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 7.2 years ago by

neeta.vanage • 880

modified 3.2 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 7.1 years ago by

neeta.vanage • 880

modified 7.0 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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