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Fig. 42.
e o
5.173. (a) p = (n — 1)2/(n + 1)2= 0.040; (b) AcI)/(1) = 1 -
- (1 — p)2N = 0.34, where N^ is the number of lenses. 5.175. (a) 0.83; (b) 0.044.
5.176. See Fig. 42, where o and e are the ordinary and extraor-
dinary rays. 5.177. S 11°. 5.178. For the right-handed system of coordinates: (1) circular anticlockwise polarization, when observed toward the incoming wave; (2) elliptical clockwise polarization, when observed toward the incoming wave; the major axis of the ellipse coincides with the
straight line y = x;
(3) plane polarization, along the straight line y = —x.
5.179. (a) 0.490 mm; (b) 0.475 mm. 5.180. X = 4dAn/(2k 1); 0.58, 0.55 and 0.51 pm respectively
at k = 15, 16 and 17.
5.181. Four. 5.182. 0.69 and 0.43 pm.
5.183. d = (k — 1/2) Xl/An = 0.25 mm, where k = 4.
5.184. An = 2J€ Ax = 0.009. 5.185. Let us denote the intensity of transmitted light by I. in the case of the crossed Polaroids, and by Iiiin the case of the parallel Polaroids. Then Il =11210 sin22y•sin2(8/2), = 11210 [1 — sin22cp•sin2(6/2)1.
The conditions for the maximum and the minimum:
Polaroids (^) ' max I min
II
A= (k + 1/2) X, cp=31/ A= k7,, for any cp
o =1cX, for any tp A = (k + 1/2) X, cp= n/
Here A is the optical path difference for the ordinary and extraor-
dinary rays, k = 0, 1, 2,...
la)
Fig. 43.
(b)
5.187. (a) The light with right-hand circular polarization (from the observer's viewpoint) becomes plane polarized on passing through a quarter-wave plate. In this case the direction of oscillations of the electric vector of the electromagnetic wave forms an angle of +45° with the axis 00' of the crystal (Fig. 43a); in the case of left- hand polarization this angle will be equal to —45° (Fig. 43b). (b) If for any position of the plate the rotation of the Polaroid (located behind the plate) does not bring about any variation in the intensity of the transmitted light, the initial light is natural;
if the intensity of the transmitted light varies and drops to zero, the initial light is circularly polarized; if it varies but does not drop to zero, then the initial light is composed of natural and cir- cularly polarized light. 5.188. (a) Ax = 1/2k (n, — no) (), (b) d (74 — n'e) = —2 (71, — no) 06x < 0. 5.189. An = aX/n = 0.71.10-4, where a is the rotational con- stant. 5.190. a = n/Ax tan 0 = 21 ang. deg./mm, /(x) cost (nx/Ax), where x is the distance from the maximum. 5.191. dmin= (1/a) arcsin = 3.0 mm. 5.192. 8.7 mm. 5.193. [a] = 72 ang. deg./(dm•g/cm3). 5.194. (a) Emir, = 111/ 4B1 = 10.6 kV/cm; (b) 2.2.108inter- ruptions per second. 5.195. An = 2cHV/0), where (^) c is the velocity of light in vacuum. 5.196. V = 1/2 (9)1 — (P2)/1H = 0.015 ang. min/A. 5.197. If one looks toward the transmitted beam and counts the positive direction clockwise, then q = (a — VNH) 1, where N is the number of times the beam passes through the substance (in Fig. 5.35 the number is N = 5). 5.198. Hmin= 1tl4V1 = 4.0 kA/m, where V is the Verdet con- stant. The direction along which the, light is transmitted changes to the opposite. 5.199. t = mcwolkI = 12 hours. Although the effect is very small, it was observed both for visible light and for SHF radiation.
5.217. / = /0 (1 p) (^) e (x 5.218. AX 5.219. I= -4:+2 e-x(b-a). 5.220. Will decrease exp (Rd) = 0.6.102times., 5.221. d = 0.3 mm. 5.222. d = (In 2)/p, = 8 mm. 5.223. N = (In 71)/In 2 = 5.6. 5.224. c = 2/z (n2— n1) = 3.0.108m/s. 5.225. First of all note that when v < c,^ the time rate is practic- ally identical in the reference frames fixed to the source and to the receiver. Suppose that the source emits short pulses with the inter- vals To. Then in the reference frame fixed to the receiver the distance between two successive pulses is equal to X = — v,-To, when measured along the observation line. Here vris the projection of the source velocity on the observation line (v,. = v cos 0). The frequency of received pulses v = clk = v 01(1 — vile),^ where vo = 1/To. Hence (v — vo)/vo = (v/c) cos 0. 5.226. AX —X V2T/mc2cos 0 = —26 nm. 5.227. T = 4nRX c6X. = 25 days, where R is the radius of the Sun. 5.228. d = (AX.IX),,ctIn = 3.107km, m = (AX/X4c1/4/27cy = 2.9.1029kg, where -v is the gravitational constant. 5.229. co =- coo(1 ± ()1(1 — (3), where^ 13 =^ V/c; -7-- coo (1 + (^) 2V /c). 5.230. v = 1/2kAv 900 km per hour. 5.231. Substituting the expressions for t' and x' (from the Lorentz transformation) into the equation cot — kx = co' t' — k' x' , we obtain
= (1 -1-13)11/ 1 —132, k = k' (1 + ()/V 1 —132,
where 13 = V/c. Here it is taken into account that co' = ck'. 5.232. From the formula co' = w V(1 —13)/(1 +13) we get = v/c = 0.26. 5.233. v= c "12-1= 7' 1.104km/s. OAT-HI 5.234. co = coo 5.235. AX = kT/moc2= 0.70 nm, where mois the mass of the atom. 5.236. (a) co coo/111-132 =5.0 .1010s-1 ; (b) Co = coo li1 — =1.8.10i0 s-1. Here 6 =^ v/c. 5.237. The charge of an electron and the positive charge induced in the metal form a dipole. In the reference frame fixed to the elec- tron the electric dipole moment varies with a period T' = d'/v, where d' = d^ yi -^ (v/c)2. The corresponding "natural" frequency 347
Fig. 44.
is v' = vld'. Due to the Doppler effect the observed frequency is
, 111—(v/02 yid
The corresponding wavelength is X, = c/v = d (c/v — cos 0). When 0 = 45° and v c the wavelength is^ 0.6 p,m. 5.238. (a) Let vsbe the projection of the velocity vector of the radiating atom on the observation direction. The number of atoms with projections falling within the interval vs, vs + dvx is
n (vx) dvx exp (—mv:12kT)•dvx.
The frequency of light emitted by the atoms moving with velocity vsis o) = (.1)0(1 -I- vx/c). From the expression the frequency distri- bution of atoms can be found: n (e) do) = n (vx) dvx. And finally it should be taken into account that the spectral radiation intensity n (co). (b) 6.o)/6)0= 2 V(2 In 2)^ kr/mc2.
5.239. u— (^) 1+ cin +17 If V V lcn' c,^ then
5.240. v = 1/2c60 = 30 km/s. 5.242. 0' = 8°. 5.243. The field induced by a charged particle moving with velocity V excites the atoms of the medium turning them into sources of light waves. Let us consider two arbitrary points A^ and^ B^ along the path of the particle. The light waves emitted from these points when the particle passes them reach the point P (Fig. 44) simultaneously and amplify each other provided the time taken by the light wave to propagate from the point A to the point C is equal to that taken by the particle to fly over the distance AB. Hence, we obtain cos 0 = v/V, where v = c/n is the phase velocity of light. It is evident that the radiation is possible only if V^ > v, i.e. when the velocity of the particle exceeds the phase velocity of light in the medium. 5.244. Tmin = (nl V n2— 1 — 1)mc2;^ 0.14 MeV and 0.26 GeV respectively. For muons. 5.245. T — n^ cos0^ 1) mcz = 0.23 MeV. I/ n2cos 5.247. T2 = bT il(b-FT jAX)= 1.75 kK. 5.248. Am= 3.4 Rm. 5.249. 5.109kg/s, about 1011years. 5.250. T =V 3cRpluM =2.107K, where^ R^ is the universal gas constant, M is the molar mass of hydrogen.
V (^) — (v/c) cos 0 — —WO cos 0
u v )
