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4.198. Since t> 7', where 7' is the period of oscillations, W = 1/21/ Ego/RE2maR2t = 5 kJ. 4.199. B = Bmsin kx• sin wt, where Bm Em, with BD, = Ernie. 4.200. Sx = 1 /4 cocEl sin 2kx• sin 2wt, (Sx ) = 0. 4.201. Wm/We =1/8calow2R2= 5.0.10-1b 4.202. We/Wm =1/86,110(02R2= 5.0.10- 4.204. 4:Ds = I2R. 4.205. S = I2 1/ mi2eU/4a2eor2. 4.207. To the left. 4.208. (ID = V I. 4.209. (0) = 1/2170/ 0cos cp. 4.211. The electric dipole moment of the system is p = /eri =- = (elm) Mrc, where M is the mass of the system, rcis the radius vector of its centre of inertia. Since the radiation power P oc p 2oc
cc r t, and in our case rc = 0, P = 0 too.
4nEo
0 6) 4.212. (P)=- =5 .10-15W.
4.213. P =
1 2 4 qe2^ \ 2 (4neo)3 3c 3 mR2 / 4.214. AW 7=4--
ile4q (4neo)3 3c3m2vb3 • 4.215. AW/T = 1/3e3B/e0c3m2 = 2.10-18. 4.216. T = Toe--a't, where a = 1/3e4B2/a80c3m3. After r 2.5 s for the electron, 1 1.6.101°^ s =^ 0.5.103years for the proton. 4.217. S1/S2 = tan2 (o)//c) = 3. 4.218. (a) Suppose that t is the moment of time when the particle is at a definite point x, y of the circle, and t' is the moment when the information about that reaches the point P. Denoting the observed values of the y coordinate at the point P by y' (see Fig. 4.40), we shall write
The sought acceleration is found by means of tiation of y' with respect to t': dy' = dy dy dt d2 y (^) d dy'
dt' dt' = dt dt" dt' 2 =dt' dt dt'
the double differen-
=- v _ 2 vIc—yIR
R (1—vy/cR)3 ' where the following relations are taken into account: x = R sin wt, y = R cos wt, and w = v/R. (b) Energy flow density of electromagnetic radiation S is pro- portional to the square of the y projection of the observed accelera- tion of the particle. Consequently, S1/S2 = (1 v/c)4/(1 — v/c)4. 4.219. (P) = 8/3ar2S0. 4.220. (w) = 3/8Po/ar2c. 4.221. P = 1 /6p26)4/neoc2.
to —=
334
4.222. (P)I (S) = (e21m)2416a.
4.223. (P)/(S) =
(e((:)/ m)20; 4)2.
4.224. R = 3P/163-tcypMcx 0.6 p.m. 5.1. (a) 3 and 9 mW; (b) cp = 1/2 (V1 + V2) clue^ =^ 1.6 lm, where A = 1.6 mW/lm, V1 and V2 are the values of relative spectral response of an eye for the given wavelengths. 5.2. (^) a = V l-LoicoAcD/23-cr2 Vx, hence^ Em = 1.1 V/m,^ Hm = 3.0 mA/m. Here A = 1.6 mW/lm, Vx is the relative spectral response of an eye for the given wavelength. 0.—(R/02 I 5.3. (a) (E)= 1/2E0; (b) (E):= (^) 1—R11 R2— 50 lx. 5.4. M = 2 /3o-cLo. 5.5. (a) cD = nLAS sin20; (b)^ M = 5.6. h R, E = LS/4R2 =^ 40 lx.
5.7. I = /0/cos30, = nI 0R21h2 = 3.102lm.
5.8. Erna, = (9/161C) pES/R,2 = 0.21 lx, at the distance R/V-3 from the ceiling. 5.9. E = 5.10. E 5.11. M = E, (1 + h2/R2,) = 7.102lm/m2. 5.12. E0 = aLR2/h,2= 25 lx. 5.13. e' = e — 2 (en) n. 5.14. Suppose n1, n2, n3are the unit vectors of the normals to the planes of the given mirrors, and e0, e1, e2, e3are the unit vectors of the incident ray and the rays reflected from the first, second, and the third mirror. Then (see the answer to the foregoing problem): el= e, — 2 (eons) n1, e2 =et— 2 (e1n2) n2, e3 = e2— 2 (e2n3) ns. Summing termwise the left-hand and right-hand sides of these expressions, it can be readily shown that e3 = 5.15. 01= arctan n = 53°. 5.16. n1/n2 = 1/1/-112 — 1 =1.25. 5.17. x= [1 --V (1— sin2 0)/(n2 — sin20)] d sin 0 = 3.1 cm. 5.18. h' = (hn2 cos30)/(n2— sin20)3/2. 5.21. 0 = 83°. 5.22. From 37 to 58°. 5.23. a = 8.7°. 2 sin (8/2) 5.24. Aa — An = 0.44°.
- 171 —^ n2sin2(0/2) 5.27. (a) f =^ 11341 —^ p2) = 10^ cm; (b) (^) f = 111132132 - = = 2.5 cm. 5.28. I' = P1012(f — sr = 2.0.103cd. 5.29. Suppose S is a point source of light and S' its image (Fig. 38). According to Fermat's principle the optical paths of all rays originating at S and converging at S'^ are equal. Let us draw
335
5.47. F < D/do = 20.
5.48. F= 60.
5.49. (a) F = 2a/o/do= 15, where /0is the distance of the best vision (25 cm); (b) r < 2a/o/do. 5.50. The principal planes coincide with the centre of the lens.
The focal lengths in air and water: f = —110 = —11 cm, f'
= noleo = +15^ cm. Here cl) = (2n — no—^ 1)1 R,^ where n and no
are the refractive indices of glass and water. The nodal points coincide
and are located in water at the distance x = f' f = 3.7 cm from
the lens. 5.51. See Fig. 39. 5.54. (a) The optical power of the system is cb =cD 2 -
- d(1)02= + 4 D, the focal length is^25 cm.^ Both principal planes
H H'
(a)
H' H
F
Fig. 39.
(c)
are located in front of the converging lens: the front one at a distance of 10 cm from the converging lens, and the rear one at a distance of
10 cm from the diverging lens (x = d(1)2/0 and x' = — dcloin);
(b) d = 5 cm; about 4/3.
5.55. The optical power of the given lens is cD = clpi+ cD2^ -
- (dln) c1302, x = d021 = 5.0 cm,^ x' —4011 nal =^ 2.5 cm, i.e. both principal planes are located outside the lens from the side of its convex surface.
5.56. f = tltz. The lens should be positioned in the front
fly- d
principal plane of the system, i.e. at a distance of x =
= -1- f 2— d) from the first lens.
5.57.^4 :13 =^ 2c13' — 20'2//no = 3.0 D, where cD' = (2n —no—1)/R,
n and noare the refractive indices of glass and water.
5.58. (a) d = nARI (n — 1) = 4.5 cm; (b) d = 3.0 cm.
5.59. (a) cD = d (n-1)21nR2 > 0, the principal planes are locat-
ed on the side of the convex surface at a distance of d from each
other, with the front principal plane being removed from the convex
surface of the lens by a distance of RI (n — 1); (b) cb = (1/R2-1/R1) X
X (n — 1)/n < 0; both principal planes pass through the common curvature centre of the surfaces of the lens. 5.60. d = 1 /2 n (R1 R2)I (n — 1) = 9.0 cm, F = R1/R2 =5.0. 5.61. (1) = 2(n2— 1)/n2R = 37 (^) D. 5.63. p = 3.107m; Oni= 1.6-10 m-1. 5.65. 1.9a. 22-
5.66. Let us represent the kth oscillation in the complex form Eh =aei(4)t+(k-1)(pi = ateixot,
where a: = aei(4-1)9 (^) is the complex amplitude. Then the complex amplitude of the resulting oscillation is N
A*. E IA) = a [1 + + 029 ei(N-1)cpi
k= = a (eiTN —1)1(elq) —1). Multiplying A* (^) by the complex conjugate value and extracting the square root, we obtain the real amplitude
al 71—cos 1— cos^ Arcp —a sin (Ny/2)sin (T/2) •
5.67. (a) cos 0 = (k — cp1231) XI d, (^) k = 0, ±1, ±2,.. .; (b) cp = It/2, d/X, = k 1/4, k = 0, 1, 2,... 5.68. AT = 231 [k — (d/k) (^) sin (cot + a)], where k = 0, ±1, ±2,... 5.69. X = 2AxAh/l — 1) = 0.6 Rm. 5.71. (a) Ax = X (b r)I2ar = (^) 1.1 mm, 9 maxima; (b) the shift is ox = (blr) 61 = (^) 13 mm; (c) the fringe pattern is still sharp when Ox < Ax/2, hence 6„,,„x= (1 + r/b)X14a = 43 p.m. 5.72. X. = 2ccAx = 0.64 fan. 5.73. (a) Ax = kfla = (^) 0.15 mm, 13 maxima; (b) the fringes are still sufficiently sharp when Ox < Ax/2, where (^) Ox is the shift of the fringes from the extreme elements of the slit, hence, 6„,ax = = ?f2/2ab = 37 Rm. 5.74. X = 2a e(n — 1) Axl (a b) = 0.6 p.m. 5.75. Ax X/28 (n — n') = 0.20 mm. 5.76. The fringes are displaced toward the covered slit over the distance Ax = hl (n — (^) 1)/d = 2.0 mm. 5.77. n' = n Nkl 1 = 1.000377. 5.78. (a) Let E, E', and E" be the electric field vectors in the incident, reflected and transmitted waves. Select the x-,^ y-axes at the interface so that they coincide in direction with E and H^ in the incident wave. The continuity of the tangential components across the interfacei yields
E + E' = E".
The minus sign before H' appears because H' It H. Rewrite the second equation taking into account that HocnE.Solving the obtained and the first equation find:
E" = 2En1/(n1 + n2).
