Question

1. (a) A laser beam enters an optical fiber with an incident angle of 6max as shown in Figure 1. Its corresponding refracted light is subsequently incident on the boundary between the core and the cladding whose refractive indices are ni and n2, respectively. The second incident angle satisfies the critical angle, 0c, for total internal reflection. Derive the Numerical Aperture (NA) of the fiber that dictates the maximum acceptance angle, amax. (6 Marks) ec Core ni 0max Cladding n2 Eigure 1

Answer 1

Refractive index of core = n_1 Refractive index of cladding = n_2 Refractive index of surrounding = n_0 The incident angle is theta_max as shown in figure The refraction angle = theta We can write theta_c = (90 degree - theta) - (i) theta_c Second incident angle For theta_c, the refraction angle at core - cladding interface is 90 degree which satisfies critical angle for total internal reflection. So, at core - cladding interface n_1 sin theta_c = n_2 sin 90 degree sin theta = n_2/n_1 - (ii) At, point of entrance of light at optical fibre n_0 sin theta_max = n_1 sin theta - (iii) using eqn (i) in (iii), we get, \r\n n_0 sin theta_Max = n_1 sin(90 degree - theta_c) = n_1 cos theta_c From eqn (ii), Sin theta_c = n_2/n_1 Now, cos theta_c = squareroot 1 - sin^2 theta_c = squareroot 1 - n^2_2/n^2_1 Numerical Aperture (N.A) indicates the light collecting efficiency of an optical fibre and denoted by term n_0 sin theta_max N. A = n_0 sin theta_max = n_1 cos theta_c = n_squareroot 1 - n^2_2/n^2_1