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 — 0_ 1
7.3°.
5.123. X (co — dl2k = 0.6 um.
5.125. 55°.
5.126. d = 2.8 pm.
5.127. X = (d sin A0)/1/ 5 — 4 cos AO = 0.54 um.
5.128. (a) 45°; (b)-64°.
5.129. x = 2R/(n —1) y ((Ja)2 -1= 8 cm.
5.130. From the condition d En sin 0 — sin (0 + Oh)] = kX we
obtain 0 0 = —18.5°, 0+1 = 0°; kmax = +6, 0 +6 = +78.5°. See
Fig. 40.
5.131. hh = X (k — 1/2)/(n — 1), where k = 1, 2,.. .;
a sin 0 1 = X/2.
5.132. v = Xvf/Ax = 1.5 km/s.
5.133. Each star produces its own diffraction pattern in the
objective's focal plane, with their zeroth maxima being separated
Fig. 40.by an angle (Fig. 41). As the distance d decreases the angle 0 be-
tween the neighbouring maxima in each diffraction pattern increases,
and when 0 becomes equal to 2, the first deterioration of visi-
bility occurs: the maxima of one system of fringes coincide with the
minima of the other system. Thus, from the condition 0 = 2p and
the formula sin 0 = Xld we obtain , = X/2d ge, 0.06".
5.134. (a) D=IcIdli 1— (kX1d) 2 = 6.5 ang. min/nm, where k= 2;
(b) D = kld V 1 —(kX1d— sin 00 2 = 13 ang. min/nm, where k= 4.
5.135. dOldX = (tan 0)/1.
5.136. AO-- 2X/Nd 1/1 — (kk/d) 2 =11".
5.139. 0 = 46°.
5.140. (a) In the fourth order; (b) 6X„,in X 2 Il = 7 pm.
5.141. (a) d = 0.05 mm; (b) 1 = 6 cm.
5.142. (a) 6 and 12 um: (b) not in the first order, yes in the se-
cond order.