An optical fiber has NA = 0.25 & cladding RI = 1.55. Determine its acceptance angle when this fiber is immersed in water having RI 1.33. Also calculate its core RI.

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written 6.9 years ago by

neeta.vanage • 870

modified 2.8 years ago by

pedsangini276 • 4.8k

Subject: Applied Physics 2

Topic: Fiber Optics

Difficulty: Medium

engineering physics

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1 Answer

1.5kviews

written 6.8 years ago by

neeta.vanage • 870

modified 6.6 years ago by

yashbeer ★ 11k

$ NA =0.25 $

$ n_2 = 1.55 $

$n_o = 1.33 $

$n_1 = ? $

$i_m = ? $

$ NA = n_1^2 - n_2^2 $

$(0.25)^2 = n_1^2 - (1.55)^2 $

$0.0625 = n_1^2 - 2.4025 $

$⸫ n_1^2 = 2.465 $

$⸫ n_1 = 1.57 $

$\text{When the fiber is in water ,}$

$ NA = 0.2497 / 1.33 \\ $

$NA = 0.1877 $

$Sin i_m= NA $

$ i_m = sin ^{-1} ( NA) $

$ i_m = sin ^{-1} ( 0.1877) $

$ i_m = 10.81° $

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