Chapter 25 – Problems 1, 5, 13, 24, 25, 31, 33, 34, 35, 41
  1. From the expression frequency = (velocity of light)/(wavelength), we find
            f = (3.0 x 108 m/s)/(633 x 10-9 m) = 4.74 x 1014 Hz
  1. One reason for this problem to be assigned is to notice the long wavelength of a typical AM radio wave.
            f = (3.0 x 108 m/s)/(200 m) = 1.50 x 106 Hz
  1. Intensity of a laser beam is calculated from the beam power and the beam area.
            Intensity = (3.0 x 10-3 W)/(pi x (0.001 m)2) = 950 W/m2
  1. This problem refers to Sec 25.14: The sun is so far away at RE = 1.5 x 1011 in that it can be treated as a joint light source. It also radiates it's energy isotropically. The intensity of the sun's radiation on earth is E = 1340 W / m2, also called the illuminance. The total energy flux Φ is related to E and R by
                

    Thus, in one hour, the sun radiates an energy of Φ (1 hour) = (3.79 x 1026W)
    (3000 s) = 1.36 x 1030J.

  2. This is a problem that makes use of the inverse square law, assuming the sun’s radiation is isotropic. Let IM and IE represent the intensities of the sun’s radiation at the position of Mercury and Earth, respectively. The ratio (IM)/(IE) is the inverse of the squares of the distances from the sun to the two planets. We know the value of IE and the distances. Therefore
             I    M/IE = (RE/R M)2 → IM = (1340 W/m2)[(1.5 x 10 11 m)/(0.58 x 1011 m)]2 = 8960 W/m2
  1. The term illuminance is a special term used for visible light, and we have not covered it. In this problem, you can change the words illuminance to intensity because the two are conceptually similar and behave the same way. The important point is that this is not an inverse square law problem because the area of the beam does not increase as the square of the distance from the source. Beams of light from a flashlight or searchlight are intentionally focused. Nevertheless, the intensity will always decrease as the beam area spreads, and the relationship is an inverse one. Let I1 and I2 represent the intensities at points 1 and 2, respectively in arbitrary units. Then,
            I    2 / I1 = A1 / A2 → I2 = (20,000 units)[(2.0 m)/(8.0 m)] = 5000 units.
  1. As in problem 31, we will change the words illuminance to intensity. Since the source is a point source, the radiation is isotropic. The logic is the same as for problem 25 above. Since we have a point source, the radiation from the source obeys the inverse square law. Since 80 cm is 8 times as far from the source as 10 cm, the intensity at 80 cm is (1/8)2 the intensity at 10 cm. As a result,
           I (at 80 cm) = (2000 units)/64 = 31 units

  2.  Gamma rays source, point-size, has intensity I independent of distance, see Sec.25.14. Thus,
               the intensity at R = 20cm is the same at R = 60cm.
  1. Once more, the magic word point source is used. This means the source is isotropic. The data tell us that the X-ray intensity at the unknown point P is 1% of the intensity at 5 cm. But 1% implies that the intensity at point P is 1/100 the intensity at 5 cm. We now set up the usual ratio.
           I(at point P)/I(at 5 cm) = 1/100 = [(5 cm)/P] 2 → P = 50 cm.

  1. This problem refers to Sec. 25.15.

    The lamp of intensity I60 = 60W givens an illuminance

    A lamp of intensity I100 = 100W gives at distance above a table R100 an illuminance

             

    We want this to be

              E100 = E60, or