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Unformatted text preview: 924 CHAPTER 34 ELECTROMAGNETIC WAVES CHAPTER 34 Answer to Checkpoint Questions 1. 2. 3. 4. 5. 6. (a) (Use Fig. 345.) On the right side of rectangle, E is in the negative y direction; on the left side, E + dE is greater and in the same direction; (b) E is downward. On the right side, B is in the negative z direction; on the left side B + dB is greater and in the same direction. positive direction of x (a) same; (b) decrease (a), (d), (b), (c) (zero) (a) (a) yes; (b) no Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. (a) positive direction of z ; (b) x into (a) same; (b) increase; (c) decrease polarized both 20 clockwise from the y axis (a) and (b) A = 1, n = 4, = 30 two (b) 30 ; (c) 60 ; (d) 60 ; (e) 30 ; (f) 60 a, b, c d, b, a, c none (a) b; (b) blue; (c) c c CHAPTER 34 ELECTROMAGNETIC WAVES 925 14. 15. 16. 1:5 (a) d; (b) b; (c) c b Solutions to Exercises & Problems (a) (b) 1E 3 0 108 f = C = (1:0 :10)(6:4m/s 6 m) = 4:7 10 3 Hz : 5 10 1 1 T = f = 4:7 10 3 Hz = 212 s = 3 min 32 s : 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) Take d to be 150 km = 150 103 m. Then 2E d = 150 103 m = 5:00 10 4 s : t = c 3:00 108 m/s
(b) At full moon the Moon and the Sun are on opposite sides of the Earth, so the distance traveled by the light is d = (1:50 108 km) + 2(3:82 105 km) = 1:508 108 km = 1:508 1011 m. The time taken by light to travel this distance is 1011 m t = d = 1::508108 m/s = 503 s = 8:38 min : c 3 00 The distances are given in Appendix C. (c) Take d to be 2(1:3 109 km) = 2:6 1012 m. Then
12 m :6 t = d = 3200 108 m/s = 8:7 103 s = 2:4 h : c : 10 (d) Take d to be 6500 ly and the speed of light to be 1:00 ly/y. Then t = d = 16500 ly = 6500 y : c :00 ly/y 926 CHAPTER 34 ELECTROMAGNETIC WAVES The explosion took place in the year 1054 6500 = 5446 or B.C. 5446. (a) (b)
3E 8 c :0 f = = 0:3067 10 m/sm = 4:5 1024 Hz : 10 15 8 c = f = 3:0 10 m/s = 1:0 107 m ; 30 Hz which is about 1.6 Earth radii.
4E (a) From Fig. 342 we nd the wavelengths in question to be about 515 nm and 610 nm. (b) Again from Fig. 342 the wavelength is about 555 nm, the frequency is c= = (3:00 108 m/s)=555 nm = 5:41 1014 Hz, and the period is (5:41 1014 Hz) 1 ' 1:85 10 15 s: Consider two wavelengths, 1 and 2 , whose corresponding frequencies are f1 and f2 . Then 1 = C=f1 and 2 = C=f2 . If 1 =2 = 10 then
5E 1 = C=f1 = f2 = 10 ; 2 C=f2 f1
i.e., The spaces are the same on both scales. Since ,
6E c f = 7P 8 :0100 9 c2 = (3:0 10 m/s)(010 9 m)210 m) = 7:49 109 Hz : (632:8 (a) Suppose that at time t1 the Moon is starting a revolution (on the verge of going behind Jupiter, say) and that at this instant the distance between Jupiter and the Earth is `1 . The time of the start of the revolution as seen on Earth is t = t1 + `1 =c. Suppose the 1 Moon starts the next revolution at time t2 and at that instant the EarthJupiter distance is `2 . The start of the revolution as seen on Earth is t = t2 + `2 =c. Now the actual period 2 of the Moon is given by T = t2 t1 and the period as measured on Earth is T = t t = t2 t1 + `c2 2 1 `2 = T + `2 `1 : c c CHAPTER 34 ELECTROMAGNETIC WAVES 927 The period as measured on Earth is longer than the actual period because the Earth moves during a revolution and light takes a nite time to travel from Jupiter to the Earth. For the situation depicted in the diagram, light emitted at the end of a revolution travels a longer distance to get to the Earth than light emitted at the beginning. Suppose the position of the Earth is given by the angle , measured from x. Let R be the radius of the Earth's orbit and d be the distance from the Sun to Jupiter. Then the law of cosines, applied to the triangle with the Sun, the Earth, and Jupiter at the vertices, yields `2 = d2 + R2 2dR cos . This expression can be used to calculate `1 and `2 . Since the Earth does not move very far during one revolution of the Moon, we may approximate `2 `1 by (d`=dt)T and T by T + (d`=dt)(T=c). Now d` = p 2Rd sin 2vd sin d 2 + R2 2dR cos dt = pd2 + R2 2dR cos ; dt d
where v = R (d=dt) is the speed of the Earth in its orbit. For = 0, (d`=dt) = 0 and T = T . Since the Earth is then moving perpendicularly to the line from the Sun to Jupiter its distance from the planet does not change much during a revolution of the Moon. On p the other hand, when = 90 , d`=dt = vd= d2 + R2 and T = T 1 + vd p2 2 : c d +R The Earth is now moving parallel to the line from the Sun to Jupiter and its distance from the planet changes during a revolution of the Moon. (b) Let t be the actual time for the Moon to make N revolutions and t the time for N revolutions to be observed on Earth. Then t = t + `2 c `2 ;
where `1 is the EarthJupiter distance at the beginning of the interval and `2 is the EarthJupiter distance at the end. Suppose the Earth is at x at the beginning of the interval and p at y at the end. Then `1 = d R and `2 = d2 + R2 . Thus (d R) : c A value can be found for t by measuring the observed period of revolution when the Earth is at x and multiplying by N . Notice that the observed period is the true period when the Earth is at x. Now measure the time interval as the Earth moves from x to y. This is t . The dierence is p t = d2 + R2 (d R) : t c If the radii of the orbits of Jupiter and the Earth are known, the value for t t can be used to compute c.
2 2 t = t + d + R p 928 CHAPTER 34 ELECTROMAGNETIC WAVES Since Jupiter is much further from the Sun than the Earth, d2 + R2 may be approximated by d and t t may be approximated by R=c. In this approximation only the radius of the Earth's orbit need be known.
8E p p c = f = 2c LC = 2(3:0 108 m/s) (0:253 10 6 H)(25:0 10 p 12 F) = 4:7 m : If f is the frequency and is the wavelength of an electromagnetic wave then f = c. The frequency is the same as the frequency of oscillation of the current in the LC circuit of the p generator. That is, f = 1=2 LC , where C is the capacitance and L is the inductance. Thus p = c: 2 LC The solution for L is (550 10 9 m)2 2 = L = 42 Cc2 42 (17 10 12 F)(3:00 108 m/s)2 = 5:0 10 21 H : This is exceedingly small. (a) The arrangement in Fig. 343 features a pair of wire terminals with alternating charges q and q each, resembling an electric dipole. The arrangement in Fig. 3441 replaces that with a wire loop through which an alternating current exists, resembling a magnetic dipole. (b) Due to the symmetry of the Maxwell's Equations in free space (where i = q = 0), to get the eld of a magnetic dipole all you need to do is to take the eld of an electric dipole, depicted in Fig. 344, and make the following transformation: E ! B; B ! E.
11E 10P 9E 4 V/m : Bm = Ecm = 3320 10 8 m/s = 1:07 10 :00 10 12 T : Since the Ewave oscillates in the z direction and travels in the x direction, we have Bx = Bz = 0 and h i 15 15 t x = 2:0 cos[10 (t x=c)] By = Bm cos 10 c 3:0 108 h i = (6:7 10 9 ) cos 1015 t x ; c 12E CHAPTER 34 ELECTROMAGNETIC WAVES 929 where B , x, c and t are all measured in their respective SI units. Dierentiate both sides of Eq. 3411 with respect to x:
13P @B = @ 2 B : @t @x@t Dierentiate both sides of Eq. 3418 with respect to t: @ @B = @ 2 B = @ @E = @ 2 E : 0 0 @t2 @t @x @x@t @t 0 0 @t
Substituting @ 2 E=@x2 = @ 2 B=@x@t from the rst equation above into the second one, we get @2E = @2E ; 0 0 @t2 @x2 or @ 2 E = 1 @ 2 E = c2 @ 2 E : @t2 @x2 @x2 Similarly, dierentiate both sides of Eq. 3411 with respect to t:
0 0 @ @E = @ 2 E = @ @x @x @x2 @x and dierentiate both sides of Eq. 3418 with respect to x: @2E = @2B ; @x@t @t2 @2B = @2E : @x2 0 0 @x@x
Combining these two equations, we get @ 2 B = 1 @ 2 B = c2 @ 2 B : @t2 0 0 @x2 @x2
(a) From Eq. 341
14P @ 2 E = @ 2 [E sin(kx !t)] = !2 E sin(kx !t) ; m @t2 @t2 m
and
2 @2 c2 @ E = c2 @x2 [Em sin(kx !t)] = k2 c2 sin(kx !t) = !2 Em sin(kx !t) : @x2 930 CHAPTER 34 ELECTROMAGNETIC WAVES So is satised. Analogously, you can show that Eq. 342 satises @ 2 E = c2 @ 2 E @t2 @x2 @ 2 B = c2 @ 2 B : @t2 @x2
(b) From E = Em f (kx !t) @ 2 E = E @ 2 f (kx !t) = !2 E d2 f m m du2 @t2 @t2 u=kxt
and c2 @ 2E @x2 = c2 Em @ 2 f (kx !t) @t2 2 2 E k2 d f = c m du2 Since ! = ck the R.H.S. of the two equations above are equal. So u=kx t : @ 2 E = c2 @ 2 E : @t2 @x2
Change E to B and repeat the derivation above to show that B = Bm f (kx !t) satises @ 2 B = c2 @ 2 B : @t2 @x2
15E Using S = (1=0 )E B, you can easily verify that S is in the direction of propagation for all these cases. If P is the power and t is the time interval of one pulse, then the energy in a pulse is
16E E = P t = (100 1012 W)(1:0 10 9 s) = 1:0 105 J : 17E 6W P 1:0 I = 4r2 = 4[(4:3 ly)(9: 10 1015 m/ly)]2 = 4:8 10 46 29 W/m2 : CHAPTER 34 ELECTROMAGNETIC WAVES 931 18E E B B = 0 S, where E = E k and S = S ( j). You can verify easily that since k ( i) = j,
has to be in the negative x direction (with unit vector i). Also V/m B = E = 3:0100108 m/s = 3:3 10 7 T : c The fraction is
2 1 6 37 106 e frac = 4R2 = 4 1::50 1011m des m 19E 2 = 4:51 10 10 : The region illuminated on the Moon is a circle with radius R = des =2, where des is the EarthMoon distance (3:82 108 m) and is the fullangle beam divergence in radians. The area A illuminated is 1 A = R2 = 4 d2 2 = 1 (3:82 108 m)2 (0:880 10 6 rad)2 = 8:88 104 m2 : es 4 Solve r from 4r2 I = P :
r
21E 20E 1:0 3 r = 4P = 4(100 10 W2 ) = 0:89 m : I W/m
22E s 1 1 S = jE Bjav = (EB )av = Eav Bav = (Em = 2)(Bm = 2) =
0 Em Bm 20 p p = 2 Em 20 c 0 = 2 cBm 20 0 0 ; where we used Eav Bav = (EB )av since E and B have the same phase, and Em = cBm :
23E 2 8 m/s)(1 10 4 2 I = S = cBm = (3:0 1026 10 :60H/m)2 T) = 1:2 106 W/m2 : 2 2(1: 0 932 CHAPTER 34 ELECTROMAGNETIC WAVES (a) (b) 24E 5 V/m Bm = Ecm = 3:00:00 108 m/s = 1:67 10 8 T : 2 00 V/m)2 I = S = 2Emc = 2(4 10 7 (5:m/A)(3:00 108 m/s) = 3:31 10 2 W/m2 : 0 T Let the distance in question be r0 . Then
2 2 P = 4I0 r0 = 4I1 r1 ; 25P where I1 = 1:5I0 and r1 = r0 150 m. Solve for r0 : p f0 = 1:5 (150 m) = 8:2 102 m : 1:5 1
The energy density of an electromagnetic wave is given by uEM = uE + uB . From the discussion in Section 344 uE = uB = 1 0 E 2 , so uEM = 2uE = 0 E 2 . Upon averaging 2 over time this becomes 2 uEM = 0 E 2 = 0 Erms : Combine this equation with Eq. 3426 in the text to obtain
2 1 2 1 I = c Erms = c uEM = c ucEM = cuEM ; 0 0 0
26P p where we used c2 = 1=0 0 .
2 Use I = Em =20 c to calculate Em :
27P Em = 20 Ic = 2(4 10 7 T m/A)(1:40 103 W/m2 )(3:00 108 m/s) = 1:03 103 V/m :
Also
4 Bm = Ecm = 1::03 10 8V/m = 3:43 10 6 T : 3 00 10 m/s p p CHAPTER 34 ELECTROMAGNETIC WAVES 933 (a) (b) (c) 28P 20 Bm = Ecm = 3:00 : V/m = 6:7 10 9 T : 108 m/s
2 (2:0 V/m)2 I = 2Emc = 2(4 10 7 T m/A)(3:00 108 m/s) = 5:3 10 3 W/m2 : 0 P = 4r2 I = 4(10 m)2 (5:3 10 3 W/m2 ) = 6:7 W : (a) The power received is 29P Pr = (1:0 10
(b) The power would be 2 12 W) [(1000 ft)(0:3048 m/ft)] =4 4(6:37 106 m)2 = 1:4 10 22 W : P = 4r2 I = 4[(2:2 104 ly)(9:46 1015 m/ly)]2 = 1:1 1015 W : (a)
30P 1:0 10 12 W 4(6:37 106 m)2 P I = dP=4 = (r)2 =4 2 3 4(3 = [(0:17 :0 10 W) m)]2 10 3 rad)(40 = 83 W/m2 : d r r (b) P 0 = 4r2 I = 4(40 m)2 (83 W/m2 ) = 1:7 106 W:
31P source (a) The average rate of energy ow per unit area, or intensity, is related to the electric eld 2 amplitude Em by I = Em =20 c, so Em = 20 cI = 2(4 10 7 H/m)(3:00 108 m/s)(10 10 6 W/m2 ) = 8:7 10 2 V/m : p q 934 CHAPTER 34 ELECTROMAGNETIC WAVES (b) The amplitude of the magnetic eld is given by 10 2 Bm = Ecm = 83::700 108 V/m = 2:9 10 m/s 10 T : (c) At a distance r from the transmitter the intensity is I = P=4r2 , where P is the power of the transmitter. Thus P = 4r2 I = 4(10 103 m)2 (10 10 6 W/m2 ) = 1:3 104 W :
32P (a) (b) (c) (d) 3 P I = 2r2 = 2180 10 3W 2 = 3:5 10 6 W/m2 : (90 10 m) Pr = IA = (3:53 10 6 W/m2 )(0:22 m2 ) = 7:8 10 7 W :
7:7 10 7 W Ir = 2Pr 2 = 2(90 103 m)2 = 1:5 10 r
17 W/m2 : Em = 20 Ir c = 2(4 10 7 T m/A)(1:5 10 = 1:1 10 7 V/m :
(e) p p 17 W/m2 )(3:00 108 m/s) 7 E Brms = p m = p1:06 10 V/m = 2:5 10 2c 2 (3:00 108 m/s) 16 T : 33E 10 W/m2 pr = I = 3:00 108 m/s = 3:3 10 8 Pa : c 34E The plasma completely re
ects all the energy incident on it, so the radiation pressure is given by pr = 2I=c, where I is the intensity. The intensity is I = P=A, where P is the power and A is the area intercepted by the radiation. Thus 2(1 5 10 W) pr = 2P = (1:00 10 6:m)(3:00 108 m/s) = 1:0 107 Pa = 10 MPa : 2 Ac
9 CHAPTER 34 ELECTROMAGNETIC WAVES 935 (a) (b) 35E 2 103 pr = I = 13:400 10W/m = 4:7 10 6 N/m2 : 8 m/s c : pr = 4:7 10 6 N/m2 = 4:7 10 p0 1:0 105 N/m2 11 : (a) 36E (b) I (R2 ) Fr = pr e e c 3 W/m2 )(6:37 6 2 = (1:4 103:00 108 m/s 10 m) = 6:0 108 N :
(R2 ) = = 3:6 1022 N ; which is much greater than Fr .
37E Fg = GM2s Me = (6:67 10 d
es 11 N m2 =kg2 )(2:0 1030 kg)(5:98 1024 kg) (1:5 1011 m)2 Since the surface is perfectly absorbing the radiation pressure is given by pr = I=c, where I is the intensity. Since the bulb radiates uniformly in all directions the intensity a distance r from it is given by I = P=4r2 , where P is the power of the bulb. Thus P W pr = 4r2 c = 4(1:5 m)250000 108 m/s) = 5:9 10 8 Pa : (3: (a) Since c = f , where is the wavelength and f is the frequency of the wave, c = 3:00 108 m/s = 1:0 108 Hz : f= 3:0 m (b) The magnetic eld amplitude is 300 Bm = Ecm = 3:00 V/m = 1:00 10 6 T : 108 m/s B must be in the positive z direction when E is in the positive y direction in order for E B to be in the positive x direction (the direction of propagation).
38P 936 CHAPTER 34 ELECTROMAGNETIC WAVES (c) The angular wave number is k = 2 = 3:2m = 2:1 rad/m : 0
The angular frequency is ! = 2f = 2(1:0 108 Hz) = 6:3 108 rad/s :
(d) The intensity of the wave is
2 2 I = 2Emc = 2(4 10 7(300 V/m) 108 m/s) = 119 W/m2 : H/m)(3:00 0 (e) Since the sheet is perfectly absorbing, the rate per unit area with which momentum is delivered to it is I=c, so dp = IA = (119 W/m2 )(2:0 m2 ) = 8:0 10 7 N : dt c 3:00 108 m/s
The radiation pressure is dp=dt = 8:0 10 7 N = 4:0 10 7 Pa : pr = A 2:0 m2
39P (a) (b) (c) (d) 5:00 10 3 I = dP=4 = [(2:00)(633 10 Wm)]2 =4 = 3:97 109 W/m2 : 2 9 109 W/m2 pr = I = 3:397 108 m/s = 13:2 Pa : c :00
2 3 Fr = d pr = Ppr = (5:00 : 10 10W)(13:2 Pa) = 1:67 10 4 I 3 97 9 W/m2 11 N : Fr 6(1:67 10 11 N) a = Fr = (d3 =6) = (5:00 103 kg/m3 )[(2:00)(633 10 9 m)]3 m = 3:14 103 m/s2 : CHAPTER 34 ELECTROMAGNETIC WAVES 937 The mass of the cylinder is m = (d2 =4)H , where d1 is the diameter of the cylinder. Since 1 it is in equilibrium
2 Fnet = mg Fr = Hd1 g 4 40P d2 1
4 2I = 0 : c Solve for H : 2I 2P 1 H = gc = d2 =4 gc = (2:60 10 3 m)2 (9:8 m/s28(4::60 W)108 m/s)(1:20 103 kg/m3 ) )(3 00 7 m: = 4:91 10 Denote the fraction by . The contribution to Pr due to the absorbed part of the radiation is Pr1 = Iabsorbed =c = I=c; while the contribution to Pr due to the re
ected part is Pr2 = 2Ire
ected =c = 2I (1 )=c. Thus 41P Pr = Pr1 + Pr2 = I + 2I (1c ) = I (2 c ) : c
42P Let f be the fraction of the incident beam intensity that is re
ected. The fraction absorbed is 1 f . The re
ected portion exerts a radiation pressure of p = 2fI0
r and the absorbed portion exerts a radiation pressure of p = (1 f )I0 ;
a c where I0 is the incident intensity. The factor 2 enters the rst expression because the momentum of the re
ected portion is reversed. The total radiation pressure is the sum of the two contributions: p = p + p = 2fI0 + (1 f )I0 = (1 + f )I0 :
total
r a c To relate the intensity and energy density, consider a tube with length ` and crosssectional area A, lying with its axis along the propagation direction of an electromagnetic wave. The c c 938 CHAPTER 34 ELECTROMAGNETIC WAVES electromagnetic energy inside is U = uA`, where u is the energy density. All this energy will pass through the end in time t = `=c so the intensity is U I = At = uA`c = uc : `
Thus u = I=c. The intensity and energy density are inherently positive, regardless of the propagation direction. For the partially re
ected and partially absorbed wave the intensity just outside the surface is I = I0 + fI0 = (1 + f )I0 , where the rst term is associated with the incident beam and the second is associated with the re
ected beam. The energy density is therefore u = I = (1 +cf )I0 ; c
the same as radiation pressure. Firstly, the intensity of light as detected on the re
ecting surface is I () = I? cos . Secondly, the change in momentum for the beam of light upon re
ection is P () = P? cos . So
43P pr () / I ()P () = (I? cos )(P? cos ) = pr? cos2 : 44P mv A vt CHAPTER 34 ELECTROMAGNETIC WAVES 939 The bullets (of mass m and speed v each) which will strike a surface of area A of the plane within time t to t + t must all be contained in the cylindrical volume shown above at time t. Since the number of bullets contained in the cylinder is N = n(Avt) and each bullet changes its momentum by pb = mv, the rate of change of the total momentum for the bullets that strike the area is F = Ptotal = Nptb = (Avt)nmv = Anmv2 ; t t where n is the number density of the bullets. The pressure is then pr = F = nmv2 = 2nK ; A
where K = 1 mv2 . Note that nK is the kinetic energy density. 2
45P If the beam carries energy U away from the spaceship (of mass m and speed v) then it also carries momentum p = U=c away. Since the total momentum of the spaceship and light is conserved, this is the magnitude of the momentum acquired by the spaceship. If P is the power of the laser, then the energy carried away in time t is U = Pt. Thus p = Pt=c and 103 W)(1 d)(8 104 s/d) p Pt v = m = mc = (10 : 103 kg)(3:00:64 8 m/s) = 1:9 10 3 m/s = 1:9 mm/s : (1 5 10
46P Let Fg = Fr or and solve for the area A: 11 2 kg2 8 1030 kg)(3 A = cGmMs = (6:67 10 N m40 )(1500 kg)(1:99 1011 m)2 :00 10 m/s) 2Id2 2(1: 103 W/m2 )(1:50 es = 9:5 105 m2 = 0:95 km2 :
47P G mMs = 2IA d2 c es (a) Let r be the radius and be the density of the particle. Since its volume is (4=3)r3 , its mass is m = (4=3)r3 . Let R be the distance from the Sun to the particle and let M be the mass of the Sun. Then the gravitational force of attraction of the Sun on the particle has magnitude 3 Fg = GMm = 4GMr : R2 3R2 940 CHAPTER 34 ELECTROMAGNETIC WAVES All of the radiation that passes through a circle of radius r and area A = r2 , perpendicular to the direction of propagation, is absorbed by the particle, so the force of the radiation on the particle has magnitude The force is radially outward from the Sun. Notice that both the force of gravity and the force of the radiation are inversely proportional to R2 . If one of these forces is larger than the other at some distance from the Sun, then that force is larger at all distances. The two forces depend on the particle radius r dierently: Fg is proportional to r3 and Fr is proportional to r2 . We expect a small radius particle to be blown away by the radiation pressure and a large radius particle with the same density to be pulled inward toward the Sun. The critical value for the radius is the value for which the two forces are equal. Equate the expressions for Fg and Fr , then solve for r. You should obtain 3P r = 16GMc : (b) According to Appendix C, M = 1:99 1030 kg and P = 3:90 1026 W. Thus, 3(3:90 1026 W) r= 16(6:67 10 11 N m2 /kg2 )(1:99 1030 kg)(1:0 103 kg/m3 )(3:00 108 m/s) = 5:8 10 7 m : (a) Negative y direction. (b) Ex = Ey = 0 and Ez = cB sin(kx + !t). You can check that, with this expression for E, E B is in the direction of propagation. (c) Since Ex = Ey = 0 the wave is plane polarized with E along the z axis.
49E 48E If P is the power output of the Sun, then at the position of the particle the radiation intensity is I = P=4R2 and since the particle is perfectly absorbing the radiation pressure on it is P pr = I = 4R2 c : c 2 2 Fr = pr A = 4Pr2 c = 4Pr2 c : R R (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 2 electric eld amplitude are related by I = Em =20 c, so Em = 20 cI = 2(4 10 7 H/m)(3:00 108 m/s)(5:0 10 3 W/m2 ) = 1:9 V/m : p q CHAPTER 34 ELECTROMAGNETIC WAVES 941 (b) The radiation pressure is pr = Ia =c, where Ia is the absorbed intensity. Thus 5:0 10 3 W/m2 = 1:7 10 pr = 3:00 108 m/s
50E 11 Pa : As the unpolarized beam of intensity I0 passes the rst polarizer, its intensity is reduced to 1 I0 . After passing through the second polarizer whose dirction of polarization is at an 2 angle from that of the rst one I = 1 I0 cos2 = 1 I0 . Thus cos2 = 2=3, which gives 2 3 = 35 :
51E I0 I 0 /2 (I 0 /2) cos 245o [(I 0 /2) cos245 o ]cos 245o Use I = Im cos2 . The diagram above shows that Inal = (I0 =2) cos4 45 = 1 : I0 I0 8
After passing through the rst 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 ( 1 2 ) = cos2 (1 + 2 ). Finally after passing through the third one it is again reduced by a factor of cos2 ( 2 3 ) = cos2 (2 + 3 ). So If = 1 cos2 ( + ) cos2 ( + ) 1 2 2 3 I0 2 = 1 cos2 (50 + 50 ) cos2 (50 + 50 ) = 4:5 10 2 % : 2 After passing through the rst 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 (1 + 2 ).
53P 52E 942 CHAPTER 34 ELECTROMAGNETIC WAVES Finally after passing through the third one it is again reduced by a factor of cos2 (2 + 3 ). So If = 1 cos2 ( + ) cos2 ( + ) 1 2 2 3 I0 2 = 1 cos2 (40 + 20 ) cos2 (20 + 40 ) = 0:031 = 3:1% : 2
54P After passing through the rst polarizer the initial intensity I0 of the light is reduced to 1 I0 . After that it is reduced by a factor of cos2 30 each time the light passes through one 2 more polarizer. So If = 1 (cos2 30 )3 = 0:21: I 2
0 After passing through the rst polarizer the initial intensity I0 of the light is reduced to I1 = I0 cos2 . Since the angle between the polarization directions of the two polarizers is 90 , the nal intensity of light is If = I1 cos2 (90 ) = I0 cos2 sin2 = (I0 =4) sin2 (2). Let If = 0:10I0 to obtain sin2 (2) = 0:40: The angle is = 20 or = 70 . The intensity of the transmitted light is
56P 55P If = (I0 cos2 70 ) cos2 (90 70 ) = (43 W/m2 )(cos2 70 )(cos2 20 ) = 4:4 W/m2 :
57P In this case we replace I0 cos2 70 by 1 I0 as the intensity of the light after passing through 2 the rst polarizer. So 1 1 If = 2 I0 cos2 (90 70 ) = 2 (43 W/m2 )(cos2 20 ) = 19 W/m2 :
58P Let I0 be the intensity of the incident beam and f be the fraction that is polarized. Thus the intensity of the polarized portion is fI0 . After transmission this portion contributes fI0 cos2 to the intensity of the transmitted beam. Here is the angle between the direction of polarization of the radiation and the polarizing direction of the lter. The intensity of CHAPTER 34 ELECTROMAGNETIC WAVES 943 Set the ratio equal to 5:0 and solve for f . You should get f = 0:67.
59P the unpolarized portion of the incident beam is (1 f )I0 and after transmission this portion contributes (1 f )I0 =2 to the transmitted intensity. Thus the transmitted intensity is I = fI0 cos2 + 1 (1 f )I0 : 2 As the lter is rotated cos2 varies from a minimum of 0 to a maximum of 1, so the transmitted intensity varies from a minimum of 1 Imin = 2 (1 f )I0 to a maximum of 1 1 Imax = fI0 + 2 (1 f )I0 = 2 (1 + f )I0 : The ratio of Imax to Imin is Imax = 1 + f : I 1 f
min (a) The rotation cannot be done with a single sheet. If a sheet is placed with its polarizing direction at an angle of 90 to the direction of polarization of the incident radiation, no radiation is transmitted. It can be done with two sheets. Place the rst sheet with its polarizing direction at some angle , between 0 and 90 , to the direction of polarization of the incident radiation. Place the second sheet with its polarizing direction at 90 to the polarization direction of the incident radiation. The transmitted radiation is then polarized at 90 to the incident polarization direction. The intensity is I0 cos2 cos2 (90 ) = I0 cos2 sin2 , where I0 is the incident radiation. If is not 0 or 90 the transmitted intensity is not zero. (b) Consider n sheets, with the polarizing direction of the rst sheet making an angle of = 90 =n with the direction of polarization of the incident radiation and with the polarizing direction of each successive sheet rotated 90 =n in the same direction from the polarizing direction of the previous sheet. The transmitted radiation is polarized with its direction of polarization making an angle of 90 with the direction of polarization of the incident radiation. The intensity is I = I0 cos2n (90 =n). You want the smallest integer value of n for which this is greater than 0:60I0 . Start with n = 2 and calculate cos2n (90 =n). If the result is greater than 0:60 you have obtained the solution. If it is less, increase n by 1 and try again. Repeat this process, increasing n by 1 each time, until you have a value for which cos2n (90 =n) is greater than 0:60. The rst one will be n = 5. (a)
60P 2 2 2 If = Ef = Ev = Ev 2 2 2 2 I0 E0 Ev + Eh Ev + (2:3Ev )2 = 0:16 : 944 CHAPTER 34 ELECTROMAGNETIC WAVES (b) Since now the horizontal component of E will pass through the glasses,
2 If = Eh = (2:3Ev )2 = 0:84 : 2 2 2 I0 Ev + Eh Ev + (2:3Ev )2 You can see easily that the angle of incidence for the light ray on mirror B is 90 . So the outgoing ray r0 makes an angle 90 (90 ) = with the vertical direction and is antiparallel to the incoming one. The angle between i and r0 is therefore 180 :
62E 61E n1 sin 1 = n2 sin 2 : Take medium 1 to be the vacuum, with n1 = 1 and 1 = 32:0 . Medium 2 is the glass, with 2 = 21:0 . Solve for n2 :
32 n2 = n1 sin 1 = (1:00) sin 21::0 = 1:48 : sin 2 sin 0 63E Use the law of refraction: Note that the normal to the refracting surface is vertical in the diagram. The angle of refraction is 2 = 90 and the angle of incidence is given by tan 1 = w=h, where h is the height of the tank and w is its width. Thus 1 = tan
The law of refraction yields 1 w = tan h 1 1:10 m = 52:31 : 0:850 m sin 90 n1 = n2 sin 2 = (1:00) sin 52:31 = 1:26 ; sin 1 where the index of refraction of air was taken to be unity. (a) The law of refraction requires that sin 1 = sin 2 = nwater = const. You can check that this is indeed valid for any given pair of 1 and 2 . For example sin 10 = sin 8 = 1:3, and sin 20 = sin 15 300 = 1:3, etc. (b) nwater = 1:3, as shown in part (a).
64E CHAPTER 34 ELECTROMAGNETIC WAVES 945 Consider a ray that grazes the top of the pole, as shown in the diagram to the right. Here 1 = 35 , `1 = 0:50 m, and `2 = 1:50 m. The length of the shadow is x + L. x is given by x = `1 tan 1 = (0:50 m) tan 35 = 0:35 m. According to the law of refraction n2 sin 2 = n1 sin 1 . Take n1 = 1 and n2 = 1:33 (from Table 341). Then 1 sin 1 = sin 1 sin 35:0 = 25:55 : 2 = sin 65P 1 l1 air water
2 n2 1:33 l2 L is given by L = `2 tan 2 = (1:50 m) tan 25:55 = 0:72 m :
The length of the shadow is 0:35 m + 0:72 m = 1:07 m.
66P L x (a) Refer to the gure to the right. The sh cannot see light coming from anywhere beyond points A and B because of total internal re
ection. Thus the diameter of the circle in question is given by d = 2h tan c . For water n = 1:33 so Eq. 3447 gives sin c = 1=1:33, or c = 48:75 . Thus A
c c air water h
c c B
c c d = 2h tan c = 2(2:00 m)(tan 48:75 ) = 4:56 m : fish (b) Since d / h, the diameter of the circle will increase as the sh descends (i.e., as h increases.) Let be the angle of incidence and 2 be the angle of refraction at the left face of the plate. Let n be the index of refraction of the glass. Then the law of refraction yields sin = n sin 2 . The angle of incidence at the right face is also 2 . If 3 is the angle of emergence there, then n sin 2 = sin 3 . Thus sin 3 = sin and 3 = . The emerging ray is parallel to the incident ray. You wish to derive an expression for x in terms of . If D is the length of the ray in the glass, then D cos 2 = t and D = t= cos 2 . The angle in the diagram equals 2 and x = D sin = D sin( 2 ). Thus
67P x = t sin( 2 ) : cos
2 946 CHAPTER 34 ELECTROMAGNETIC WAVES If all the angles , 2 , 3 , and 2 are small and measured in radians then sin , sin 2 2 , sin( 2 ) 2 , and cos 2 1. Thus x t( 2 ). The law of refraction applied to the point of incidence at the left face of the plate is now n2 , so 2 =n and = (n 1)t : xt n n (a) { (c) Use the law of refraction to determine the paths of various light rays. The index of refraction for fused quartz can be found in Fig. 3419. The paths traversed by rays representing these lights are shown below. Here 61 for red light, 62 for yellowgreen light and 63 for blue light. 68P 35 o 23 o 37o red
60 o blue (a) Suppose there are a total of N transparent layers (N = 5 in our case). Label these layers from left to right with indeces 1, 2, . . . , N . Let the index of refraction of the air be n0 . Denote the initial angle of incidence of the light ray upon the airlayer boundary as i and the angle of the emerging light ray as f . Note that, since all the boundaries are parallel to each other, the angle of incidence upon the boundary between the j th and the (j + 1)th layers, j , is the same as the angle between the trasmitted light ray and the normal in the j th layer. So for the rst boundary (the one between the air and the rst layer) n1 = sin i ; n sin for the second boundary
0 1 69P etc. Finally for the last boundary n2 = sin 1 ; n1 sin 2 n0 = sin N : nN sin f CHAPTER 34 ELECTROMAGNETIC WAVES 947 Multiply these equations to obtain n1 n0 n2 n1 n3 n0 = sin i n2 nN sin 1 sin 1 sin 2 sin 2 sin N : sin 3 sin f Obviously the L.H.S. of the equation above can be reduced to n0 =n0 while the R.H.S. is equal to sin i = sin f . Equate these two expressions to obtain n sin f = n0 sin i = sin i ; 0
which gives i = f . 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, i.e., f = 0 for ray a and f = 20 for ray b. (b) In this case all you need to do is to change the value of n0 from 1.0 (for air) to 1.5 (for glass). This obviously does not change the result above. 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. (a) A ray diagram is shown to the right. The incident ray is normal to the water surface and so is not refracted. The angle of incidence at the rst mirror is 1 = 45 . According to the law of re
ection the angle of re
ection is also 45 . This means the ray is horizontal after re1 2 2 1 ection and the angle of incidence at the second mirror is 2 = 45 . Since the angle of re
ection at the second mirror is also 45 the ray leaves that mirror normal to the water surface. Again there is no refraction at the water surface and the emerging ray is parallel to the incident ray. (b) A ray diagram is shown in the next page. On incidence the ray makes the angle 1 with the normal to the water surface. The angle of refraction 2 can be found using the law of refraction: sin 1 = n sin 2 , where n is the index of refraction of the water. The normal to the water surface and the normal to the rst mirror make an angle of 45 . If the normal to the water surface is continued downward until it meets the normal to the rst mirror the triangle formed has an interior angle of 180 45 = 135 at the vertex formed by the normal. Since the interior angles of a triangle must sum to 180 , the angle of incidence at the rst mirror satises 3 + 2 + 135 = 180 , so 3 = 45 2 . The angle of re
ection at the rst mirror is also 45 2 .
70P 948 CHAPTER 34 ELECTROMAGNETIC WAVES 1 2 3 3 6 5 4 4 Now look at the triangle formed by the ray and the normals to the two mirrors. It is a right triangle, so 3 + 4 + 90 = 180 and 4 = 90 3 = 90 45 + 2 = 45 + 2 . The angle of re
ection at the second mirror is also 45 + 2 . Now continue the normal to the water surface downward from the exit point of the ray to the second mirror. It makes an angle of 45 with the mirror. Consider the triangle formed by the second mirror, the ray, and the normal to the water surface. The angle at the intersection of the normal and the mirror is 180 45 = 135 . The angle at the intersection of the ray and the mirror is 90 4 = 90 (45 + 2 ) = 45 2 . The angle at the intersection of the ray and the water surface is 5 . These three angles must sum to 180 , so 135 + 45 2 + 5 = 180 . This means 5 = 2 . Now use the law of refraction to nd 6 : sin 6 = n sin 5 . But since 5 = 2 , this means sin 6 = n sin 2 . Finally, since sin 1 = n sin 2 , we conclude that sin 6 = sin 1 and 6 = 1 . The exiting ray is parallel to the incident ray. From the diagram to the right we nd + 90 + (90 1 ) = 180 and = 1 . Solve for : 2 2 = 1 ( + ). The law of refraction then gives 2 sin n = sin = sin sin( 1 ) 2 1 ( + ) sin = 2 1 : sin 2 90o 90o /2 /2
71P /2 Use the result of the last problem (71P) to solve for . Note that = 60:0 in our case. 72P CHAPTER 34 ELECTROMAGNETIC WAVES 949 Thus from we get sin 1 ( + ) n= 2 1 sin 2 : sin 1 ( + ) = n sin 1 = (1:31) sin 6020 = 0:655 ; 2 2 which gives 1 ( + ) = sin 1 (0:655) = 40:9 . Thus = 2(40:9 ) = 2(40:9 ) 60:0 = 2 21:8 .
Use the result obtained in 71P: : sin 1 ( + ) sin 1 (60:0 + 300 ) 2 = 1:41 : n= 2 1 = sin 2 sin 1 (60:0 ) 2
74E 73P c = sin
75E 1 1 = sin n 1 1 = 34 : 1:8 Let 1 (= 45:0 ) be the angle of incidence at the rst surface and 2 be the angle of refraction there. Let 3 be the angle of incidence at the second surface. The condition for total internal re
ection at the second surface is n sin 3 1. You want to nd the smallest value of the index of refraction n for which this inequality holds. The law of refraction, applied to the rst surface, yields n sin 2 = sin 1 . Consideration of the triangle formed by the surface of the slab and the ray in the slab tells us that 3 = 90 2 . Thus the condition for total internal re
ection becomes 1 n sin(90 2 ) = n cos 2 . Square this equation and use sin2 2 + cos2 2 = 1 to obtain 1 n2 (1 sin2 2 ). Now substitute sin 2 = (1=n) sin 1 to obtain 2 2 1 sin 1 = n2 sin2 : 1n n2 1 The largest value of n for which this equation is true is the value for which 1 = n2 sin2 1 . Solve for n: q p n = 1 + sin2 1 = 1 + sin2 45:0 = 1:22 : (a) No refraction occurs at the surface ab, so the angle of incidence at surface ac is 90 . For total internal re
ection at the second surface, ng sin(90 ) must be greater than na .
76E 950 CHAPTER 34 ELECTROMAGNETIC WAVES Here ng is the index of refraction for the glass and na is the index of refraction for air. Since sin(90 ) = cos , you want the largest value of for which ng cos na . Recall that cos decreases as increases from zero. When has the largest value for which total internal re
ection occurs, then ng cos = na , or = cos 1 na = cos ng 1 1 = 48:9 : 1:52 The index of refraction for air was taken to be unity. (b) Replace the air with water. If nw (= 1:33) is the index of refraction for water, then the largest value of for which total internal re
ection occurs is = cos 1 nw = cos ng 1 1:33 = 29:0 : 1:52 This problem is similar to 66P. The diameter D of the circle in question is 77E D = 2h tan c = 2h tan sin 1 nw 1 = 2(80:0 cm) tan sin 1 1 1:33 = 182 cm : 78P (a) The diagram on the right shows a cross section, through the center of the cube and parallel to a face. L is the length of a cube edge and S labels the spot. A portion of a ray from the source to a cube face is also shown. Light leaving the source at a small angle is refracted at the face and leaves the cube; light leaving at a suciently large angle is totally re
ected. The light that passes through the cube face forms a circle, the radius r being associated with the critical angle for total internal re
ection. If c is that angle then sin = 1 ;
c r L/2 L
c S where n is the index of refraction for the glass. As the diagram shows, the radius of the circle is given by r = (L=2) tan c . Now sin 1 tan c = cos c = p sin c 2 = p 1=n 2 = p 2 c n 1 1 (1=n) 1 sin c n CHAPTER 34 ELECTROMAGNETIC WAVES 951 and the radius of the circle is r = p L = p10 mm = 4:5 mm : 2 n2 1 2 (1:5)2 1
If an opaque circular disk with this radius is pasted at the center of each cube face the spot will not be seen (provided internally re
ected light can be ignored). (b) There must be six opaque disks, one for each face. The total area covered by disks is 6r2 and the total surface area of the cube is 6L2 . The fraction of the surface area that must covered by disks is
2 2 2 f = 6r2 = r2 = (4:47 mm) = 0:63 : 6L L (10 mm)2 (a) and (b) The index of refraction n for fused quartz is slightly higher on the blusih side of the visible light spectrum (with shorter wavelength). Since sin c = 1=n, when this equation rst becomes valid for the reddish light the bluish light has not yet reached its critical angle so it will not have total internal re
ection, leaving the re
ected light appear reddish. So (b) is possible while (a) is not. (c) The angle is about c = sin 1 (1=n) = sin 1 (1=1:47) = 43 :
80P 79P 90 o 1 2
90o 3 4 (a) A ray diagram is shown above. Let 1 be the angle of incidence and 2 be the angle of refraction at the rst surface. Let 3 be the angle of incidence at the second surface. The angle of refraction there is 4 = 90 . The law of refraction, applied to the second surface, yields n sin 3 = sin 4 = 1. As shown in the diagram, the normals to the surfaces at P and Q make an angle of 90 with each other. The interior angles of the triangle formed by the ray and the two normals must sum to 180 , so 3 = 90 2 and sin 3 = sin(90 2 ) = p p cos 2 = 1 sin2 2 . According to the law of refraction, applied at Q, n 1 sin2 2 = 1. 952 CHAPTER 34 ELECTROMAGNETIC WAVES The law of refraction, applied to point P , yields sin 1 = n sin 2 , so sin 2 = (sin 1 )=n and
2 n 1 sin 21 = 1 : n
s Square both sides and solve for n. You should get n = 1 + sin2 1 :
(b) The greatest possible value of sin2 1 is 1, so the greatest possible value of n is nmax = p 2 = 1:41. (c) For a given value of n, if the angle of incidence at the rst surface is greater than 1 , the angle of refraction there is greater than 2 and the angle of incidence at the second face is less than 3 (= 90 2 ). That is, it is less than the critical angle for total internal re
ection, so light leaves the second surface and emerges into the air. (d) If the angle of incidence at the rst surface is less than 1 , the angle of refraction there is less than 2 and the angle of incidence at the second surface is greater than 3 . This is greater than the critical angle for total internal re
ection, so all the light is re
ected at Q. (a) When the incident ray is at the minimum angle for which light exits 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 is the critical angle for total internal re
ection. Let 1 be the angle of incidence and 2 be the angle of refraction at the rst face and let 3 be the angle of incidence at the second face. You want to solve for 1 . A ray diagram is shown below. The law of refraction, applied to point C , yields n sin 3 = 1, so sin 3 = 1=n = 1=1:60 = 0:625 and 3 = 38:68 . The interior angles of the triangle ABC must sum to 180 , so + = 120 . Now = 90 3 = 51:32 , so = 120 51:32 = 69:68 . Thus, 2 = 90 = 21:32 . The law of refraction, applied to point A, yields sin 1 = n sin 2 = 1:60 sin 21:32 = 0:5817. Thus 1 = 35:6 .
B
60 o
81P q 1 A 2 3 C CHAPTER 34 ELECTROMAGNETIC WAVES 953 (b) 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 3 = sin 1 . Now + = 120 , = 90 3 , and = 90 2 , as before. This means 2 + 3 = 60 . Thus the law of refraction becomes sin 1 = n sin(60 2 ), or sin 1 = n sin 60 cos 2 n cos 60 sin 2 , where the trigonometric identity sin(A B ) = sin A cos B cos A sin B was used. Now apply the law of refraction to point A: sin 1 = n sin 2 . This means sin 2 = q p (1=n) sin 1 and cos 2 = 1 sin2 2 = 1 (1=n2 ) sin2 1 . Thus sin 1 = n sin 60 1 (1=n)2 sin2 1 cos 60 sin 1 or Square both sides and solve for sin 1 . You should get n sin 60 1:60 sin 60 sin 1 = q =q = 0:80 )2 + sin2 60 )2 + sin2 60 (1 + cos 60 (1 + cos 60 and 1 = 53:1 . (a) The light that passes through the surface of the lake is within a cone of apex angle 2c , where c is the critical angle for total internal re
ection. Imagine a sphere with the light source at its center and suppose that surface area A of the sphere is inside the cone. Since the intensity is the same at every point on the sphere, the fraction of the total energy emitted that passes through the surface is frac = A=4R2 , where R is the radius of the sphere. You now need to develop an expression for A in terms of the critical angle c . Consider a ring of r angular thickness d on the sphere surface. As you can see on the diagram, its radius is r = R sin , so R c its circumference is 2R sin . Its thickness is R d, so its area is dA = 2R2 sin d. Integrate from zero to c to obtain
82P q (1 + cos 60 ) sin 1 = sin 60 n2 q sin2 1 : A = 2R2 Z c 0 = 2R2 (1 sin d = 2R2 cos cos c ) : 2R2 (1 c 0
p p The critical angle is given by sin c = 1=n, so cos c = 1 sin2 c = 1 1=n2 and frac = 4R2 cos c ) = 1 1 2
r 1 1 n2 ! : 954 CHAPTER 34 ELECTROMAGNETIC WAVES (b) For n = 1:33, frac = 1 1 2 " s 1 1 (1:33)2 = 0:170 : # From the diagram to the right sin = n ; sin 1 > : cos sin c = n2 : n
8 > < c c 83P Solve for : 1 sin = n1 sin = n1 1 or = sin (b) Now
1 pn2 1 p s cos2 = n1 1 n2 n1 2 = n2 n2 ; 1 2 q n2 : 2 = sin
1 p(1:58)2 (1:53)2 = 23:2 : (a) Consider the two light rays, a and b, as shown. Obviously the distance traveled by ray a following the zigzag path is longer than that of ray b by 1 L = sin c 84P L b a c c L b a 1 L; So (b) Now L 1 t = c=n = Ln1 sin c 1 c 1 n 1 = Ln1 n =n 1 = L n1 (n1 n2 ) : c c 2 2 1 n t = L n1 (n1 n2 ) = 3:00 300 m m/s c 2 108
85E 1:58 (1:58 1:53) = 5:16 10 8 s : 1:53 (a) Use Eq. 3449: B = tan 1 nw = tan 1 (1:33) = 53:1 : CHAPTER 34 ELECTROMAGNETIC WAVES 955 (b) Yes, since nw depends on the wavelength of the light. The angle of incidence B for which re
ected light is fully polarized is given by Brewster's law, Eq. 3448 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 B = tan 1 (n2 =n1 ) = tan 1 (1:53=1:33) = 49:0 . From Fig. 3419 we nd nmax = 1:470 for = 400 nm and nmin = 1:456 for = 700 nm. The corresponding Brewster's angles are B,max = tan 1 nmax = tan 1 (1:470) = 55:77 and B,min = tan 1 (1:456) = 55:52 : (a) and (b) At Brewster's angle incident + refracted = B + 32:0 = 90:0 ; so B = 58:0 and nglass = tan B = tan 58:0 = 1:60:
88P 87E 86E ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.
 Fall '07
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