Chapter 35, example problems: (35.02) Two radio antennas A and B radiate in phase. B is 120 m to the right of A. Point Q along extension of line AB. 40 m to the right of B. Frequency and wavelength can be varied. A B Q (a) Destructive interference at Q. Longest wavelength? AQ – BQ = (1/2)λ ⇒ 120 m = (1/2)λ ⇒ λ = 240 m. (Note: If you change (1/2) to (3/2), (5/2), etc., you will get other possible shorter wavelengths. But the problem asks for the longest wavelength.) (b) Constructive interference at Q. Longest wavelength? Since AQ – BQ can not be zero, we demand AQ – BQ = λ. ⇒ 120 m = λ ⇒ λ = 120 m. (35.16) Coherent light. Frequency 6.32 × 1014 Hz. Pass through two thin slits. Fall on a screen 85.0 cm away. 3rd bright fringe occurs at ±3.11 cm on either side of center bright fringe. (a) Slit separation d. d sin θ = mλ = 3 × (3 × 108 m/s / 6.32 × 1014 Hz) tan θ = 3.11 cm / 85 m. But this θ is clearly very small. So sin θ ≅ tan θ (≅ θ in rad.) ∴ d = 3 × (3 × 108 m/s / 6.32 × 1014 Hz) / (3.11 cm / 85 cm) = 3.89 × 10-5 or d = 3.89 × 10-3 cm. (Note: The unit Hz is just s−1.) (b) Location of the third dark fringe? The first dark fringe occurs midway between the center (zeroth) bright fringe and the first bright fringe. So the third dark fringe occurs midway between the second and third bright fringes. Hence we solve dθ = 2.5 × (3 × 108 m/s / 6.32 × 1014 Hz) to get θ = 0.0305. Multiplying it by 85 cm gives 0.0259 m, or 2.59 cm being the distance on the screen between the third dark fringe and the center of the center bright fringe. (35.26) Two antennas, 9.00 m apart. Radiate in Phase at 120 MHz. Receiver, placed 150 m from both antennas measured an intensity I0. Receiver moved so that it is now 1.8 m closer to one antenna than to the other. (a) phase difference: φ = 2π × 1.8 m / λ = 2π × 1.8 m / (3.00 × 108 m/s / 120 × 106 Hz) = 4.524 rad. (b) Intensity at new position: I = I0 cos2 (φ /2) = 0.4064 I0 . Why I = I0 cos2 (φ /2)? Recall the phasor diagram at right: φ φ / 2 m(35.30) Glass plate, 9.00 cm long, placed in contact with a second plate. Held at a small angle by a metal strip 0.0800 mm thick under one end. Air between the plates. Illuminated from above by light with wavelength λ = 656 nm. Number of interference fringes per centimeter? The interference is between light reflected from the bottom surface of the top plate, and the top surface of the bottom plate. Other surfaces are not relevant. (If you want to know why it is so, come to class and listen to the lectures.) The left end is the center of a dark fringe, because the lower reflection gets a 180° phase shift ⎯ due to reflection by a denser medium, but not the upper Reflection, which is a reflection by a lighter medium. The difference in path lengths between the two paths is zero on the left edge, and it increases conti- nuously to the maximum value of 2 × 0.0800 mm = 0.1600 mm on the right edge for light coming down essentially perpendicular to the plates (called normal incidence). The phase difference between the two paths on the right edge of the plates is 2π × (path length difference / λ) = 2π × (0.1600 mm / 656 nm) = 574.68 rad., but for this problem, it is only necessary to look at the part (0.1600 mm / 656 nm) = 243.9, which is slightly below 244. It means that there are almost 244 wavelengths in this maximum path-length difference. Thus the center of the 244th dark fringe besides the one on the left edge of the plates is just outside the right edge of the plates and therefore can not be seen. But the bright fringe just to the left of this not-fully-observable dark fringe needs only a path-length difference of 243.5 wavelengths, and is therefore already fully developed. Since there is a bright fringe to the left of each dark fringe, including the first one, it is clear that the total number of bright fringes that can be seen is 244. Dividing it by 9.00 cm gives 27.11 bright fringes per centimeter that can be seen. (Note that the inverse of 27.11 number per centimeter is 0.0369 cm. This is the spacing between the neighboring bright fringes that can be seen from above.) (35.38) Michelson interferometer. Jan uses λ1 = 606 nm. He moved the movable mirror away from him by some distance, and counted 818 bright fringes. Then Linda changes to λ2 = 502 nm, and moved the movable mirror toward her by some distance, and counted also 818 bright fringes. The fringes moved in opposite directions across the line. Find the distances moved by the movable mirror in each case. (a) Jan moved the mirror 818 × 606 nm / 2 = 2.479 ×10-4 m away from him, and Linda moved the mirror 818 × 502 nm / 2 = 2.053 ×10-4 m toward her. (Note: Light travels to the movable mirror and back, so the distance moved by the mirror should be counted twice.) (b) The resultant displacement of the mirror = 2.479 ×10-4 m - 2.053 ×10-4 m = 0.426 ×10-4 m , or 4.26 ×10-5 m .(35.44) Two radio antennas, 200 m apart, at A and B, radiate in phase. Frequency 5.80 MHz. Receiver at C. BC is perpendicular to AB. C B A x Find location of C for destructive inter- ference. We need AC – BC = (1/2) λ. ⇒ √x2 + (200 m)2 – x = (1/2) (3 × 108 m/s / 5.80 × 106 Hz) = 25.86 m, or, after adding x to both sides and then squaring both sides, we obtain: x2 + (200 m)2 = x2 + 25.86 m × x + (25.86 m)2. After simplifying, we obtain 25.86 m × x = 39331 m2, giving x = 1521 m. (35.50) Two-slit interference. The two slits are of different widths. Distant screen. Amplitude from the first slit = E. Amplitude from the second slit = 2E. (a) I (φ ) = ? The intensity is proportional to the net amplitude squared. A phasor diagram tells you how to find the net amplitude. φ For φ = 0, the net amplitude is 3E. For φ ≠ 0, the net amplitude follows the law of cosine (as suggested by the diagram): √ (2E)2 + (E)2 − 2 (2E) (E) cos (180° − φ) = √ 5E 2 + 4E 2 cos φ . The ratio of the two net amplitudes is √(5/9) + (4/9) cos φ . Therefore, I (φ ) = I0 [(5/9) + …