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Unformatted text preview: Lecture-22 Examples Solutions HW 22-4. (HRW 33-51) In Fig. 33-53, a 2.00-m-long vertical pole extends from the bottom of a swimming pool to a point 50.0 cm above the water. Sunlight is incident at . What is the length of the shadow of the angle pole on the level bottom of the pool? Figure 33-53
Solution: 51. Consider a ray that grazes the top of the pole, as shown in the diagram that follows. . Here 1 = 90 = 35, l1 = 0.50 m, and l 2 = 150 m. The length of the shadow is x + L. x is given by x = l1 tan 1 = (0.50 m) tan 35 = 0.35 m. According to the law of refraction, n2 sin 2 = n1 sin 1. We take n1 = 1 and n2 = 1.33 (from Table 33-1). Then, 2 = sin -1
L is given by FG sin IJ = sin FG sin 35.0 IJ = 2555 . H 133 K . . H n K
1 -1 2 . L = l 2 tan 2 = (150 m) tan 25.55 = 0.72 m. The length of the shadow is 0.35 m + 0.72 m = 1.07 m. HRW 33-83 During a test, a NATO surveillance radar system, operating at 12 GHz at 180 kW of power, attempts to detect an incoming stealth aircraft at 90 km. Assume that the radar beam is emitted uniformly over a hemisphere. (a) What is the intensity of the beam when the beam reaches the aircraft's location? The aircraft reflects radar waves as though it has a cross-sectional area of only 0.22 m2. (b) What is the power of the aircraft's reflection? Assume that the beam is reflected uniformly over a hemisphere. Back at the radar site, what are (c) the intensity, (d) the maximum value of the electric field vector, and (e) the rms value of the magnetic field of the reflected radar beam? SSM Solution: 83. (a) The area of a hemisphere is A = 2r2, and we get I = P/A = 3.5 W/m2. (b) Our part (a) result multiplied by 0.22 m2 gives 0.78 W. (c) The part (b) answer divided by the A of part (a) leads to1.5 10-17 W/m2. (d) Then Eq. 33-26 gives Erms = 76 nV/m Emax = 2 Erms = 1.1 10-7 V/m. (e) Brms = Erms/c = 2.5 10-16 T = 0.25 fT. ...
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- Spring '08