Irodov – Problems in General Physics

5.143. According to Rayleigh's criterion the maximum of the
line of wavelength k must coincide with the first minimum of the
line of wavelength X, + 8k. Let us write both conditions for the least
deviation angle in terms of the optical path differences for the
extreme rays (see Fig. 5.28):

bn — (DC CE) = 0, b (n 8n) — (DC CE) = 2 +
Hence, bon k. What follows is obvious.
5.144. (a) X/8% = 2bB/2 3 ; 1.2.10 4 and 0.35.10 4 (b) 1.0 cm.
5.145. About 20 cm.
5.146. R = 7.10 4 , 4,„„ 1 „, '.. 4 cm.
5.147. About 50 m.
5.148. Suppose 6,1) and AV are the minimum angular separations
resolved by the telescope's objective and the eye respectively
1.22k/D, = 1.22k/d 0 ). Then the sought magnification
of the telescope is 1.'„iin = AV/Alp = Did, = 13.
5.149. d min = 0.61X,/sin a = 1.4 pm.
5.150. Suppose dmin is the minimum separation resolved by the
microscope's objective, Alp is the angle subtended by the eye at
the object over the distance of the best visibility to (25 cm), and
AV is the minimum angular separation resolved by the eye (Alr =
= 1.22X/d 0 ). Then the sought magnification of the microscope is
rmin = AcAlp =^2 (1^0 14) sin a = 30.
5.151. 26, 60, 84, 107 and 134°.
5.152. a = 0.28 nm, b = 0.41 nm.
5.153. Suppose a, 3, and y are the angles between the direction
to the diffraction maximum and the directions of the array along the
periods a, b, and c respectively. Then the values of these angles can
be found from the following conditions: a (1 — cos a) =
b cos 13 = k 2 A., and c cos y = k,%. Recalling that'costa cos 2 [3
cos 2 y = 1, we obtain

— 2k^1 /a

(k ila) 2 ±(k 2 lb) 2 ±(k 3 1c) 2

5.154. X=^2 V m

7,- —2psin a = 244 pm, where 1c= 2, m is the mass

of a NaC1 molecule.

5.155. d=2 si (a/2) V

2

k 2 ±k 2 - 21c, k

2

cos (a/2) = 0.28 pm, where

ki and k 2 are the orders of reflection.

5.156. r = 1 tan 2a = 3.5 cm, where a is the glancing angle

found from the condition 2d sin a = k2n ,.

5.157. / 0 /4.

5.158. (a) / 0 ; (b) 2/ 0.

5.159. E = (1),/o) = 0.6 mJ.

5.160. T1 = i/ 2 (cos

5.161. 10//— Ts cos,

5.162. I pot [Inca = PI(1 — P) = 0.3.

(P)2(N-1) = 0.12.