Link download solution manual optical fiber communications 4th edition by gerd keiser

Gerd Keiser, Optical Fiber Communications, McGraw-Hill, 4th ed., 2011 Problem Solutions for Chapter E  00co s 2 1 08 t  0 ex  0co s 21 08t  0 ey 2.1 2.2  0co s 21 08 t  10 ez The general form is: y = (amplitude) cos(t - kz) = A cos [2(t - z/)] Therefore (a) amplitude = m (b) wavelength: 1/ = 0.8 m (c)  = 2(2) = 4 -1 so that  = 1.25 m (d) At time t = and position z = m we have -1 y = cos [2(-0.8 m )(4 m)] = cos [2(-3.2)] = 2.472 2.3 x1 = a1 cos (t - 1) and x2 = a2 cos (t - 2) Adding x1 and x2 yields x1 + x2 = a1 [cos t cos 1 + sin t sin 1] + a2 [cos t cos 2 + sin t sin 2] = [a1 cos 1 + a2 cos 2] cos t + [a1 sin 1 + a2 sin 2] sin t Since the a's and the 's are constants, we can set a1 cos 1 + a2 cos 2 = A cos  (1) a1 sin 1 + a2 sin 2 = A sin  (2) provided that constant values of A and exist which satisfy these equations To verify this, first we square both sides and add: 2 A (sin  + cos ) = a sin 1  cos 1  2 2 2 + a2 sin 2  cos 2  + 2a1a2 (sin 1 sin 2 or A = a  a + 2a a cos ( -  ) 12 + cos 1 cos 2) Dividing (2) by (1) gives tan  = sin   a sin  2 a cos   a cos  a 1 2 Thus we can write x = x1 + x2 = A cos  cos t + A sin  sin t = A cos(t - ) 2.4 First expand Eq (2.3) as Ey E0 y = cos (t - kz) cos  - sin (t - kz) sin  (2.4-1) Subtract from this the expression Ex cos  = cos (t - kz) cos  E0 x to yield Ey E 0y - Ex E cos  = - sin (t - kz) sin  (2.4-2) 0x 2 Using the relation cos  + sin  = 1, we use Eq (2.2) to write  E   x  2  sin (t - kz) = [1 - cos (t - kz)] =  E    0x (2.4-3) Squaring both sides of Eq (2.4-2) and substituting it into Eq (2.4-3) yields E y  E x  E 0x  y  E 2 cos  =  E     sin    x   E  0x Expanding the left-hand side and rearranging terms yields E   E    x E   cos  = sin2  -2  E 0x E 0y    y + E 0x  E 0y  E    x 2.5 Plot of Eq (2.7) 2.6 Linearly polarized wave y 2.7 Air: n = 1.0 33  33  90  Glass (a) Apply Snell's law n1 cos 1 = n2 cos 2 where n1 = 1, 1 = 33, and 2 = 90 - 33 = 57  n2 = cos 33 = 1.540 cos 57 (b) The critical angle is found from n glass sin glass = n sin air air with air = 90 and nair = 1.0 critical = arcsin n = arcsin 1.540 g lass = 40.5 2.8 Air r Water  12 cm Find c from Snell's law n1 sin 1 = n2 sin c = When n2 = 1.33, then c = 48.75 r Find r from tan c = , which yields r = 13.7 cm 12 cm 2.9 45  Using Snell's law nglass sin c = nalcohol sin 90 where c = 45 we have n = 1.45 glass sin 45 = 2.05 2.10 critical= arcsin n pure n = arcsin 1.450 = 83.3 1.460 doped 2.11 Need to show that n1 cos 2  n cos 1  Use Snell’s Law and the relationship tan  sin  cos  1/ 2  = 0.242 2.12 (a) Use either NA = n1  n2 or 2(n1  n2 ) NA  n1 2= n1 = 0.243 n (b) A = arcsin (NA/n) = arcsin 0.242 = 14  1.0  c 2.13 (a) From Eq (2.21) the critical angle is    sin 1  n       n  1  sin  1.00   41 1.50  (c) The number of angles (modes) gets larger as the wavelength decreases. 2 1/ 2.14 NA = n1  n2  = n )2 1/ 2  n1 (1 1/2 = n1 2  Since 