Chapter 28 Solutions - 3, 5, 7, 9, 11, 12, 14, 17, 20, 21, 26, 28

  1. This problem refers to Sec 28.1, interference between waves. The waves of the two identical speakers, on at x1 = 0 and the other at x2 > 0, can add constructively (waves are in phase with each other) or destructively (waves are out of phase with each other). The person at xp = 20m  >>x2 hears loudest sounds when x2 - x1 = λ, 2λ, 3λ …… Thus, with λ = 70 cm = 0.7 m and x2  < 5m, the second speaker positions can be 0.0m, 0.7m; 1.4m; 2.1m; 2.8m, 3.5m, 4.2m; 4.9m .

  1. Again, this problem refers to Sec 28.1, recognizing now that the distance between the two speakers x2 - x1 = 1.4 m should be such that an odd number of half-wave lengths should fit between the two speakers:

    Thus, λ1 = 2.8m; λ2 = 0.93m; λ3 = 0.56 m, λ4 = 0.4m

  1. This problem refers to Sec. 28.3, (see Fig 28.3 and 28.4). The slits are a distance d = 0.070mm = 7 x 10-5m apart. The screen  is at h = 2m.
    1. The zeroth order fringe is the center of the pattern of fringes: by definition it is at x =0, and it is directly below the middle of the two slits. Therefore, the distances from the zeroth order fringe to the slits are the same: the difference is 0.
    2. Difference for the first bright fringe x is such that
                
    3. x2 = 2x1 = 3.12 cm
    4. x3 = 3x1 = 4.68 cm
  1. Again, as above, reference to Sec. 28.3. Slit separation d = 0.100 mm = 10-4m. Slit-to-screen distance h=1.5m . Yellow light wavelength λ = 589 mm = 5.89 x 10-7m.
    1. Distance x from zeroth order bright fringe to third bright fringe is such that path length difference ΔS = 3 λ, thus, with
                
    2. To third dark fringe, ΔS = λ/2, thus
                
  1. Once more, see Sec. 28.3. With 8000 lines, d = 1/8000 cm = 1.25 x 10-6 m.
    1. The angle θ2 = 72o for second order maximum, for which ΔS = 2λ. Thus, with

                     
    2. For third order maximum we will have
                     [Third order maximum is invisible.]
  1. See Sec 28.4. A helium-neon laser has λ = 633nm = 6.33 x 10-7 m. First order maximum occurs at θ1 = 52o, on each side of central maximum.
    1. Using Eq. (28.2) we can express the distance d between centers of grating lines:

                
    2. The number of grating lines per centimeter is simply
                     

  1. Diffraction grating has 9000 lines/cm, thus
                   
    Light's wavelength λ = 582 mm = 5.82 x 10-7 m.

    1. Angle θ1, for first order maximum occurs when, λ = d sin θ1
                

    2. Second order maximum at θ2 such that,

                
                [sinθ2 = 1.049 > 1]

  1. For a single slit case, see Fig 28.9.  Wavelength of the light is λ = 5.79 x 10-7m, and the slit width w = 0.020 mm = 2 x 10-5 m. Thus, the half-width angle of the bright spot is sin θc = λ / w, thus

              
  1. This problem refers to Sec. 28.6. The angular resolution θc, with which the Yerkes telescope can see spots on the Moon is given by

                   

    The linear resolution is now
                   

  2. Again, Sec 28.6 is involved. Now lens diameter D = 0.25 cm = 2.5 x 10-3m, and the light's wavelength is λ =500 mm = 5 x 10-7m. If we want millimeter lines to still be clearly discernible, then   d = 1mm = 10-3m. Now, what should be distance h?  It is obvious that,
              
  1. Refer to Fig 28.11. The separation x, given the strongest reflections, should be such that x = n λ / 2 (n = 1, 2, 3, 4, …) with λ the light's wavelength, λ = 4.4 x 10-7m.

              For the first four strongest reflections, x = (2.2, 4.4, 6.6, 8.8) x 10-7 m.
  1. Again, Fig 28.11. Now, for near zero reflection ("near" became reflection not transmission perfect from second glass plate!) x = (n + 1/2) λ / 2 gap widths, λ = 5.46 x 10-7m.


    2.           For the first four zero reflection distances, x = (1.1, 3.3, 5.5, 7.7) x 10-7 m