Consider a fiber like the one sketched below. Take the index of rwef4raction of

Question

Consider a fiber like the one sketched below. Take the index of rwef4raction of the core to be n_1 = 1.56 and the index of refraction of the cladding to be n_2 = 1.48. What is the critical angle for the n_1 to n_2 interface? If the input angle from air is theta_0 = 10 degree Calculate theta_1. If the input angle from air is theta_0 = 10 degree Calculate phi. Will this ray be guided with low loss? Justify your answer. If the input angle from air is theta_0 = 40 degree Calculate theta_1. If the input angle from air is theta_0 = 40 degree Calculate phi. Will this ray be guided with low loss? Justify your answer. What is the maximum input angle from air for a ray that will be guided with low loss by this fiber?

Explanation / Answer

a) for critical angle,

refracted angle = 90 deg for i_c incidence angle.

so, n1 sin@c = n2 sin90

1.56 sin@c = 1.48 x 1

@c =71.57 deg

b) using snell's law,

ni sini = nr sinr

1 * sin10 = 1.56 * sin@1

sin@1 = 0.111

@1 = 6.39 deg

C) beta = 90 - @1 = 90 - 6.39 = 83.61 deg

this is more than critical angle. so very low loss.

d) 1 * sin40 = 1.56 * sin@1

sin@1 = 0.412

@1 = 24.33deg

e) beta = 90 - @1 = 90 - 24.33 = 65.67 deg

this is less than critical angle. so there will be loss as ligt will not go total interna; reflection.

F) @1 = 90 - @c = 90 - 71.57 = 18.43 deg

using snell law,

1 x sin@o = 1.56 x sin18.43

@o = 29.55 deg

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