This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 33 12. The intensity of the signal at Proxima Centauri is I= P 10 × 106 W . = 2 4 πr 4 π 4.3 ly 9.46 × 1015 m / ly b gc h 2 = 4.8 × 10−29 W / m2 . 17. (a) The average rate of energy flow per unit area, or intensity, is related to the electric field 2 amplitude Em by I = Em / 2 μ 0 c , so Em = 2 μ 0cI = 2 4 π × 10−7 H / m 2.998 × 108 m / s 10 × 10−6 W / m2 = 8.7 × 10 −2 V / m. (b) The amplitude of the magnetic field is given by c hc hc h Bm = Em 8.7 × 10−2 V / m = = 2.9 × 10−10 T. 8 c 2.998 × 10 m / s (c) At a distance r from the transmitter, the intensity is I = P / 2πr 2 , where P is the power of the transmitter over the hemisphere having a surface area 2π r 2 . Thus P = 2πr 2 I = 2π (10×103 m ) 2 (10 ×10 −6 W/m 2 ) = 6.3×103 W. 20. The radiation pressure is pr = I 10 W / m2 = = 3.3 × 10−8 Pa. 8 c 2.998 × 10 m / s 34. After passing through the first polarizer the initial intensity I0 reduces by a factor of 1/2. After passing through the second one it is further reduced by a factor of cos2 (π – θ1 – θ2) = cos2 (θ1 + θ2). Finally, after passing through the third one it is again reduced by a factor of cos2 (π – θ2 – θ3) = cos2 (θ2 + θ3). Therefore, If 1 1 = cos 2 (θ1 + θ 2 )cos 2 (θ 2 + θ 3 ) = cos 2 (50° + 50°)cos 2 (50° + 50°) I0 2 2 = 4.5 ×10−4. Thus, 0.045% of the light’s initial intensity is transmitted. 37. (a) Since the incident light is unpolarized, half the intensity is transmitted and half is absorbed. Thus the transmitted intensity is I = 5.0 mW/m2. The intensity and the electric field amplitude are related by 2 I = Em / 2 μ 0c, so Em = 2 μ 0cI = 2(4 π × 10 −7 H / m)(3.00 × 108 m / s)(5.0 × 10−3 W / m2 ) = 19 V / m. . (b) The radiation pressure is pr = Ia/c, where Ia is the absorbed intensity. Thus pr = 5.0 × 10−3 W / m2 = 17 × 10−11 Pa. . 8 3.00 × 10 m / s 45. The law of refraction states n1 sin θ 1 = n2 sin θ 2 . We take medium 1 to be the vacuum, with n1 = 1 and θ1 = 32.0°. Medium 2 is the glass, with θ2 = 21.0°. We solve for n2: n2 = n1 sin θ 1 sin 32.0° = (100) . = 148. . sin θ 2 sin 210° . FG H IJ K 51. Consider a ray that grazes the top of the pole, as shown in the diagram that follows. Here θ1 = 90° – θ = 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 θ 1 = (0.50 m) tan 35° = 0.35 m. According to the law of refraction, n2 sin θ2 = n1 sin θ1. We take n1 = 1 and n2 = 1.33 (from Table 33‐1). Then, θ 2 = sin −1
L is given by FG sinθ IJ = sin FG sin 35.0° IJ = 2555° . H 133 K . . Hn K
1 −1 2 L = l 2 tan θ 2 = (150 m) tan 25.55° = 0.72 m. . The length of the shadow is 0.35 m + 0.72 m = 1.07 m. 58. The critical angle is θ c = sin −1 66. (a) We refer to the entry point for the original incident ray as point A (which we take to be on the left side of the prism, as in Fig. 33‐57), the prism vertex as point B, and the point where the interior ray strikes the right surface of the prism as point C. The angle between line AB and the interior ray is β (the complement of the angle of refraction at the first surface), and the angle between the line BC and the interior ray is α (the complement of its angle of incidence when it strikes the second surface). When the incident ray is at the minimum angle for which light is able to exit the prism, the light exits along the second face. That is, the angle of refraction at the second face is 90°, and the angle of incidence there for the interior ray is the critical angle for total internal reflection. Let θ1 be the angle of incidence for the original incident ray and θ2 be the angle of refraction at the first face, and let θ3 be the angle of incidence at the second face. The law of refraction, applied to point C, yields n sin θ3 = 1, so sin θ3 = 1/n = 1/1.60 = 0.625 ⇒ θ3 = 38.68°. The interior angles of the triangle ABC must sum to 180°, so α + β = 120°. Now, α = 90° – θ3 = 51.32°, so β = 120° – 51.32° = 69.68°. Thus, θ2 = 90° – β = 21.32°. The law of refraction, applied to point A, yields sin θ1 = n sin θ2 = 1.60 sin 21.32° = 0.5817. FG 1 IJ = sin FG 1 IJ = 34° . H nK H 18K .
−1 Thus θ1 = 35.6°. (b) We apply the law of refraction to point C. Since the angle of refraction there is the same as the angle of incidence at A, n sin θ3 = sin θ1. Now, α + β = 120°, α = 90° – θ3, and β = 90° – θ2, as before. This means θ2 + θ3 = 60°. Thus, the law of refraction leads to sin θ1 = n sin ( 60° − θ 2 ) ⇒ sin θ1 = n sin 60° cos θ 2 − n cos 60° sin θ 2 where the trigonometric identity sin(A – B) = sin A cos B – cos A sin B is used. Next, we apply the law of refraction to point A: sin θ1 = n sin θ 2 ⇒ sin θ 2 = (1/ n ) sin θ1 which yields cosθ 2 = 1 − sin 2 θ 2 = 1 − 1 / n 2 sin 2 θ 1 . Thus, ch sin θ 1 = n sin 60° 1 − 1 / n sin 2 θ 1 − cos 60° sin θ 1 2 bg or b1 + cos 60°g sinθ = sin 60°
1 n 2 − sin 2 θ 1 . Squaring both sides and solving for sin θ1, we obtain sin θ 1 = b n sin 60° 1 + cos 60° + sin 2 60°
2 g = b 160 sin 60° . 1 + cos 60° + sin 2 60°
2 g = 0.80 and θ1 = 53.1°. 69. The angle of incidence θB for which reflected light is fully polarized is given by Eq. 33‐48 of the text. If n1 is the index of refraction for the medium of incidence and n2 is the index of refraction for the second medium, then θ B = tan −1 (n2 / n1 ) = tan −1 (1.53 /1.33) = 49.0°. 76. (a) Suppose there are a total of N transparent layers (N = 5 in our case). We label these layers from left to right with indices 1, 2, …, N. Let the index of refraction of the air be n0. We denote the initial angle of incidence of the light ray upon the air‐layer boundary as θi and the angle of the emerging light ray as θf. We note that, since all the boundaries are parallel to each other, the angle of incidence θj at the boundary between the j‐th and the (j + 1)‐th layers is the same as the angle between the transmitted light ray and the normal in the j‐th layer. Thus, for the first boundary (the one between the air and the first layer) n1 sin θ i = , n0 sin θ 1 for the second boundary n2 sin θ 1 = , n1 sin θ 2 and so on. Finally, for the last boundary n0 sin θ N = , n N sin θ f Multiplying these equations, we obtain FG n IJ FG n IJ FG n IJ L FG n IJ = FG sin θ IJ FG sin θ IJ FG sin θ IJ L F sin θ I . H n K H n K H n K H n K H sin θ K H sin θ K H sin θ K GH sin θ JK
1 0 2 1 3 0 i 1 2 2 N f 2 N 1 3 We see that the L.H.S. of the equation above can be reduced to n0/n0 while the R.H.S. is equal to sinθi/sinθf. Equating these two expressions, we find sin θ f = FG n IJ sin θ Hn K
0 0 i = sin θ i , which gives θi = θf. So for the two light rays in the problem statement, the angle of the emerging light rays are both the same as their respective incident angles. Thus, θf = 0 for ray a, (b) and θf = 20° for ray b. (c) In this case, all we need to do is to change the value of n0 from 1.0 (for air) to 1.5 (for glass). This does not change the result above. That is, we still have θf = 0 for ray a, (d) and θf = 20° for ray b. Note that the result of this problem is fairly general. It is independent of the number of layers and the thickness and index of refraction of each layer. 77. The time for light to travel a distance d in free space is t = d/c, where c is the speed of light (3.00 × 108 m/s). (a) We take d to be 150 km = 150 × 103 m. Then, t= d 150 × 103 m = = 5.00 × 10−4 s. c 3.00 × 108 m / s (b) At full moon, the Moon and Sun are on opposite sides of Earth, so the distance traveled by the light is d = (1.5 × 108 km) + 2 (3.8 × 105 km) = 1.51 × 108 km = 1.51 × 1011 m. The time taken by light to travel this distance is d 1.51×1011 m t= = = 500 s = 8.4 min. c 3.00 ×108 m/s (c) We take d to be 2(1.3 × 109 km) = 2.6 × 1012 m. Then, d 2.6 × 1012 m t= = = 8.7 × 103 s = 2.4 h. 8 c 3.00 × 10 m / s (d) We take d to be 6500 ly and the speed of light to be 1.00 ly/y. Then, t= d 6500 ly = = 6500 y. c 1.00 ly / y The explosion took place in the year 1054 – 6500 = –5446 or 5446 b.c. ...
View
Full
Document
This note was uploaded on 06/11/2010 for the course PHYSICS 207 taught by Professor Ellis during the Spring '10 term at SUNY Buffalo.
 Spring '10
 Ellis
 Energy

Click to edit the document details