo = 633 nm) in normally reflected light, what is its minimum thickness there? Solution: If the film appears red, it is because the thickness is such that interference creates a minimum in intensity for red light at wavelength λo in vacuum. These minima occur for a film of thickness d and index n at d cos θt = (2m + 1) λf 4 (33) where the wavelength of light in the film λf relates to the vacuum wavelength via λf = λo /n. If we consider the intensity at normal incidence (θt = 0) and the minimum thickness (corresponding to m = 0), d= λf λo = ≈ 118 nm 4 4n (34) 6. Hecht 9.36 One of the mirrors in a Michelson interferometer is moved, and 1000 fringe pairs shift past the hairline in a viewing telescope during the process. If the device is illuminated with 500 nm light, how far was the mirror moved? Solution: A fringe pair will shift pass the hairline whenever one of the mirrors moves by λ/2. If N fringe pairs move past the hairline, one of the mirrors must have moved by a distance 1 ∆d = Nλ 2 (35) With N = 1000 and λo = 500 nm, ∆d = 250 µm. 7. Hecht 9.47 A glass camera lens with an index of refraction of 1.55 is to be coated with a cryolite film (n ≈ 1.30) to decrease the reflection of normally incident green light (λo = 500 nm). What thickness should be deposited on the lens? Solution: Reflectivity minima occur for a film of thickness d and index n when d cos θt = (2m + 1) λf 4 (36) where the wavelength of light in the film λf relates to the vacuum wavelength λo via λf = λo /n. If we consider the intensity at normal incidence (θt = 0), the minimum thickness must be d = (2m + 1) λf λf λo = = ≈ 96 nm 4 4 4n (37) 8. Bekefi & Barrett 8.9 A plane electromagnetic wave of wavelength λo is incident on two long, narrow slits, each having width 2a and separated by a distance 2b, with b ￿ a. One of the slits is covered by a thin dielectric plate of thickness d, and dielectric coefficient κ, with d chosen so that √ ( κ − 1)d/λo = 5/2. The interference pattern due to the slits is observed in a plane a distance L from the slits, where L is large enough so that the far field approximations may be used, that is, the pattern depends only on the angle θ from the normal to the slits, as shown. (a) Consider effects due to interference only. What is the condition for a maximum in the pattern? Sketch the interference pattern. 2a θ 2b 2a dielectric slab (b) Now include effects due to both interference and diffraction. How is the intensity distribution modified from that obtained in (a)? Let b/a = 10, sketch the resulting interference-diffraction pattern. (Assume that all angles involved are small enough so that cos θ ≈ 1, and hence that the optical path through the dielectric is independent of angle.) Solution: Consider the revised figure below, which places a screen a distance L ￿ {a, b} from the slits. An observation point on the screen is then a distance r1 from one slit and and r2 from the other, as well as a distance R from the midpoint of the two slits. L r1 2a R θ 2b r2 2a dielectric slab The law of cosines allows us to find the distances r1 and r2 in terms of b, R and θ r2 = b2 + R2 − 2bR cos (θ + 90) = b2 + R2 + 2bR sin (θ) 2 (38) r2 = b2 + R2 − 2bR sin (θ) 1 (39) The difference of the squared differences is then r2 − r2 = 4bR sin (θ) = (r1 + r2 ) (r2 − r1 ) 2 1 r2 − r1 = 4bR sin (θ) r1 + r2 (40) (41) One can solve this analytically for r2 in terms of r1 and then compute r2 − r1 exactly.iii However, since we are explicitly encouraged to use the far field approximation, we may as well save a bit of tedium. If we make the approximation that θ is small, amounting to the condition L ≈ R, then iii This results in r2−r1 ≈ ￿ 2r2 − 8bR sin θ, which does not depend only on θ as the far field approximation requires. 2 r1 ≈ r2 and thus r1 + r2 ≈ 2R. We cannot do the same for r2 − r1 , however, since we are interested in difference between r2 and r1 on the scale of the incident wavelength. This leads us to r2 − r1 ≈ 4bR sin (θ) 2R (42) This is the geometric path difference between the two beams, exactly as we have previously derived for the two slit problem. Ignoring the finite width of the slits for the moment, this would be sufficient to calculate the two slit interference pattern. In the present case, we must also take into account the propagation delay suffered by the second beam resulting from its transit through the dielectric. In crossing a distance d through the dielectric, the second beam requires a time t2 = d dn = v c (43) while the first transiting only vacuum through the same distance d requires a time t1 = d c (44) The second beam therefore suffers a delay of δt2 = ￿ d d ￿√ (n − 1) = κ−1 c c (45) √ where we have used the result n = κ in a medium where µ ≈ µo . This time delay implies a phase delay for the second beam relative to the first of ￿ d ￿√ ￿ ωd ￿√ κ−1 = κ − 1 = 5π (46) c λ √ Here we have used λ = c/f = 2πc/ω and the given relationship ( κ − 1)d/λo = 5/2. We must add this phase delay to the geometric phase delay k (r2 − r1 ) to find the total phase difference between the two beams: ωδt2 = (47) δtot = 2kb sin θ + 5π This gives the intensity on the screen asiv I = 4Io cos 2 ￿ δtot 2 ￿ ￿ ￿ 5π = 4Io cos kb sin θ − 2 2 (48) We will have maxima when the argument of sin2 is an integer multiple of π: 5π kb sin θ − = mπ 2 or λ 2b sin θ = 2 ￿ ￿ 5 m+ 2 (49) Note that whether we pick +5π/2 or −5π/2 makes no difference here, it physically only amounts to asking whether we are considering points above or below the midpoint of the screen, and ultimately makes no difference at all. iv Since m is just an integer, what this formula is really telling us is that the path difference 2b sin θ must be a half-integer multiple of λ/2. We could just as well re-index the starting point of m and write λ r2 − r1 ≈ 2b sin θ = 2 ￿ 1 m+ 2 ￿ (50) The gist of the matter is that the interference pattern now has a minimum at θ = 0 as a result of the dielectric slab, rather than a maximum. Otherwise, it looks just like the usual two-slit problem. The inclusion of diffraction essentially turns each slit into an array of point sources rather than a single point source. When we take into account the finite slit width, in the far-field regime we can imagine each point within a slit acts as a new source of spherical waves which will interfere with each other. We can model this as an array of point sources, as in problem 3, letting N → ∞. Moreover, now the constant phase difference between neighboring oscillators makes sense - since each successive evenly-spaced source is slightly farther from the observation point, this phase offset is accounting for the propagation delay due to the spacing between oscillators. If we have a slit of width 2a made up of N oscillators, then the spacing between oscillators is d = 2a/N, or N = 2a/d. Our slit is the case that d → 0 with fixed 2a. From problem 3, the total field of N oscillators in an array, each of intensity A, was ￿ ￿ ￿￿ sin Nα/2 N−1 E = (NA) cos ωt + α N sin α/2 2 (51) The phase difference α for oscillators spaced a distance d from each other can be written (in analogy to our work above) α = kd sin ψ, where ψ ≈ θ is the angle to the point of interest. Adding 2a = Nd and squaring to find the intensity (and noting k = 2π/λ), ￿ ￿ sin2 d 2πa sin ψ λ ￿ ￿ I=A 2 2πa sin sin ψ λN (52) 2 Taking the limit that N → ∞, with ψ ≈ θ and defining Io ≡ A2 , the intensity of a slit of finite width 2a is 2 I = N Io ￿ sin β β ￿2 with β= 2πa sin θ λ (53) In the end, accounting for the finite width of the slits merely modulates the intensity by a factor sin2 (β)/β. Thus, for the problem at hand, I = 4Io ￿ sin ￿ 2πa λ sin θ 2πa λ sin θ ￿ ￿2 cos 2 ￿ 2πb 5π sin θ − λ 2 ￿ (54) The intensity plot is left as an exercise to the reader . . . ...
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