Assignment 9, solutions:

Problem 1:

NA = n_isinq_cone = (n_core² - n_cladding²)^1/2 = 0.56, q_cone = 34.1 deg.

For a ray incident at 45 deg the condition for total internal reflection is not satisfied and the light will escape the fiber after a short distance.

Problem 2:

10^{(A(total)/10)} = (P_in(0)/P_out(L).  10^-(0.2L/10) = 0.5.

L(km) = 10 log₁₀(2)/0.2 = 15.

Problem 3:

The number of guided modes in a step-index multimode fiber is given by V²/2.

V = k_f a NA = (2p/l₀) a NA.  NA = (n_core² - n_cladding²)^1/2 = 0.232.  a = 25 mm.

V= 42.8.  # of modes = 917.

Problem 4:

If a ray  propagating along the axis of the fiber travels a distance d, then a ray incident at the critical angle q_c travels a distance L = d/sinq_c.

The respective travel times are t_d = dn_core/c  and t_L = dn_core/(sinq_c c).

sinq_c = n_cladding/n_core.  q_c = 81.9 deg.

For d = 1000 m we have t_d = 5000 ns  and t_L =5050.51 ns.

The difference in travel time is therefore 50.51 ns/km.

Problem 5: (B)