Irodov – Problems in General Physics

4.198. Since t> 7', where 7' is the period of oscillations, W = 1 /21/ Ego/RE^2 maR^2 t = 5 kJ. 4.199. B = Bm sin kx• sin wt, ...

4.222. (P)I (S) = (e 2 1m) 2 416a. 4.223. (P)/(S) = (e ((:)/ m)20;)2.^4 4.224. R = 3P/163-tcypMc x 0.6 p.m. 5.1. (a) 3 and 9 mW; ...

circles with the centres at S and S' and radii SO and S'M. Conse- quently, the optical paths (DM) and (OB) must be equal: n•DM = ...

5.47. F < D/do = 20. 5.48. F= 60. 5.49. (a) F = 2a/o/do = 15, where /^0 is the distance of the best vision (25 cm); (b) r < ...

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 comp ...

Hence, we see that E" and E are collinear, that is. cophasal. (b) E' = E(n, — n 2 )/(n + n 1 7 ), that is at n 2 > n 1 and E' ...

5.96.(a) kmax = 2d/? = 1.0.10 5 , (b) AX, = X/k = X 2 /2d = 5 pm. 00 5.97. (^) $ I(r)r dr. o 5.98. b = ar 2 /(k?.a — r 2 ). 2.0 ...

Fig. 41. 5.122. (a) AO = arcsin (n sin 0) — 0 = 7.9 0 ; (b) from the con- dition b (sin 0 1 — n sin 0) = ±X we obtain AO = 0+1 — ...

5.143. According to Rayleigh's criterion the maximum of the line of wavelength k must coincide with the first minimum of the lin ...

5.163. P = (1 — 1)1(1 — cos 2p) = 0.8. 5.164. (a) Let us represent the natural light as a sum of two mutual- ly perpendicular co ...

Fig. 42. e o 5.173. (a) p = (n — 1) 2 /(n + 1) 2 = 0.040; (b) AcI)/(1) = 1 - (1 — p) 2 N = 0.34, where N is the number of lense ...

0' , la) Fig. 43. (b) 5.187. (a) The light with right-hand circular polarization (from the observer's viewpoint) becomes plane p ...

5.200. (a) a = eEolinw 2 =5.10-' 6 cm, where E 0 = V-2//sac, v = = aw =_.- 1.7 cm/s; (b) Fm1Fe = 2.9.10-11. 5.201. (a) e = 1 —n ...

5.217. / = / 0 (1 p) (^) e (x2 5.218. AX 5.219. I= -4:+2 e-x(b-a). 5.220. Will decrease exp (Rd) = 0.6.10 2 times., 5.221. d = 0 ...

Fig. 44. is v' = vld'. Due to the Doppler effect the observed frequency is , 111—(v/0 2 yid The corresponding wavelength is X, = ...

5.251. t = (i 3 — 1) cpd/18crT: = 3 hours, where c is the spe- cific heat capacity of copper, p is its density. 5 252. T2 == T1 ...

5.281. Let us write the energy and momentum conservation laws in the reference frame fixed to the electron for the moment preced ...

6.15. AN/N = 4T2 71e4 (0.7 - (^) Af i + 0.3 t) pdNAi 2 A cot^22 2 = 1.4.10-3, where Z 1 and Z2 are the atomic numbers of copper ...

6.46. Eb = Re 4 /2h^2 , R = μe 4 /2h 3 , where It is the reduced mass of the system. If the motion of the nucleus is not taken i ...

dEldr = 0 we find reff h 2 Ime 2 = 53 pm, Enii,, —me 4 /2h 2 = -= —13.6 eV. 6.75. The width of the image is A 6 + A' 6 -I- hl/p6 ...